问:
Calculate the following series --- need to understand why I have it incorrect
Okay I know I have the answer incorrect but not sure why...

Calculate the following series:

$$\sum_{j=0}^n 2^j \text{ for } n = 1,2,3,4$$

so I think it would be $2^1 = 2$ but the answer is $3$ and the rest are $7,9,32$... I do not understand how they got $3$. Any guidance will help.

Body:
$$
\sum_{j=0}^12^j=2^0+2^1=3
$$
$$
\sum_{j=0}^22^j=2^0+2^1+2^2=7
$$
$$
\sum_{j=0}^32^j=2^0+2^1+2^2+2^3=15 \qquad[\text{not }9]
$$
$$
\sum_{j=0}^42^j=2^0+2^1+2^2+2^3+2^4=31 \qquad [\text{not }32]
$$


问:
If $A$ is an infinite set and $b \notin A$, is the equivalence $A \sim A \bigcup \{b\}$ provable without the axiom of choice (i.e. in ZF)?

A set $A$ is said to be infinite if there is a surjection from $A$ to $\mathbb N_0$.

Let $A$ be infinite, and denote by $b$ some set that is not included in $A$. (Such a $b$ exists by the specification axiom). To prove the equivalence

$$A \sim A \bigcup \{b\} \tag{
1}$$

in elementary set theory, one usually uses the fact that $A$ has a countably infinite subset (whose universal existance depends on some kind of choice).

Q: But can the equivalence in $(1)$ be proven in some other way, without the axiom of choice? More specifically, in the context of ZF without regularity?

I assume it cannot be done but I am not sure... I am still quite the novice when it comes to axiom of choice considerations, having begun only yesterday. If if it cannot be done, should I expect the impossibility to have an elementary proof? (Again, I think not but am unsure).

答:
No.

The definition you're using is also known as weakly Dedekind-infinite, and the property you are trying to prove is equivalent to Dedekind-infinite.

If there is an infinite Dedekind-finite set, then there is one which is weakly Dedekind-infinite. To see this, note that if $A$ is Dedekind-finite, either it is weakly Dedekind-infinite, or else its power set can be mapped onto $\Bbb N_0$ (by mapping finite sets to their size and infinite sets to $0$).

Now we can use a theorem of Kuratowski that $A$ is weakly Dedekind-infinite if and only if its power set is Dedekind-infinite. Since $A$ was not weakly Dedekind-infinite, its power set is Dedekind-finite, but we also proved that it is weakly Dedekind-infinite.

ContentLicense: CC BY-SA 4.0


问:
Determine which of the following spaces are separable
A topological space $(X,\tau)$ is said to be seperable if it has a  dense subset which is countable.Determine which of the following spaces are seperable.

(i) the set $\mathbb{R}$ with the usual topology:

$\mathbb{Q} \subset \mathbb{R}$ is a dense subset and is countable,So $\mathbb{R}$ is seperable.

(ii)a countable set with discrete topology :

All proper subset of $X$ will be close ,so its closure is itself.So they aren't dense.$X \in \tau$ So $X$ is open and $\phi \in \tau$ so $X$ is close also.$X$ is closed,so its closure is $X$ itself.So is dense and $X$ is countable.So $(X,\tau)$ is seperable.

(iii)a countable set with the finite closed topology:

Similarly here $X$ is dense and being countable,the topological space is seperable.

(iv)$(X,\tau)$ where $X$ is finite.

Similarly here $X$ is dense in itself and being finite is countable.So Topological space is seperable.

(v)$(X,\tau)$ where $\tau$ is finite.

Taking $\tau$ as indiscrete topology.So only open and close sets are $X$ and $\phi$. $X$ is dense but unless it is countable we cannot say $\tau$  is seperable.

(vi)a uncountable set with the discrete topology.

only dense set is $X$ but $X$ is not countable so Topology is not seperable.

(vii)an uncountable set with the finite closed topology

Here $X$ is dense set but not countable.I am unable to conclude whether any countable subset of $X$ can be dense or not?

Please check i believe my arguments are correct for first six topology.And give me a hint for seventh one.

答:
Your (i-iv),(vi) seem fine to me.

For (v), why do you assume $\tau$ is indiscrete? Even in that case, what about some intermediate countable set $A \subsetneq X$?

For both questions (v) and (vii), I offer one HINT: can you create a set which is not a subset of any closed set?


问:
Graph automorphism groups: reference request
I would like to find something like dictionary that contains info about graph automorphisms groups. Recently have found "Algebraic Graph Theory" by Godsil but I cannot find something similar in this book. I assume that good references exist. I will be grateful for any comments.</p>

答: The book "Introduction to Graph Theory" by Behzad and Chartrand has a chapter on Automorphism groups of Graphs. Among other things they prove Frucht's theorem, which states that every finite group is the full automorphism group of some graph. Perhaps you would find a good introduction there.</p>


问:
Improper Integral Question $ \int^{\infty}_0 \frac{\mathrm dx}{1+e^{2x}}$
I want to check if it's improper integral or not
$$ \int^{\infty}_0 \frac{\mathrm dx}{1+e^{2x}}.$$
What I did so far is :

set $t=e^{x} \rightarrow \mathrm dt=e^x\mathrm dx \rightarrow \frac{\mathrm dt}{t}=dx
$ so the new integral is:
 $$ \int^{\infty}_0 \frac{\mathrm dt}{t(1+t^2)} =  \int^{\infty}_0 \frac{\mathrm dt}{t}-\frac{\mathrm dt}{1+t^{2}}$$
now how I calculate the improper integral, I need to right the $F(x)$ of this integral and then to check the limit?

Thanks!

答:
Does this solution make sense too?

Let   $\displaystyle t=1+e^{2x},$   $x\in(0,\infty), t\in(2,\infty)$

So    $\displaystyle dx = \frac{1}{2(t-1)}dt$,

Then,

      $\displaystyle\int^{\infty}_0\frac{1}{1+e^{2x}}dx$

  $\displaystyle= \frac{1}{2}\int^{\infty}_2\frac{1}{t(t-1)}dt$
  (Please pay attention to the changing interval)

  $\displaystyle= \frac{1}{2}\int^{\infty}_2(\frac{1}{t-1} - \frac{1}{t})dt$

  $\displaystyle= \frac{1}{2}\left(\left[\ln{\left(t-1\right)}\right]_2^{\infty} - \left[\ln{t}\right]_2^{\infty}\right)$

  $\displaystyle= \frac{1}{2}\lim_{t\to\infty}\ln{(t-1)} - \frac{1}{2}\lim_{t\to\infty}\ln{t} + \frac{1}{2}\ln{2}$

  $\displaystyle= \frac{1}{2}\lim_{t\to\infty}\ln{\frac{t-1}{t}} + \frac{1}{2}\ln{2}$

  $\displaystyle= \frac{1}{2}\ln{2}$

ContentLicense: CC BY-SA 3.0


问:
Which book is better for learning calculus Stewart, Larson or Thomas
I want to master calculus in every possible way, I'm working in my bases like algebra and trigonometry (Precalculus) since I haven't had a good start in calculus, I want to read books like Calculus by Spivak, Calculus by Apostol and Courant books from Calculus and analysis.

I want to know which books of calculus those 3 authors (Stewart, Larsom, Thomas) could help me to make a good aproach to calculus, if they are any substantial differences, if you think they are others best books please tell me

答:
My teacher recommended this. It's great, trust me!

https://www.springer.com/gp/book/9783319127712


问:
Compact expression for the coefficients of the Gauss-Weingarten equations
In the book Differential Geometry by Lipschutz, the Gauss-Weingarten equations are written as follows:

On a patch $\mathbf{x}=\mathbf{x}(u^1,u^2)$ on a surface of class $\geq2$ the vectors $\mathbf{x}_i=\frac{\partial\mathbf{x}}{\partial u^i}$, $\mathbf{x}_{ij}=\frac{\partial^2\mathbf{x}}{\partial u^i \partial u^j}$ and $\mathbf{N}_i=\frac{\partial\mathbf{N}}{\partial u^i}$, with $i,j=1,2$,  satisfy:
$$\mathbf{x}_{1 1}=\Gamma_{11}^{1} \mathbf{x}_{1}+\Gamma_{11}^{2} \mathbf{x}_{2}+b_{11} \mathbf{N} $$
$$\mathbf{x}_{1 2}=\Gamma_{12}^{1} \mathbf{x}_{1}+\Gamma_{12}^{2} \mathbf{x}_{2}+b_{12} \mathbf{N} $$
$$\mathbf{x}_{2 2}=\Gamma_{22}^{1} \mathbf{x}_{1}+\Gamma_{22}^{2} \mathbf{x}_{2}+b_{22} \mathbf{N} $$
$$\mathbf{N}_{1}=\beta_{1}^{1} \mathbf{x}_{1}+\beta_{1}^{2} \mathbf{x}_{2}$$
$$\mathbf{N}_{2}=\beta_{2}^{1} \mathbf{x}_{1}+\beta_{2}^{2} \mathbf{x}_{2}$$

With $b_{ij}$ the second-order fundamental coefficients.
Using the Einstein summation convention, with $i,j,\alpha=1,2$, I see that it can be expressed more compatly as:

$$\mathbf{x}_{i j}=\Gamma_{i j}^{\alpha} \mathbf{x}_{\alpha}+b_{i j} \mathbf{N}$$
$$\mathbf{N}_i=\beta^\alpha _i \mathbb{x}_\alpha$$

However, in the book the coefficients $\Gamma_{i j}^{\alpha}$ and $\beta^\alpha _i$ are written in a way that gives you a headache:

$$\Gamma_{11}^1=\frac{g_{22}g_{11,1}-2g_{12}g_{12,1}+g_{12}g_{11,2}}{2\left(g_{11}g_{22}-g_{12}^2\right)},\quad \Gamma_{12}^1=\frac{g_{22}g_{11,2}-g_{12}g_{22,1}}{2\left(g_{11}g_{22}-g_{12}^2\right)}, \quad \Gamma_{22}^1=\frac{2g_{22}g_{12,2}-g_{22}g_{22,1}-g_{12}g_{22,2}}{2\left(g_{11}g_{22}-g_{12}^2\right)}$$
$$\Gamma_{11}^2=\frac{2g_{11}g_{12,1}-g_{11}g_{11,2}+g_{12}g_{11,1}}{2\left(g_{11}g_{22}-g_{12}^2\right)}, \quad \Gamma_{12}^2=\frac{g_{11}g_{22,1}-g_{12}g_{11,2}}{2\left(g_{11}g_{22}-g_{12}^2\right)}, \quad \Gamma_{22}^2=\frac{g_{11}g_{22,2}-2g_{12}g_{12,2}+g_{12}g_{22,1}}{2\left(g_{11}g_{22}-g_{12}^2\right)}$$
$$\beta_1^1=\frac{b_{12}g_{12}-b_{11}g_{22}}{g_{11}g_{22}-g_{12}^2},\quad \beta_1^2=\frac{b_{11}g_{12}-b_{12}g_{11}}{g_{11}g_{22}-g_{12}^2},\quad \beta_2^1=\frac{b_{22}g_{12}-b_{12}g_{22}}{g_{11}g_{22}-g_{12}^2},\quad \beta_2^2=\frac{b_{12}g_{12}-b_{22}g_{11}}{g_{11}g_{22}-g_{12}^2}$$

Where $\{g_{ij}\}$ and $\{b_{ij}\}$ are the first and second-order fundamental coefficients, respectively, and where $g_{ij,k}=\frac{\partial g_{ij}}{\partial u^k}$.

Is there a compact way to express the coefficients $\Gamma_{i j}^{\alpha}$ and $\beta^\alpha _i$ in terms of the first and second-order fundamental coefficients, $g_{ij}$ and $b_{ij}$?

答:
I answer myself: the Christoffel symbols can be expressed in terms of the first fundamental coefficients as

$$\Gamma_{i j}^{\alpha} = \frac{1}{2} g^{\alpha \gamma} \left[g_{j \gamma,i}+{g_{\gamma i,j}} - {g_{i j,\gamma}} \right] \tag{1}$$

with $i,j,\alpha, \gamma=1,2$, and the coefficients $\beta^\alpha _i$ can be expressed in terms of the second fundamental coefficients as

$$\beta^\alpha _i = - g^{\alpha \gamma}b_{i \gamma} \tag{2}$$

So the Gauss-Weingarten equations and their coefficients can be written compactly as

$$\mathbf{x}_{i j}=\Gamma_{i j}^{\alpha} \mathbf{x}_{\alpha}+b_{i j} \mathbf{N}$$
$$\mathbf{N}_i=- g^{\alpha \gamma}b_{i \gamma} \mathbb{x}_\alpha$$

With the coefficients $\Gamma_{i j}^{\alpha}$ given by $(1)$.


答:
Among the modern literature, I haven't seen anything more comprehensive, application-connected, and thorough than Zorich's 2 volume.

Maybe even precede a study of Zorich by studying the first 3 chapters of Kolmogorov & Fomin, which is likely to save you a lot of time.

Title: Which book is better for learning calculus Stewart, Larson or Thomas

Body:
I want to master calculus in every possible way, I'm working in my bases like algebra and trigonometry (Precalculus) since I haven't had a good start in calculus, I want to read books like Calculus by Spivak, Calculus by Apostol and Courant books from Calculus and analysis.

I want to know which books of calculus those 3 authors (Stewart, Larsom, Thomas) could help me to make a good aproach to calculus, if they are any substantial differences, if you think they are others best books please tell me


问:
$\forall x (\forall y \exists z (y = xz) \implies (x \neq 0))$ is True but $\forall x \forall y \exists z ((y = xz) \implies (x \neq 0))$ is False
Consider two logical statements for real numbers $x,y,z$

$\forall x (\forall y \exists z (y = xz) \implies (x \neq 0))$

$\forall x \forall y \exists z ((y = xz) \implies (x \neq 0))$

In some course notes on logic I am instructed that 1) is true and 2 is false

For 1) I believe there is two cases. Suppose $x=0$, then the statement $\forall y \exists z (y=xz)$ is false for $y=10$, and so we get False implies True, which is True. If $x \neq 0$ then we get True implies True, which is True.

For 2) I'm not sure why I can't simply use the same reasoning, or what the real difference is between 1 and 2)

Any insights appreciated.

答:
For 2, the statement is: for all $x, y$, there exists $z$ such that [$y = xz \implies x \ne 0$ is true].

Consider when $x= y= 0$. We can choose $z=0$, then $y=xz$ is true while $x\ne 0$ is false, and [true implies false] is false. This shows that 2 is false.


问:
Maximizing log(x)
For what functions is maximizing $\log f(x)$ the same as maximizing $f(x)$?

I've read this thread. I'm looking for a rigorous, calculus based answer.

$$\frac{d \log f(x)}{dx} = \frac{d f(x)}{dx} \frac{d \log f(x)}{d f(x)} = \frac{d f(x)}{dx} \frac{1}{f(x)}$$

How do we find the maxima from this, and what exactly implies that if and only if the function is increasing, the maxima are at the same values of $x$?

答:
Your second paragraph seemed to indicate that this holds even if $f$ is decreasing, or whatever.

I think what MPW means is that 'this' holds for any possible $f$. Such $f$ must be strictly positive and all that other stuff.

But vacuously, I guess you could say that

If $f$ is not strictly positive, then maximising $\ln f$ is equivalent to maximising $f$. $\tag{1}$

whether or not it's increasing,

I think there's some confusion with the preposition here. Perhaps MPW thought 'it' referred to $g$. (I think similar confusion may arise from the aforementioned 'this'.)

Anyhoo, to be explicitly, I indeed think that

maximising $\ln f$ is equivalent to maximising $f$ $\tag{2}$

for

any such $f: A \to \mathbb R, A \subseteq \mathbb R$ such that (2) makes sense $\tag{3}$

I shall attempt to prove this using 2nd derivatives, which MPW doesn't seem to use. (Oh drat. It's just occurring to me that we might talk about maximisation even if 2nd derivative of $f$ doesn't exist.)

Proof:

Step 1: 1A. Observe that $f$ is strictly positive. Let $g=\ln(f)$, $g: A \to \mathbb R$. Observe that 1B. both $g$ and $f$ are twice differentiable, 1C. that $g'(x)=0$ iff $f'(x)=0$, 1D. that $g''(x) (f(x))^2= f(x)f''(x)-(f'(x))^2$ and 1E. that $g''(x) < 0$ iff $ f(x)f''(x) < (f'(x))^2$.

If direction:

Step 2: Let $x \in \mathbb R$ maximise $f$, i.e. $x \in A, f'(x)=0$ and $f''(x) < 0$.

Step 3: By Steps 1A, 1C, 1E and 2, $x \in A, g'(x) = 0$ and $g''(x) < 0$, i.e. $x \in \mathbb R$ maximises $g$ because we get that  $(\text{positive number}) \ f''(x) = (\text{positive number}) \ (\text{negative number})$ $= \text{negative number} < 0 = 0^2 = (f'(x))^2$.

Only if direction:

Step 4: Let $x \in \mathbb R$ maximise $g$, i.e. $x \in A, g'(x) = 0$ and $g''(x) < 0$.

Step 5: By Steps 1A, 1C, 1E and 4, $x \in A, f'(x)=0$ and $f''(x) < 0$, i.e. $x \in \mathbb R$ maximises $f$ because $\text{positive number} \ f''(x)  < 0^2 = 0$ implies $f''(x) < 0$.

QED

Edit: I honestly had no idea that 5 years ago I actually had answered the 7 year old question that OP links to. LOL. What a small world: The reason I came to OP's question is because OP commented on my questions/answers or replied to my comments on chess se.

Title: Maximizing log(x)


问:
Find all real polynomial solutions of a functional equation
Let f be a real polynomial

a)

$f(x)f(x+1)=f(x^2+x+1)$

b)

$f(x)f(2x^2)=f(2x^3+x)$

答:
a) $$f(x)f(x+1)=f(x^2+x+1)$$

Note that $f(x)=0$ works. If $f(x) \not =0$, note that $f(x)=x^2+1$ works. Thus if we write $f(x)=(x^2+1)^ng(x)$, where $n \geq 0, n \in \mathbb{Z}$, and $x^2+1$ is not a factor of the polynomial $g(x)$, then $g(x)$ also satisfies the equation $g(x)g(x+1)=g(x^2+x+1)$. Note that $g(x) \not =0$.

If $g(x)=c \not =0$ is a constant, then $c^2=c$, so $c=1$. In fact, $g(x)=1$ works.

Otherwise $g(x)$ has a root $\alpha$.

Note that if $\alpha$ is a root of $g(x)$, then $g(\alpha)=0$, so $g(\alpha^2+\alpha+1)=g(\alpha)g(\alpha+1)=0$ and $g(\alpha^2-\alpha+1)=g(\alpha-1)g(\alpha)=0$, so $\alpha^2+\alpha+1$ and $\alpha^2-\alpha+1$ are also roots of $g(x)$. If $\pm i$ is a root of $g(x)$, then $(\pm i)^2+ \pm i+1=\pm i$ and $(\pm i)^2-(\pm i)+1= \mp i$ are roots of $g(x)$, so $x^2+1$ is a factor of $g(x)$, a contradiction. Therefore $i, -i$ are not roots of $g(x)$.

We shall show that at least one of $|\alpha^2+\alpha+1|, |\alpha^2-\alpha+1|$ is $>|\alpha|$. Assume on the contrary that $|\alpha^2+\alpha+1| \leq |\alpha|$ and $|\alpha^2-\alpha+1| \leq |\alpha|$. Then by triangle inequality $2|\alpha| \geq |\alpha^2+\alpha+1|+|\alpha^2-\alpha+1|=|\alpha^2+\alpha+1|+|-\alpha^2+\alpha-1| \geq |2\alpha|=2|\alpha|$. Thus equality holds, so $|\alpha^2+\alpha+1|=|-\alpha^2+\alpha-1|=|\alpha|$ and $\alpha^2+\alpha+1, -\alpha^2+\alpha-1, \alpha$ have the same argument, so $\alpha^2+\alpha+1=-\alpha^2+\alpha-1=\alpha$. Thus $\alpha^2+1=0$, so $\alpha= \pm i$, a contradiction.

Therefore at least one of $|\alpha^2-\alpha+1|, |\alpha^2+\alpha+1|$ is $>|\alpha|$. We now construct an infinite sequence of roots of $g(x)$ with increasing modulus as follows: $\alpha_1=\alpha$, and $\alpha_{n+1}=\alpha_n^2 \pm \alpha_n+1$, where the sign is chosen to make $|\alpha_n^2 \pm \alpha_n+1|>|\alpha|$. (If both signs work, randomly choose one of them) This implies that $g(x)$ has infinitely many roots, a contradiction.

Therefore $g(x)=1$, so $f(x)=(x^2+1)^n$. To conclude, $f(x)=0$ and $f(x)=(x^2+1)^n$ are the only polynomial solutions for the given equation.

b) $$f(x)f(2x^2)=f(2x^3+x)$$

Note that $f(x)=0$ works. If $f(x) \not =0$, note that $f(x)=x^2+1$ works. Thus if we write $f(x)=(x^2+1)^ng(x)$, where $n \geq 0, n \in \mathbb{Z}$, and $x^2+1$ is not a factor of the polynomial $g(x)$, then $g(x)$ also satisfies the equation $g(x)g(2x^2)=g(2x^3+x)$. Note that $g(x) \not =0$.

If $g(x)=c \not =0$ is a constant, then $c^2=c$, so $c=1$. In fact, $g(x)=1$ works.

Otherwise $g(x)$ has a root $\alpha$.

Note that substituting $x=0$ gives $g(0)^2=g(0)$, so $g(0)=0, 1$.

If $g(0)=0$, then write $g(x)=x^ah(x), a \in \mathbb{Z}^+, h(0) \not =0$, then $x^ah(x)(2x^2)^ah(2x^2)=(2x^3+x)^ah(2x^3+x)$. Thus $2^ax^{2a}h(x)h(2x^2)=(2x^2+1)^ah(2x^3+x)$, so substituting $x=0$ gives $h(0)=0$, a contradiction.

Thus $g(0)=1$, so $0$ is not a root of $g(x)$.

If $\alpha$ is a root of $g(x)$, then $g(\alpha)=0$, so $g(2\alpha^3+\alpha)=0$, so $2\alpha^3+\alpha$ is also a root of $g(x)$. If $\pm i$ is a root of $g(x)$, then $2(\pm i)^3 \pm i=\mp i$ is also a root of $g(x)$, so $x^2+1$ is a factor of $g(x)$, a contradiction. Thus $\pm i$ are not roots of $g(x)$.

If $\alpha$ is a root of $g(x)$ for some $|\alpha| \geq 1$, then note that by triangle inequality $|2\alpha^2+1|+1=|2\alpha^2+1|+|-1| \geq |2\alpha^2| \geq 2$, so $|2\alpha^3+\alpha|=|\alpha||2\alpha^2+1| \geq |\alpha| \geq 1$. If $|2\alpha^3+\alpha|=|\alpha|$, then we must have $|2\alpha^2+1|=1$, so equality must hold in the triangle inequality above, so $2\alpha^2+1$ must be a negative real number, so $\alpha=\pm i$, a contradiction. Therefore $|2\alpha^3+\alpha| \not =|\alpha|$, so $|2\alpha^3+\alpha|>|\alpha| \geq 1$.

We can thus construct an infinite sequence of roots of $g(x)$ with increasing modulus as follows: $\alpha_1=\alpha, \alpha_{n+1}=2\alpha_n^3+\alpha_n$, so $|\alpha_{n+1}|>|\alpha_n|>1$. Therefore $g(x)$ has infinitely many roots, a contradiction.

Thus all roots $\alpha$ of $g(x)$ satisfy $0<|\alpha|<1$. Finally note that $1=|f(0)|=\prod_{g(\alpha)=0}{|\alpha|^{m(\alpha)}}<1$ (where $m(\alpha)$ is the multiplicity of the root $\alpha$), a contradiction.

Therefore the only solution is $g(x)=1$, so $f(x)=(x^2+1)^n$. To conclude, $f(x)=0$ and $f(x)=(x^2+1)^n$ are the only polynomial solutions for the given equation.


问:
$\exists x \forall y (P(y) \implies Q(x))$ and $\forall y P(y) \implies \exists x Q(x)$ are not logically equivalent
I am trying to show that the following are not logically equivalent (according to a practice question)

$\exists x \forall y (P(y) \implies Q(x))$ and $\forall y P(y) \implies \exists x Q(x)$

In the first case I am trying to find some kind of statement where $x,y$ are integers (something like $P(x)$ is the is even predicate and $Q(x)$ is odd predicate, or maybe that $Q(x)$ implies $x$ divides $y$).

I am imagining I need a scenario where one statement is True implies False, and the other is True, for the same values of $x,y$.

If $\forall y P(y)$ is false, then both implications will be true, so suppose that $\forall y$ P(y) is true. I'm not sure how to proceed from here. Hints/Clarifications appreciated.

答: With false and true respectively $0$ and $1$, the statements are$$\max_yP(y)\le\max_xQ(x)$$ and$$\min_yP(y)=1\implies\max_xQ(x)=1.$$These statements differ in truth value if$$\min_yP(y)=0,\,\max_yP(y)=1,\,\max_xQ(x)=0.$$


问:
Finding a contradiction in equations and inequalities.
Question:-In the figure shown, Coefficient of friction between the blocks C and B is $0.4$. There is no friction between block C and ground.The system of blocks is released from rest in the shown situtation.Find the accelerations of masses.(Given $m_{B}=5kg , m_{C}=10kg , m_{A}=3kg$ )

I made two cases

$1)$ Block B and C move together.

$2)$ Block B and C will not move together.

When I solve equations for Case $1$, I find that block will move together.This implies that Case $1$ is true.

$$30-T=3a_{3}$$

$$T-f=5a_{5}$$

$$f=10a_{10}$$

$$f \leq 20$$

Note that if blocks B and C will move together then $a_{3}=a_{5}=a_{10}=a$

On solving we get $a=\frac {5}{3}m/s^2$ and $f=16.67N$

As case $1$ is true. So contradiction condition must arrive during solving equations for case $2$.

I got the following equations for case $2$.  Despite my all efforts I was unable to find how equations will contradict.

\begin{align}
30-T&=3a_{3}\\
T-f&=5a_{5}\\
f&=10a_{10}\\
f&\leq {20}\\
a_{3}&>a_{10}\\
\end{align}

Note that $a_{3}=a_{5}$

My attempt

Adding equation $1$ and $2$, we get

$$30-f=8a_{3}$$

So, $30-8a_{3}\leq 20$ and $10a_{10}\leq 20$

This gives $a_{3}\geq \frac54$ and $a_{10}\leq 2$

Also, $$T-f \geq \frac{25}{4}$$

$$30-T\geq \frac{15}{4}$$

This gives $T\leq \frac{105}{4}$

Adding eqation $1$ and $3$ we get

$$30+f-T=10a_{10}+3a_{3}$$

Using $f-T\leq \frac {-25}{4}$

$$12a_{3}+40a_{10}\leq {95}$$

Adding equation $2$ and $3$,we get

$$T=5a_{3}+10a_{10}$$

Using $T\leq \frac {105}{4}$

$$4a_{3}+8_{10}\leq 21$$

So We got following inequalities involving $a_{3}$ and $a_{10}$

$$a_{3}>0$$
$$a_{10}>0$$
$$a_{3}>a_{10}$$
$$a_{3}\geq \frac54$$
$$a_{10}\leq 2$$
$$12a_{3}+40a_{10}\leq {95}$$
$$4a_{3}+8a_{10}\leq 21$$

On plotting them on graph where x-axis represents $a_{3}$ and y-axis represents $a_{10}$ I got quadrilateral having coordinates $(1.25,0), (1.25,1.25), (1.75,1.75), (5.25,0)$.

So, $1.25\leq a_{3} \leq 5.25$ and $0\leq a_{10}\leq 1.75$

Can anybody help to find contradiction.

答:
I belive you have made a mistake. In either case we can calculate the frictional force directly, since we have friction at a maximum as we have motion:
$$F_\text{MAX}=\mu R=0.4\times5g=2g$$

I will calculate all accelerations below:

For B and A, let $a_1$ be their common acceleration. Resolve for A and then B:
$$\begin{align}
& 3g-T=3a_1\\
& T-2g=5a_1
\end{align}$$
Add the two equations together to eliminate $T$:
$$g=8a_1\implies a_1=\frac{g}{8}$$

This is actually equivalent to treating A and B as a single particle:

Using $F=ma$, it is clear that as the tensions cancel out we have
$$3g-2g=(3+5)a_1\implies g=8a_1\implies a_1=\frac{g}{8}$$

Now for box C. We have
$$2g=10a_2\implies a_2=\frac{g}{5}$$
where $a_2$ is the acceleration of C.

As an aside, we can prove that there is motion as follows:

First, note that the tension must be less than $3g$; otherwise A would move upwards, which is absurd. Now suppose that $T\le2g$. Then A will still accelerate downwards. But $F_\text{MAX}=2g$, so B will not move. This is a contradiction as B is connected to A. Hence $2g<T<3g$ and there is movement since $T>F_\text{MAX}$ as required.


问:
Solve non-homogeneous recurrence $x_{n+2} - 2x_{n+1} + x_n = 8$
I'm trying to solve this recurrence $x_{n+2} - 2x_{n+1} + x_n = 8$, but the solution of the homogeneous is $a_n=b+cn$, and I don't know how to change it to find the non-homogeneous solution.

Body: Try the Ansatz $x_n=kn^2$ so$$8=k[n^2+4n+4-2n^2-4n-2+n^2]=2k\implies k=4.$$


问:
Clarification on necessity
I am currently just being introduced to the concept of logic in mathematics and my lecturer was talking about "if...then" statements.

For example, for some statements $P$ and $Q$, "if $P$ then $Q$".

I think sufficiency is quite intuitive, but why do we say that $Q$ is a necessary condition for $P$? Wikipedia provides one reason to be that it is impossible to have $P$ without $Q$, but why can't I have $P$ standalone? I mean, it is a statement in itself after all, right? Would it not make sense if I just wrote a $P$ down as a sentence? Why does $Q$ have to exist for $P$ to exist?

Perhaps if I can be provided with a detailed explanation, as well as an intuitive one, that will be nice :)

答:
We are told $P\implies Q$.

There are $4$ hypothetical possibilities.

$P$ is not true and $Q$ is not true.

This could happen.

$P$ is not true and $Q$ is true.

The could happen.

$P$ is true and $Q$ is not true.

This can not happen.  We we told $P\implies Q$ so if we have $P$ we must have $Q$ be true. This is just not possible.

$P$ is tru and $Q$ is true.

Now in every single case where $P$ is true we had to have $Q$ also be true.  So.... $Q$ is necessary for $P$.

That's all.

.....

but why can't I have P standalone?

Well, what does $Q$ do, while we have $P$ stand by itself?  We can ignore $Q$ and not look at $Q$ but and we can choose to not invite $Q$ to our party, but we can't make $Q$ ... disappear.  Now matter what we do with $P$, $Q$ is still going to exist and either be true or false even if you don't look at it.

So we have $P$ at our party and there it is... being true.  That's fine but someone might wonder "Hmm, I wonder what $Q$ is doing" and everyone else can ignore $Q$ but, $Q$ will be sitting on the other side of the hill doing it's $Q$ thing of being true or false.  So someone wonders over to check on $Q$ and what does she see?

Well,  $P\implies Q$ so, lo and behold, $Q$ is being true.  It doesn't matter if we look at $Q$.  So long as $P\implies Q$ then if we ever have $P$, it will always be that $Q$ is true, whether we look at it or not.

And since every time $P$ is true, $Q$ is true, we can't have $P$ be true without $Q$ being true (even if we stick $Q$ is a little cottage three thousand miles away and pass a law that no-one is allowed to check in on him-- $Q$ will just sit there being true all by himself unnoticed and unloved), so $Q$ is necessary for $P$ to be true.

Title: Clarification on necessity


答:
I can suggest "Godsil, Chris, and Gordon F. Royle. Algebraic graph theory. Vol. 207. Springer Science & Business Media, 2001" which contains a complete chapter about graphs and groups, covering automorphisms of graphs.

For a nice overview on Algebraic Graph Theory in general, I have enjoyed "Knauer, Ulrich, and Kolja Knauer. Algebraic graph theory: morphisms, monoids and matrices. Vol. 41. Walter de Gruyter GmbH & Co KG, 2019."

I've also found the following book, which I haven't read yet but it looks promising: "Lauri, Josef, and Raffaele Scapellato. Topics in graph automorphisms and reconstruction. Vol. 432. Cambridge University Press, 2016."

Title: Graph automorphism groups: reference request

Body: I would like to find something like dictionary that contains info about graph automorphisms groups. Recently have found "Algebraic Graph Theory" by Godsil but I cannot find something similar in this book. I assume that good references exist. I will be grateful for any comments.</p>


问:
Solving state-space function with using of Runge-Kutta method
I need to implement my own integration routine that will take state space function $f$, free variable $t$, and initial state $x(0)$ as input and produce the solution $x(t)$ as output. I thought that using Runge-Kutta method will be great. But I cannot understand how to apply it to the state-space function matrix.
$$\dot{x}=f(t,x)$$
$$\dot{x}=A\cdot{x}$$
I have only $A$ matrix. How can I apply Runge-Kutta method?
For example, could you provide step-by-step solution for:
$$
\begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
-10 & -5 & -2 \\
\end{bmatrix}
\quad
$$
Thanks in advance!

答:
You have a system $\dot{x} = f(x)$ with

$$
f(x) = A x = \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
-10 & -5 & -2
\end{pmatrix} \begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix} = \begin{pmatrix}
x_2 \\
x_3 \\
-10 x_1 - 5 x_2 - 2 x_3
\end{pmatrix}
$$

The formula for Runge Kutta with step size $h$ is:

$$
\begin{align}
k_1 &= f(x(t)) \\
k_2 &= f(x(t) + \frac{h}{2}k_1) \\
k_3 &= f(x(t) + \frac{h}{2}k_2) \\
k_4 &= f(x(t) + h k_3) \\
x(t + h) &= x(t) + \frac{h}{6}(k_1 + 2 k_2 + 2 k_3 + k_4)
\end{align}
$$

So all you need to do is choose a step size $h$ and initial condition $x(0) = x_0$.

For example use $h = 0.01$ and

$$
x_0 = \begin{pmatrix}
5 \\
-2 \\
3
\end{pmatrix}
$$

Insert this:

$$
\begin{align}
k_1 &= f(x_0) = \begin{pmatrix}
-2 \\
3 \\
-46
\end{pmatrix} \\
k_2 &= f(x_0 + \frac{h}{2}k_1) = \begin{pmatrix}
-1.985 \\
2.77 \\
-45.515
\end{pmatrix} \\
k_3 &= f(x_0 + \frac{h}{2}k_2) = \begin{pmatrix}
-1.98615 \\
2.772425 \\
-45.51485
\end{pmatrix} \\
k_4 &= f(x_0 + h k_3) = \begin{pmatrix}
-1.97227575 \\
2.5448515 \\
-45.02970925
\end{pmatrix} \\
\end{align}
$$

And so:

$$
x(0.01) = x(0) + \frac{0.01}{6}(k_1 + 2 k_2 + 2 k_3 + k_4) = \begin{pmatrix}
4.98014237375 \\
-1.972283830833333 \\
2.544850984583333
\end{pmatrix}
$$

If you repeat this 500 times you get:

If you simulate even longer you can see this is an undamped oscillation which makes sense because the three eigenvalues of $A$ are $-2, \sqrt{5}i, -\sqrt{5}i$.


问:
Infinite Collection of Sets with These Properties
I passed by this problem in Abbott's "Understanding Analysis", page 11:

Produce an infinite collection of sets $A_1, A_2, A_3,...$ with the
property that every $A_i$ has an infinite number of elements, $A_i ∩ A_j = \phi$ for all $i \not= j$ , and
$\bigcup_{i=1}^{\infty} A_i = \mathbf N$

Finding the solution was not difficult conceptually (that is, if my attempt is correct), but I have no idea how to formulate it.
My idea is as follows:

If the question asked for two sets, an answer would be the sets of even and odd numbers. If they're $3$ sets, I would then give $A_1$ all the numbers starting from $1$ and increment $3$ to get $A_1 = \{1,4,7,...\}$, similarly, for $A_2$, starting from $2$ and incrementing by $3$, etc... For $n$ sets, $A_i = \{i, i+n,i+2n,...\}$.

However, for an infinite number of sets, formulating a more formal solution is confusing to me.

My other question is this: Is this the only way to solve this problem, and why?

答:
There are many ways to do this. My favorite is to give $A_1$ every second natural number,
$$
A_1=\{1,3,5,7,\dots\},
$$
then give $A_2$ every second of the remaining numbers,
$$
A_2=\{2,6,10,14,\dots\},
$$
then give $A_3$ every second of the (still) remaining numbers,
$$
A_3=\{4,12,20,28,\dots\}
$$
and so on. Equivalently,
$$
A_n=\{2^{n-1}k:k\text{ is odd}\}.
$$
Another equivalent formulation is that $A_n$ consists of those positive integers whose binary expression ends with exactly $n-1$ zeros.


Title: Solve non-homogeneous recurrence $x_{n+2} - 2x_{n+1} + x_n = 8$

Body: I'm trying to solve this recurrence $x_{n+2} - 2x_{n+1} + x_n = 8$, but the solution of the homogeneous is $a_n=b+cn$, and I don't know how to change it to find the non-homogeneous solution.

Body:
$$x_{n+2} - 2x_{n+1} + x_n = 8\tag{1}$$
Set $n=n+1$
$$x_{n+3}-2x_{n+2}+x_{n+1}=8$$
Subtract the two equations
$$x_{n+3}-2x_{n+2}+x_{n+1}-x_{n+2} + 2x_{n+1} - x_n=0$$
$$x_{n+3}-3x_{n+2}+3x_{n+1}-x_n=0$$
Characteristic equation is
$$\lambda^3-3\lambda^2+3\lambda-1=0\to (\lambda-1)^3=0\to \lambda=1$$
and finally
$$x_n=an^2+bn+c\tag{2}$$

edit

from $(1)$ we get $x_0=8+2x_1-x_2$

from $(2)$ we have $x_0=c,\;x_1=a+b+c;\;x_2=4a+2b+c$

putting all together
$$\begin{cases}
c=8+2x_1-x_2\\
a+b+c=x_1\\
4a+2b+c=x_2\\
\end{cases}
$$
$$\begin{cases}
a+b+8+2x_1-x_2=x_1\to a+b=x_2-x_1-8\\
4a+2b+8+2x_1-x_2=x_2\to 4a+2b=2x_2-2x_1-8\\
\end{cases}
$$
$$\begin{cases}
a+b=x_2-x_1-8\\
2a+b=x_2-x_1-4\\
\end{cases}
$$
subtracting the two equations, we get $a=4$ so the $(2)$ becomes
$$x_n=4n^2+bn+c$$


问:
uniformly convergence of $f_{n}(x)=\frac{n}{2}\int_{x-\frac{1}{n}}^{x+\frac{1}{n}} f(t)dt$?
Let $f:[-1,2]\rightarrow \mathbb{R}$ continuous. Define a function sequences:
\begin{eqnarray*}
f_{n}(x)=\frac{n}{2}\int_{x-\frac{1}{n}}^{x+\frac{1}{n}} f(t)dt
\end{eqnarray*}
for all $x\in [0,1]$ and for all $n\in \mathbb{N}$.

Show that $(f_{n})$ converges to $f$ uniformly in $[0,1]$.

My work:

By hypotesis $f$ is continuous in $[-1,2]$ so, By first theorem of  first fundamental theorem of calculus $f'_{n}$ exist and :
\begin{eqnarray*}
f'_{n}=\frac{n}{2}(f(x+\frac{1}{n})-f(x-\frac{1}{n}))
\end{eqnarray*}
so in this case $x\in[0,1]$ and for all $n\in \mathbb{N}$.

So I think to analysis the uniformly convergence by differential theorem and study:
\begin{eqnarray*}
|f'_{n}(x)-f'(x)|&=&|\frac{n}{2}(f(x+\frac{1}{n})-f(x-\frac{1}{n}))-f'(x)|
\end{eqnarray*}
But I don't know if exist some point $x_{0}\in(0,1)$ such that $(f_{n}(x_{0}))$ converges and if $f'(x)$ exist.

答:
HINT

$$|f_n(x)-f(x)|=\left|\frac{n}{2}\int_{x-1/n}^{x+1/n}\left(f(t)-f(x)\right)dt\right| \le \frac{n}{2}\int_{x-1/n}^{x+1/n}|f(x)-f(t)|dt$$

Now you can use that $f$ is uniform continuous on $[-1,2]$ to finish.


问:
Interpretation of generating function infinite product
Let $P$ denote the set of primes and let $s\in\{-1,1\}$. How can you interpret the coefficient of $x^n$ in the power series expansion of $$\prod_{p\in P} (1+sx^p)^s$$

for either choice of $s$?
I found how to give an interpretation to $s=1$ but have no idea with $-1$.
Thank you for helping

EDIT: for s=1 i said that i can interpret the coeficient as the number of ways to pay the amount k cents with only coins of prime cents (with only one coin for each prime).

答:
Community wiki answer based on the comments so that the question can be marked as answered:

For $s=1$, the coefficient counts the number of ways of writing $n$ as a sum of distinct primes.

For $s=-1$, the coefficient counts the number of ways of writing $n$ as a sum of primes.


答:
hint 1:

Your statement, that the first statement is true if $\forall y P(y)$ is false,
was a mistake.
In your comment it seems like you thought that
if $\forall yP(y)$ does not hold, then $P(y)$ is always false.
It can also happen that $P(y)$ is false for some $y$ and true for some $y$.

You actually should assume, that $\forall y P(y)$ is false,
but there is an $y$ such that $P(y)$ is true.

hint 2:

Consider the two cases for $\exists x Q(x)$.
What happens if it is false? what if it is true?

Title: $\exists x \forall y (P(y) \implies Q(x))$ and $\forall y P(y) \implies \exists x Q(x)$ are not logically equivalent

Body:
I am trying to show that the following are not logically equivalent (according to a practice question)

$\exists x \forall y (P(y) \implies Q(x))$ and $\forall y P(y) \implies \exists x Q(x)$

In the first case I am trying to find some kind of statement where $x,y$ are integers (something like $P(x)$ is the is even predicate and $Q(x)$ is odd predicate, or maybe that $Q(x)$ implies $x$ divides $y$).

I am imagining I need a scenario where one statement is True implies False, and the other is True, for the same values of $x,y$.

If $\forall y P(y)$ is false, then both implications will be true, so suppose that $\forall y$ P(y) is true. I'm not sure how to proceed from here. Hints/Clarifications appreciated.


问:
How do we get the formula of the sequence?
A woman is walking home with distance $d$ and speed $v$.

The dog is happy and runs at speed $\frac{3v }{ 2}$ always between woman and House back and forth.

(i) At what distances $(d_n)_{n\geq 1}$ do women and dogs meet?

(ii) Determine the total path length of the dog with the help of $(d_n)_{n\geq 1}$.

$$$$

EDIT:

Let $x$ be the distance between the woman and the house and let $s$ be the distance between the dog and the house.

At the beginning we have $x_0=d$ and $s_0=0$.

It holds that $x=d-vt$ and $s=\frac 32v t$.

The woman and the dog meet at $x=s=d_1$ for $t=t_1$.

From $x=s$ we get $d-vt=\frac{3}{2}vt \Rightarrow \frac{5}{2}vt=d \Rightarrow t=\frac{2d}{5v}$, so $t_1=\frac{2d}{5v}$.

So we have that $d_1=x(t_1)=d-v\cdot \frac{2d}{5v}=d- \frac{2d}{5}=\frac{3d}{5}$.

After the meeting of the woman and the dog, the dog goes back to the house and then again to the woman.

At time $t_1$ the dog is at $s_1=\frac{3d}{5}$.

The dog goes in direction of the house, so $s=\frac{3d}{5}-\frac{3v}{2}t$.

When $s=0$ the dog arrived at the house, so $t=\frac{2d}{5v}$.

Then the dog goes back in direction of the woman, so $s= \frac{3}{2}vt$ with $t>\frac{2d}{5v}$.

At time $t_1$ the woman is also at $x_1=\frac{3d}{5}$.

The woman continues in diection to the house, so $x=\frac{3d}{5}-vt$.

For the second meeting, it holds $x=s=d_2$.

We get \begin{equation*}\frac{3d}{5}-vt=\frac{3}{2}vt \Rightarrow \frac{3d}{5}=\frac{5}{2}vt \Rightarrow t=\frac{6d}{25v}\end{equation*} So  they meet at \begin{equation*}d_2=x(t_2)=\frac{3d}{5}-v\cdot \frac{6d}{25v}=\frac{9d}{25}\end{equation*}.

So the $n$-th meeting is at $d_n =\left (\frac{3}{5}\right )^nd$.

Is that correct so far?

For the second question do we take the sum of all the terms $d_n$ ? Or should we take twice each term $d_n$ since the dog goes back and forth?

答:
If the woman and dog start at distance $d$ from the house and re-meet when the woman is at distance $\delta$ from the house, the woman will have walked a distance $d-\delta$ and the dog will have run a distance $d+\delta$. Now since the dog runs fifty percent faster than the woman walks, it goes fifty percent further in any given amount of time, meaning

$$d+\delta={3\over2}(d-\delta)$$

and thus $\delta={1\over5}d$. So the sequence of distances (from the house) that the woman and the dog meet up is $d_n=\left(1\over5\right)^nd$. The total distance the dog runs is

$$\begin{align}
(d+d_1)+(d_1+d_2)+(d_2+d_3)+\cdots
&=d+2d\sum_{n=1}^\infty\left(1\over5\right)^n\\
&=d+2d{1/5\over1-1/5}\\
&=d+2d{1\over5-1}\\
&=d+{2d\over4}\\
&={3\over2}d
\end{align}$$

which agrees with what we said earlier: the dog goes fifty percent further than the woman.


问:
C given A in a Bayesian network where $A\to B\to C$?
Let's say I have a Bayesian network with

```<code>A-->B-->C where  A,B,C have a Bernouilli distribution
</code>```

How do I calculate $P(C=1|A=1)$? Is it $P(C=1|B=1∩A=1) + (C=1|B=0∩A=1)$ or $P(C=1∩B=1|A=1) + (C=1∩B=0|A=1)$ or something else?

Thanks

答:
In general it is $$P(C=1\mid A=1) \\=P(C=1\cap B=1\mid A=1) + P(C=1\cap B=0\mid A=1)$$  which you can extend to $$ = P(C=1\mid B=1\cap A=1)P(B=1\mid A=1) \\+ P(C=1\mid B=0\cap A=1)P(B=0\mid A=1)$$

but if your particular model of the relationship is correct, this suggests you could go further and say

$$ = P(C=1\mid B=1)P(B=1\mid A=1) + P(C=1\mid B=0)P(B=0\mid A=1)$$


问:
Line Integral - Why are these two integrals the same?
I am starting to study calculus III and I came across the following situation

Given the following form $$ydx - xdy$$
Why the integral along the semicircle
$$P(t) = cos(t)\vec{i} + sin(t)\vec{j},\:0 \leq t \leq \pi$$
It is the same when using the following parameterization
$$y = \sqrt{1-x^2}, -1 \leq x \leq 1$$
Since the parameterizations are reversed?

When computing the first integral, I got $ - \pi $.

$$\int_c ydx - xdy$$
$$x = cos(t) \Rightarrow \frac{dx}{dt} = -sin(t)$$
$$y = sin(t) \Rightarrow \frac{dt}{dt} = cos(t)$$
$$\int_0^\pi ( sin(t)(-sin(t)) - cos(t)cos(t)) dt = -\pi$$

How do I compute the second one that has an inverted sense of integration than the first one?

Thanks in advance!

答:
You must know the orientation of the curve. In this case whether it is anti-clockwise or clockwise. That decides whether you go from $0$ to $\pi$ or $\pi$ to $0$ (similarly $-1$ to $1$ or from $1$ to $-1$).

$y = \sqrt{1-x^2}$

$dy = -\frac{x}{\sqrt{1-x^2}} \ dx$

Please note that if we have anti-clockwise orientation as in your first integral then we go from $x = 1$ to $x = -1$. So the line integral is

$\displaystyle \int_{1}^{-1} (\sqrt{1-x^2} + \frac{x^2}{\sqrt{1-x^2}}) \ dx = -\pi$


问:
Prove a coin game is fair

Suppose $2n$ identical, equal valued fair coins are flipped. If no more than half land head, you win all the heads. If more than half land head, you lose all the heads. Prove the game is fair with combinatorial argument only.

This is derived from a recently closed question, which was linked to Prove that $\sum\limits_{i=1}^n 2i\binom{2n}{n-i}= n\binom{2n}{n}$.

If the game is fair then
$$\sum_{i=0}^{n} i \binom{2n}{i} = \sum_{j=n+1}^{2n} j \binom{2n}{j}\\
\implies n\binom{2n}{n} + \sum_{i=0}^{n-1} i \binom{2n}{i} = \sum_{i=0}^{n-1} (2n-i) \binom{2n}{2n-i}\\
\implies n\binom{2n}{n} = \sum_{i=0}^{n-1} 2(n-i)\binom{2n}{i} = \sum_{i=0}^{n-1} 2i \binom{2n}{n-i}$$

I found that it reduced to proving this simple symmetry:

$$(n-i) \binom{2n}{n-i} = (n+i+1) \binom{2n}{n+i+1}, \forall 0 \le i \le n-1 \tag 1$$
i.e., your gain with $2n$ heads is offset by your loss with $2n+1$ heads, gain with $2n-1$ heads is offset by loss with $2n+2$ heads, and so on, and when all land tail you breakeven.

Algebraically $(1)$ is trivially true but is there a pure combinatorial argument? Or better yet, an argument to prove the game is fair without breaking it down?

答:
Are you familiar with the combinatorial proof that
$$
k\binom{n}k=n\binom{n-1}{k-1}?
$$
Both sides answer the question: "How many ways are there to make a committee of $k$ people from a pool of $n$ people, and then make one of the committee members president?"

Applying this to each of the equation $(n-i) \binom{2n}{n-i} = (n+i+1) \binom{2n}{n+i+1}$, you get
$$
2n\binom{2n-1}{n-i-1}=2n\binom{2n-1}{n+i},
$$
which is obviously true by symmetry.

Title: Prove a coin game is fair


问:
How can I prove $\mathbf{1}_{Pa}=P\{\cdot\}_a\circ\{\cdot\}_{Pa}$
We have an elementary topos $\mathbf{C}$ with subobject classifier $\Omega$ and powerset functor $P:\mathbf{C}^{op}\to\mathbf{C}$, i.e, for all $\mathbf{C}$-objects $x,y$ we have
$$\mathbf{C}(x\times y,\Omega)\cong\mathbf{C}(y,Px)$$
and the bijection is natural in both variables. There are "canonical" morphisms
$$\{\cdot\}_a:a\longrightarrow Pa$$
for every object $a$. In $\mathbf{Set}$, given a set $X$, the map $\{\cdot\}_X$ maps the element $x\in X$ to the subset $\{x\}\subseteq X$ i.e the element $\{x\}\in PX$.

Every map $f:x\to x'$ transforms to a map $Pf:Px'\to Px$ by contravariance, so we have

$$a\xrightarrow{\quad\{\cdot\}_a\quad}Pa\xrightarrow{\quad\{\cdot\}_{Pa}\quad}PPa$$
$$Pa\xleftarrow{\quad P\{\cdot\}_a\quad}PPa$$

and this begs the question: is $P\{\cdot\}_a\circ\{\cdot\}_{Pa}$ the identity on $Pa$? And is $\{\cdot\}_{Pa}\circ P\{\cdot\}_a$ the identity on $PPa$?

I would like an "elementary" proof (that doesn't use adjoints or dinatural transformations or weird existence theorems). I've tried almost everything that comes to mind (writing down the diagrams takes too much time). If a dinatural transformation is unavoidable, then I don't mind it, but not adjoints by any means.

The ultimate goal of this is to prove that
$$\in_a(f\times \mathbf{1}_{Pa})=\in_{x}(\mathbf{1}_x\times Pf)$$
for any objects $a,x$ and map $f:x\to a$. The arow $\in_x:x\times Px\to\Omega$ is given by the ismorphism above. All of this is outlined in Sheaves in Geometry and Logic: A First Introduction to Topos Theory by Mac Lane and Moerdijk.

Thanks!

答:
No, even in $\mathbf{Set}$, neither composition is the identity in general.  First, if $a$ is a set, then for $S \in Pa$, $\{\cdot\}_{Pa}$ would send $S \mapsto \{ S \}$.  Then,
$$P\{\cdot\}_a ( \{ S \} ) = \{ b \in a \mid \{ b \} \in \{ S \} \} = \{ b \in a \mid \{ b \} = S \}.$$
Therefore, if $S$ has two or more elements, then $P\{\cdot\}_a \circ \{\cdot\}_{Pa}$ sends $S \mapsto \emptyset$.  Thus $P\{\cdot\}_a \circ \{\cdot\}_{Pa} \ne \mathbf{1}_{Pa}$.

(I think in fact, it should be possible to prove in a general topos that $P\{\cdot\}_a \circ \{\cdot\}_{Pa} = \mathbf{1}_{Pa}$ if and only if $a$ is a subobject of 1, i.e. if and only if the unique morphism $a \to 1$ is a monomorphism.  Any proof I can think of, however, would probably not qualify as being "elementary".  For a nontrivial example where this condition can be satisfied: in the category of sheaves on a topological space $X$, the subobjects of 1 are exactly $h_U$ for open $U \subseteq X$ -- where by definition $h_U(V) = \begin{cases}1, & V \subseteq U; \\ 0, & V \not\subseteq U.\end{cases}$  Or, if you don't believe that the base universe satisfies excluded middle, then in general $h_U(V) = (\mathcal{T}_X)(V, U) = \{ x \in 1 \mid V \subseteq U \}$ where in the middle term we're using the partial order category on the collection of open subsets of $X$.)

As for the other composition, $\{\cdot\}_{Pa}$ is certainly never surjective (for instance $\emptyset$ is never in the image), so it is impossible to have $\{\cdot\}_{Pa} \circ P\{\cdot\}_a = \mathbf{1}_{PPa}$.

As for the ultimate question of how to show
$${\in}_a(f \times \mathbf{1}_{Pa}) = {\in}_x(\mathbf{1}_x \times Pf)$$
for $f \in \mathbf{C}(x, a)$: as much as you don't want to use bifunctors, I don't see any way around using the naturality of the isomorphism of bifunctors $\mathbf{C}(x \times y, \Omega) \simeq \mathbf{C}(y, Px)$.  If we let $\Phi$ be this isomorphism, then expanding the definitions, this naturality says exactly that for $f \in \mathbf{C}(x, x')$ and $g \in \mathbf{C}(y, y')$, and $h \in \mathbf{C}(x' \times y', \Omega)$, we have:
$$\Phi_{x,y}(h \circ (f \times g)) = Pf \circ \Phi_{x',y'}(h) \circ g.$$
Now, by definition of ${\in}_a$, we have that $\Phi_{a,Pa}({\in}_a) = \mathbf{1}_{Pa}$ and similarly for ${\in}_x$.  Therefore,
$$\Phi({\in}_a \circ (f \times \mathbf{1}_{Pa})) = Pf \circ \mathbf{1}_{Pa} \circ \mathbf{1}_{Pa} = Pf.$$
Similarly,
$$\Phi({\in}_x \circ (\mathbf{1}_x \times Pf)) = P\mathbf{1}_x \circ \mathbf{1}_{Px} \circ Pf = Pf.$$
Since $\Phi$ is an isomorphism, and $\Phi$ of both sides of the desired identity evaluates to the same thing, we conclude that the desired identity is true.


问:
Finding the center of mass of a composite lamina
The following lamina is given and I have to find the distance from center of mass from AB and AC

I started by splitting the lamina into two shapes, one rectangle of width $4a$ height $5a$ and one of height $3a$ and width $3a$ which is a square.

$$\because \bar{x}=\frac{\sum_{i=0}^n \widehat{x}_i A_i}{\sum_{i=0}^n A_i}$$

$$\therefore \bar{x}=\frac{(\frac{4a}{2}((4a)(5a))+(\frac{3a}{2}(3a)^2)}{(3a)^2+(4a)(5a)}=\frac{107a}{58}$$

And similarly

$$\bar{y}=\frac{(\frac{5a}{2}((4a)(5a))+(\frac{3a}{2}(3a)^2)}{(3a)^2+(4a)(5a)} =\frac{127a}{58}$$
Hence the coordinates of the center of mass are
$$<\frac{107a}{58}, \frac{127a}{58}>$$

However the actual answer is

$\frac{179a}{58}$ from AB and $\frac{127a}{58}$ from AC

I am entirely baffled on how to get this answer and have little to no clue what Ive done wrong in solving this.

答:
Try

$$\therefore \bar{x}=\frac{(\frac{4a}{2}((4a)(5a))+((\frac{3a}{2}+4a)(3a)^2)}{(3a)^2+(4a)(5a)}= ? $$

Where $4a$ is a shift from the left side. The right rectangle does not start at the origin so you must take this shift into account


问:
$2$-Sylow of a subgroup of $A_n$
I don't know how to start to prove this:

Let $H$ be a subgroup of $S_n$ and let $T$ be a subgroup of $H$ such that $T$ is a $2$-Sylow of $H$.

Show that $H$ is conntained in $A_n\iff$ $T\le A_n$.

Maybe this is trivial but I'm having problems to solve this exercise...

答:
The forward implication is immediate.

For the reverse implication, suppose $T\le A_n$.

We want to show $H\le A_n$.

Suppose instead that $h\in H$, where $h$ is an odd permutation.

Our goal is to derive a contradiction.

Since $h$ is an odd permutation, and since cycles of odd order are even permutations, at least one of the cycles in the disjoint cycle decomposition of $h$ must have even order, hence $h$ also has even order.

Since $h$ has even order, some odd power of $h$, say $h^m$, has order which is a power of $2$, so $h^m$ must be an element of some $2$-Sylow subgroup of $H$.

Since the $2$-Sylow subgroups of $H$ are conjugate to each other, it follows that $h^m$ is conjugate to some element of $T$.

Since $T\le A_n$, all elements of $T$ are even permutations, hence since conjugation preserves parity, $h^m$ is an even permutation.

But $h$ is an odd permutation, hence since $m$ is odd, $h^m$ is an odd permutation, contradiction.

Title: $2$-Sylow of a subgroup of $A_n$


问:
Does there exist a continuous surjection $\mathbb R\to\mathbb R\times[0,1]$?
Study if there is a continous and surjective function, $f: \mathbb{R} \longrightarrow \mathbb{R}\times [0, 1]$ both with usual topolgy at $\mathbb{R}$ and at $\mathbb{R}\times [0,1]$, as i mentioned before i couldn't find a counter example and a surjective function neither, i will really appreciate any kind of help, thank you.

答:
The Cantor function is a continuous function from a Cantor set $C$ onto $[0,1]$.

As $C \times C \simeq C$ we in fact have a continuous function from $C$ onto $[0,1] \times [0,1]$ as well.

Take such a function $f_n, n \in \Bbb Z$ for a homeomorphic copy $C_n$ of $C$ in $[n,n+1]$ onto $[n,n+1] \times [0,1]$.

Now union up all such functions and extend linearly (or by Tietze) to get a continuous onto map from $\Bbb R$ to $\Bbb R \times [0,1]$.


问:
How to calculate the average time it takes to get to a value given an array of values
Im sorry if this is a confusing question. I'm terrible at math so please let me know if I need to elaborate on anything.

I have an array of values V

I have a target value that I would like to hit T

There are N "cycles" that happen where a random number is picked from V and added to the total amount. What is the average (A) number of times we hit T.

For example:

```<code>V = [8,9,10,11,12,13,14,15,16,17,18,19,20]
N = 10
T = 100
A = ???
</code>```

So, to explain. I have 10 random picks from my array V. On average, how likely am I that my 10 picks will >= T

Again, Im so sorry if this doesn't make sense.

Thanks for your time,

答:
There's no way to do this without a bunch of calculation.  The first thing to notice is that it's very unlikely that we won't reach $100$ in $10$ trials; even if we pick $8$ every time, we get close.  So it will be easy to calculate the probability that we don't reach $100$ and subtract from $1$.

Consider the expression $$p(x)=(x^8+x^9+\cdots+x^{19}+x^{20})^{10}$$ For any sequence of $10$ picks, take the terms in the $10$ factors corresponding to the picks and multiply them.  That is, if the first pick is $11$ we take $x^{11}$ from the first factor.  If the second pick is $14$, we take $x^{14}$ from the second factor.  In the product, the coefficient of any term $x^k$ is the number of ways to get a sum of $k$.

Now $$\begin{align}p(x)&=x^{80}(1+x+x^2+\cdots+x^{12})^{10}\\&=x^{80}\frac{(x^{13}-1)^{10}}{(x-1)^{10}}\\&=(x^{13}-1)^{10}(x-1)^{-10}\\
&=x^{80}(x^{13}-1)^{10}\sum_{k=0}^\infty(-1)^k\binom{-10}{k}x^k\\
&=x^{80}(x^{13}-1)^{10}\sum_{k=0}^\infty\binom{k+9}{k}
\end{align}$$ by the binomial theorem.
Now of course, we could expand the product and pick out the terms of degree less than $100$ manually, but I used a CAS at this point.  Notice that we don't really need an infinite sum,  We can cut it off at $k=19$, since we are only interest in terms of degree less than $100.$

```<code>from sympy import binomial, Symbol

x = Symbol('x')

p = sum(binomial(9+k,k)*x**k for k in range(20)).as_poly()
q = ((x**3-1)**10).as_poly()
r = p*q
a = r.coeffs()[-20:]
print(a)
print(1-sum(a)/13**10)
print(float(1-sum(a)/13**10))

from random import choices

trials = 1000000
count = 0
pop = list(range(8,21))
for _ in range(trials):
    c = choices(pop, k=10)
    count += sum(c) >= 100
print(count/trials)
</code>```

This produced the output

```<code>[10, 55, 210, 615, 1452, 2850, 4740, 6765, 8350, 8953, 8350, 6765, 4740, 2850, 1452, 615, 210, 55, 10, 1]
137858432801/137858491849
0.9999995716767301
0.999857
</code>```

Sympy write the coefficients in order of significance, so that $1$ is the coefficient of $x^{80}$, $10$ the coefficient of $x^{81}$ and so on.  You can check the first few and see that they are correct.  The exact probability is $$\frac{137858432801}{137858491849}\approx .9999995716767301$$

I also simulated $1,000,000$ drawings, with $999,857$ successes.

Body:
There's no way to do this without a bunch of calculation.  The first thing to notice is that it's very unlikely that we won't reach $100$ in $10$ trials; even if we pick $8$ every time, we get close.  So it will be easy to calculate the probability that we don't reach $100$ and subtract from $1$.

Consider the expression $$p(x)=(x^8+x^9+\cdots+x^{19}+x^{20})^{10}$$ For any sequence of $10$ picks, take the terms in the $10$ factors corresponding to the picks and multiply them.  That is, if the first pick is $11$ we take $x^{11}$ from the first factor.  If the second pick is $14$, we take $x^{14}$ from the second factor.  In the product, the coefficient of any term $x^k$ is the number of ways to get a sum of $k$.

Now $$\begin{align}p(x)&=x^{80}(1+x+x^2+\cdots+x^{12})^{10}\\&=x^{80}\frac{(x^{13}-1)^{10}}{(x-1)^{10}}\\&=(x^{13}-1)^{10}(x-1)^{-10}\\
&=x^{80}(x^{13}-1)^{10}\sum_{k=0}^\infty(-1)^k\binom{-10}{k}x^k\\
&=x^{80}(x^{13}-1)^{10}\sum_{k=0}^\infty\binom{k+9}{k}
\end{align}$$ by the binomial theorem.
Now of course, we could expand the product and pick out the terms of degree less than $100$ manually, but I used a CAS at this point.  Notice that we don't really need an infinite sum,  We can cut it off at $k=19$, since we are only interest in terms of degree less than $100.$

from sympy import binomial, Symbol

x = Symbol('x')

p = sum(binomial(9+k,k)*x**k for k in range(20)).as_poly()
q = ((x**3-1)**10).as_poly()
r = p*q
a = r.coeffs()[-20:]
print(a)
print(1-sum(a)/13**10)
print(float(1-sum(a)/13**10))

from random import choices

trials = 1000000
count = 0
pop = list(range(8,21))
for _ in range(trials):
    c = choices(pop, k=10)
    count += sum(c) >= 100
print(count/trials)

This produced the output

[10, 55, 210, 615, 1452, 2850, 4740, 6765, 8350, 8953, 8350, 6765, 4740, 2850, 1452, 615, 210, 55, 10, 1]
137858432801/137858491849
0.9999995716767301
0.999857

Sympy write the coefficients in order of significance, so that $1$ is the coefficient of $x^{80}$, $10$ the coefficient of $x^{81}$ and so on.  You can check the first few and see that they are correct.  The exact probability is $$\frac{137858432801}{137858491849}\approx .9999995716767301$$

I also simulated $1,000,000$ drawings, with $999,857$ successes.


问:
Determine which of the following spaces are separable
A topological space $(X,\tau)$ is said to be seperable if it has a  dense subset which is countable.Determine which of the following spaces are seperable.

(i) the set $\mathbb{R}$ with the usual topology:

$\mathbb{Q} \subset \mathbb{R}$ is a dense subset and is countable,So $\mathbb{R}$ is seperable.

(ii)a countable set with discrete topology :

All proper subset of $X$ will be close ,so its closure is itself.So they aren't dense.$X \in \tau$ So $X$ is open and $\phi \in \tau$ so $X$ is close also.$X$ is closed,so its closure is $X$ itself.So is dense and $X$ is countable.So $(X,\tau)$ is seperable.

(iii)a countable set with the finite closed topology:

Similarly here $X$ is dense and being countable,the topological space is seperable.

(iv)$(X,\tau)$ where $X$ is finite.

Similarly here $X$ is dense in itself and being finite is countable.So Topological space is seperable.

(v)$(X,\tau)$ where $\tau$ is finite.

Taking $\tau$ as indiscrete topology.So only open and close sets are $X$ and $\phi$. $X$ is dense but unless it is countable we cannot say $\tau$  is seperable.

(vi)a uncountable set with the discrete topology.

only dense set is $X$ but $X$ is not countable so Topology is not seperable.

(vii)an uncountable set with the finite closed topology

Here $X$ is dense set but not countable.I am unable to conclude whether any countable subset of $X$ can be dense or not?

Please check i believe my arguments are correct for first six topology.And give me a hint for seventh one.

答:
For v : pick one element from each $O \in \tau\setminus\{\emptyset\}$ and convince yourself all such points together are dense.

So in fact any space where $\tau$ is at most countable is separable.

(vii): any countably infinite subset cannot be closed (it's not finite) so the only closed superset is $X$ and so any countably infinite subset (which exists) is dense.

The other arguments are OK.


答: Your mistake is the $x$ cordinate of the square that you have written. If you are taking the origin at A, so the $x$ coordinate would be $4a+\frac{3a}{2}=\frac{11a}{2}$. Now if you do the calculation, you will get the right answer.

Title: Finding the center of mass of a composite lamina

Body:
The following lamina is given and I have to find the distance from center of mass from AB and AC

I started by splitting the lamina into two shapes, one rectangle of width $4a$ height $5a$ and one of height $3a$ and width $3a$ which is a square.

$$\because \bar{x}=\frac{\sum_{i=0}^n \widehat{x}_i A_i}{\sum_{i=0}^n A_i}$$

$$\therefore \bar{x}=\frac{(\frac{4a}{2}((4a)(5a))+(\frac{3a}{2}(3a)^2)}{(3a)^2+(4a)(5a)}=\frac{107a}{58}$$

And similarly

$$\bar{y}=\frac{(\frac{5a}{2}((4a)(5a))+(\frac{3a}{2}(3a)^2)}{(3a)^2+(4a)(5a)} =\frac{127a}{58}$$
Hence the coordinates of the center of mass are
$$<\frac{107a}{58}, \frac{127a}{58}>$$

However the actual answer is

$\frac{179a}{58}$ from AB and $\frac{127a}{58}$ from AC

I am entirely baffled on how to get this answer and have little to no clue what Ive done wrong in solving this.


问:
Limit of $(x\cot(x)-1)/(-(17x^2))$ when $x \to 0$
I need to find the limit:
$$\lim_{x\to0}\frac{x\cot(x)-1}{-17x^2}$$

I've been tried L'Hopital and Taylor but it's not going so well..
thank in advance

答:
Welcome to MSE: $$\lim_{x\to0}\frac{x\cot(x)-1}{-17x^2}=\\
\lim_{x\to0}\frac{x\cot(x)-1}{-17x^2}\times \frac{\tan x}{\tan x}=\\
\lim_{x\to0}\frac{x-\tan x}{-17x^2\tan x}$$ then use...$$x\to 0 \\\tan x \sim x+ \frac {x^3}{3}$$


问:
Surfaces: behavior of Pluricanonical maps
Let $S$ be an algebraic smooth surface (over $\Bbb{C}$).

Suppose we have some fibration $p\colon S\rightarrow C$ onto a smooth curve. Let $f$ be a fiber, a curve, say.

Let $K$ be a canonical divisor on $S$ and assume $K\cdot f=0$. My question is: can we conclude that each $m$-canonical map $\varphi_{mK}\colon S\rightarrow\Bbb{P}^N$, when defined, will contract $f$ to a point ?

答:
If $|mK|$ is base point free and if $f$ is irreducible, then $\varphi_{mK}(f)$ is a point.

Otherwise, there exists $x$, $y\in f$ s.t. $\varphi_{mK}(x)\neq \varphi_{mK}(y)$. By the definition of $\varphi_{mK}$, it is equivalent to say that there exists $s\in H^0(S,\mathcal{O}_S(mK))$, such that $s(x)=0$ and $s(y)\neq 0$.

Let $D:=div(s)$, then $D$ is an effective divisor, linearly equivalent to $mK$, and satisfying $x\in \operatorname{Supp}(D)$ while $y\notin \operatorname{Supp}(D)$. Therefore $f$ is not a component of $D$ (since $y\notin \operatorname{Supp}(D)$), hence $D\cdot f\geq 1$ (since $D$, $f$ have no components in common and $x\in \operatorname{Supp}(D)\cap f$). So $mK\cdot f=D\cdot f\geq 1$, we get a contradiction.


Title: Limit of $(x\cot(x)-1)/(-(17x^2))$ when $x \to 0$

Body:
I need to find the limit:
$$\lim_{x\to0}\frac{x\cot(x)-1}{-17x^2}$$

I've been tried L'Hopital and Taylor but it's not going so well..
thank in advance

Body:
Hint

$$\frac{x}{\sin(x)}=\frac{x}{x-\frac{x^3}{6}(1+\epsilon(x))}$$

$$=\frac{1}{1-\frac{x^2}{6}(1+\epsilon(x))}$$

$$=1+\frac{x^2}{6}(1+\epsilon(x))$$

So,

$$x \;cot(x)=\frac{x}{\sin(x)}\cos(x)$$
$$=(1+\frac{x^2}{6})(1-\frac{x^2}{2})+x^2\epsilon(x)$$

$$=1-\frac{x^2}{2}+\frac{x^2}{6}+x^2\epsilon(x)$$

You will find $\boxed{ \frac{1}{51}}$.


问:
$\exists x \forall y (P(y) \implies Q(x))$ and $\forall y P(y) \implies \exists x Q(x)$ are not logically equivalent
I am trying to show that the following are not logically equivalent (according to a practice question)

$\exists x \forall y (P(y) \implies Q(x))$ and $\forall y P(y) \implies \exists x Q(x)$

In the first case I am trying to find some kind of statement where $x,y$ are integers (something like $P(x)$ is the is even predicate and $Q(x)$ is odd predicate, or maybe that $Q(x)$ implies $x$ divides $y$).

I am imagining I need a scenario where one statement is True implies False, and the other is True, for the same values of $x,y$.

If $\forall y P(y)$ is false, then both implications will be true, so suppose that $\forall y$ P(y) is true. I'm not sure how to proceed from here. Hints/Clarifications appreciated.

答:
As per supinf's hints. How about this argument:

Suppose that $\exists y$ such that $P(y)$ is true, but $\forall y P(y)$ is false. Also suppose that $Q(x)$ is false for any $x$.

Then the first logical statement is false since for such a $y$ where $P(y)$ is true, we would have there exists an $x$ such that $P(y) \implies Q(x)$ is true, which is false, and so the overall statement is false.

For the second statement, since the premise $\forall y P(y)$ is false, then the statement is vacuously true. Hence the statements aren't equivalent.


问:
Sum of Fibonacci reciprocals converges
There are already proofs for this on Stackexchange, but I am seeking a specific one I am interested in.

Let $\{u_k\}^\infty_{k=0}$ be the Fibonacci sequence, where $u_0=u_1=1$ and $u_{k+1}=u_k+u_{k-1}$. Then $\sum_{k=1}^{\infty} \tfrac{1}{u_k}$ is convergent. I want to prove it with the following two statements:

For $a_k≠0, \forall k \in \mathbb{N}$, if $$\limsup_{k\to\infty}\left|\frac{a_{k+2}}{a_k}\right|<1$$ then the series $\sum_{k=1}^\infty a_k$ is absolutely convergent.

$\limsup_{k\to\infty}\left|\frac{u_{k-1}}{u_{k+1}}\right|<1$

Can anyone help me?

答:
First notice that $u_k \geq 1$ for all $k$ and $\{u_k\}$ is increasing monotonously. From $u_{k+1}=u_k+u_{k-1}$ and $u_k \geq u_{k-1}$ follows, that $$u_{k+1} \geq 2u_{k-1} $$ and thus $$\frac{a_{k+1}}{a_{k-1}}=\frac{u_{k-1}}{u_{k+1}} \leq \frac{1}{2},$$ whereas $a_k=\frac{1}{u_k}$. From this we can deduce that

$$\limsup_{k\to\infty}\left|\frac{a_{k+1}}{a_{k-1}}\right|\leq \frac{1}{2}.$$

From the fact that $\sum a_k$ converges if
$\limsup_{k\to\infty}\left|\frac{a_{k+2}}{a_{k}}\right|<1$, we can see that $\sum \frac{1}{u_k}$ converges.

Title: Sum of Fibonacci reciprocals converges


问:
Convergence of Sequences in a Topology Space
Let $X$ be the real line with the topology $\tau$ in which the open sets are the sets of the form $\left(a,\infty\right)$, $a\in\mathbb{R}$. My question is, what should go in the " " to make the following true:

Say $\left(x_n\right)_{n=1}^{\infty}$ is a sequence in $X$. We claim that $\left(x_n\right)$ converges in $\left(X,\tau\right)$ if and only if " ".

I know that constant sequence converge. Take $x_n:=(1,2,3,4,5,6,...)$. It appears that such a sequence converges to every value in $\Bbb R$. For instance, $0\in U\in \tau$, then $a<0<\infty$. Clearly, at some point $x_n>a$ and hence $x_n\in U$. I thought possibly the " " should be something about $x_n$ being increasing. However, taking $x_n:=1/n$, we see $(x_n)$ converges to $0$.

Any help is appreciated.

答:
If $(x_n)_n$ is bounded below, say by $m$, then the sequence has at least $m$ as a limit: any open neighbourhood of $m$ is of the form $(a,\infty)$ where $a < m$. Bu then all $x_n$, which by definition obey $m \le x_n$ so that $x_n \in (a,\infty)$ for all $n$.

If the sequence converges to some $p\in \Bbb R$ then all but finitely many terms $x_n$ lie in $(p-1,\infty)$ so it follows easily that then the sequence is bounded below (it's easy to take care of the finitely many outliers).

So the condition is both necessary and sufficient.


问:
Maximum of weighted sum of exponentials
Given a function
$$f(t) := \sum_k a_k e^{b_k t} $$
for some $b_k < 0$, we want to find maximum of $f(t)$ over $t > 0$ algorithmically.

A solution would be to calculate $f'(t)$ and and then convert $e^{-t}$ to a new variable, say $z$, and find roots of a polynomial with (potentially) non-integer exponents. Is there any better algorithm in terms of computational complexity?

答: Newton’s algorithm is the way to go.</p>

Title: Maximum of weighted sum of exponentials

Body: Newton’s algorithm is the way to go.


问:
What is the motivation behind the Factorization Systems of Model Categories?
I am currently trying to get a grasp of higher category theory, being promised to get a nice framework to do homotopy theory (which I currently understand to be a theory of dealing with categories with weak equivalences generated by / interacting with some notion of interval/path-object/cylinder-object).

When reading about quasicategories it is unavoidable to learn about model categories, especially when following the approach of Prof. Cisinski's Higher Categories and Homotopical Algebra. I do appreciate the fact that model categories give very nice computational tools (e.g. for computing the homotopy category and some notions of homotopy limits), but the motivation behind their definition is elusive to me. Let me make precise what I mean:

Consider the following settings of homotopy theory we might try to put into a common framework:

For topological spaces the standard interval $I=[0,1]$ lets us generate a notion of weak equivalence called homotopy equivalence and we would like to study spaces up to homotopy equivalence.

Introducing homotopy groups, we might want to classify topological spaces up to their homotopy groups and the notion of weak homotopy equivalence is born. Yet for good spaces (e.g. CW-complexes) the notion of weak homotopy equivalence coincides with the notion of homotopy equivalence in (1), so the interval comes into play again.

In homological algebra we consider the category of chain complexes and there is a notion of homotopy of maps of complexes, which again can be generated by some interval object. We want to study chain complexes up to their homology groups, so again the notion of homotopy equivalence of chain complexes is to broad and we need to use quasiisomorphisms as weak equivalences.

Category theory usually works up to equivalence of categories, which is strictly weaker than isomorphisms of categories. One can show that this notion of weak equivalence is yet again generated by the interval object given by the free category with an isomorphism.

So it becomes clear that a nice 1-categorical framework should incorporate a well behaved notion of weak equivalence and (maybe) a notion of interval generating or at least being related to this.

But where do the weak factorization systems come from?

In the case of (1) one might have the idea that having the homotopy lifting property (Hurewicz-fibration) or the homotopy extension property (Hurewicz-cofibration) is a useful thing to have. To me it is already slightly less clear, why one might have the idea in (2) to consider Serre-fibrations, and completely unclear, where Serre-cofibrations arise. Similarly it is unclear to me why I should have the idea to look at similar lifting problems in (3) or (4) and to expect them to behave as in the axioms of a model category.

It is likely that there is no general pattern to notice here and that model categories are modelled after (1) and then were observed to work in most other settings as well. Or they may have evolved slowly by picking up on tricks and facts used here and there. But I still have hope for a good general reason to consider these lifting properties, similar to how trying to figure out what associativity of $n$ morphisms up to homotopy leads to something simplicial.

TLDR: What is the intuitive (besides they get the job done) reason to consider weak factorization systems as used in the definition of a model category?

As always thank you very much for your time, considerations and answers!

答:
One thing you get from studying (the homotopy theory of) topological spaces, simplicial sets or chain complexes, is that not all maps behave well with respect to homotopical notions.

For instance, one very "annoying" thing is that pullback along arbitrary maps might not preserve weak equivalences. But you notice that some maps behave well with respect to homotopical notions, for instance pullback along certain maps do preserve weak equivalences.

There are two very different situations in which maps behave well : when they look like very neat inclusions, or very neat surjections. Of course the meaning of "very neat" varies from one context to another, but these are the kinds of maps that behave correctly with respect to homotopy theory. With hindsight, it's pretty reasonable : if a map is a very neat inclusion, you have some "wiggle room" for your homotopies (in fact this is exactly what happens in the homotopy theory of spaces)

Then you also notice that in your classical settings, every map can be factored as a very neat inclusion followed by a very neat surjection (and in fact you can choose one of them to be a weak equivalence) - this is motivated by the very classical constructions of cones and path spaces in the homotopy theory of spaces, which existed beforehand.

What this tells you in particular, is that you can reduce many things to these nicely behaved maps, and since they are nicely behaved, you can do lots of things even with arbitrary maps - in particular it is clear how this factorization property is desirable. If these maps are to behave well, it's good to know that up to homotopy, any map looks like one that behaves well !

One thing that's important to spot here is that these notions came from the practice of homotopy theory - Quillen (and others) saw how people did homotopy theory and the commonalities in those contexts, and what they were used for. Now this idea of approximating arbitrary maps by nice maps not only appeared in the contexts he looked at , but also seemed essential and pretty practical.

Note, in particular, that it's not at all the case that you need a whole factorization system in general. The main point is the approximation of maps by nice maps - for instance since you're reading Cisinski's book, you'll see later that he has chapters about ($\infty$-)categories with weak equivalences and fibrations. In that setting you don't have a whole factorization system, but you still have those nice maps (the fibrations) that carry your computations and also help you control your localization - since not every map is nice, it's good to have nice maps that approximate things.

tldr : Factorization systems aren't the conceptual thing. The conceptual thing underlying these notions is that of having maps that behave nicely with respect to your "homotopy theory"; and being able to approximate arbitrary maps by nicely behaved maps.

In the classical settings, there are two ways to be well-behaved : being a neat inclusion, or being a neat surjection. In many settings, you have both, and you can factor your maps in the way you probably know, and this is a huge technical convenience.
In some settings you only have one kind - this is still very good, as long as you can approximate arbitrary maps by nice map, but it's less technically convenient.


问:
Find the smallest m for which this is possible
Problem:
Eight singers participate in an art festival where m songs are performed. Each song is performed by 4 singers, and each pair of singers performs together in the same number of songs. Find the smallest m for which is possible.

My solution :

Let $k$ is the number songs that each pair of singers performs together.

Now if singer A and B sing together one song we count 1

Now each song performs by 4 singers so . There is all $m.C_4^2$.

$$kC_2^8=mC_2^4$$

So the smallest m is 14 and at that time $k=3$

答:
Your logic is correct, if hard to follow.  I would say it more like there are $C^4_2=6$ pairs of singers on the stage for each song, which means $6m$ pairs perform in total.  There are $C^8_2=28$ pairs of singers, so we need $6m$ to be a multiple of $28$, which gives $m=14$ as the smallest solution.  This leads to each pair being on the stage $3$ times.

This proves that the number of songs must be a multiple of $14$, but you have not shown that there is a solution with $m=14$.  It might not be possible to arrange the singers properly.  To find a solution we think about the seven songs that singer $1$ sings.  She is paired three times with each other singer, so each other singer gets paired with six others besides $1.$  It is appealing to make each pair appear once together during the seven songs.  Then we can take the opposite group for the other seven songs.  This leads to $$\begin {array}&1234&5678\\1256&3478\\1278&3456\\1357&2468\\1368&2457\\1458&2367\\1467&2358 \end{array}$$ which I believe satisfies the requirement.


问:
Function from $\mathbb R^2$ to $\mathbb R$ and connection subset
I'm trying to figure it out for a really long time. Does there exist a function from $\mathbb R^2$ to $\mathbb R$ that while  $ f \le 1 $ there is infinitely many connected components and at the same time $ f < 1 $ is connected.

答:
Consider the following sets:
$$H = \{ (x,y) : y<0\}$$
$$S= \bigcup_{n \in \Bbb N} \{ (x,y): |y-n| < 1/3  \ \ ; \ \ |x| <1\}$$

Here is a sketch of the two sets. Note that $S$ is made by infinitely many disjoint rectangles.

Now define $f: \Bbb R^2 \to \Bbb R$
$$f(p)=\begin{cases}0 & \mbox{ if } p \in H \\ 1 & \mbox{ if } p  \in S \\ 2 & \mbox{ otherwise}
\end{cases}$$
$\{ f \le 1\} = H \cup S$ has infinitely many disjoint components.

$\{ f < 1\} = H$ is connected.


Title: Function from $\mathbb R^2$ to $\mathbb R$ and connection subset

Body: I'm trying to figure it out for a really long time. Does there exist a function from $\mathbb R^2$ to $\mathbb R$ that while  $ f \le 1 $ there is infinitely many connected components and at the same time $ f < 1 $ is connected.

Body: Define $f(0,0)=1/2,$ $f(n,0)=1, n=1,2\dots,$ and $f=2$ everywhere else. Then $f^{-1}(-\infty,1) = \{(0,0)\}$ and $f^{-1}(-\infty,1]$ $ = \{(n,0):n=0,1,\dots\}.$


问:
Can somebody provide me with an example for a sequence which has all integer numbers as limits points.
Can somebody provide me with an example for a sequence which has as limits points the set $Z$.

Because $Z$ is a closed set than it can be the set of all the limits points.

答:
Basically if $\varphi$ is any bijection between $\mathbb N^2$ and $\mathbb N$ then calling $\varphi^{-1}(n)=(n_1,n_2)$

We get that $x_n=n_1+\dfrac 1{n_2}$ will have all natural numbers as accumulations points.

It is not too hard to adapt it to negative numbers (split for instance odd/even $\mapsto$ negative/positive).

For instance $\varphi^{-1}(n)=\bigg(\min(n-\lfloor\sqrt{n}\rfloor^2,\lfloor\sqrt{n}\rfloor)\quad,\quad2\lfloor\sqrt{n}\rfloor-\max(n-\lfloor\sqrt{n}\rfloor^2,\lfloor\sqrt{n}\rfloor)\bigg)$

--> see this post How to show that $\max(a,b)^2+\max(a,b)+a-b$ is bijective?

But there are others like Cantor-pairing function and such, it just happens I had this one above at hand...


问:
$x|\mu \sim N(\mu, \sigma^2)$ and $\mu|x_0 \sim N(x_0, \sigma^2)$ what is $x|x_0$
I am trying to calculate the distribution of some univariate variable $X$ that follows a Normal distribution with mean $\mu$ but $\mu$ itself is also random. For simplicity I assume Normal distributions with the same known variance. Hence:

$$ p(x | \mu) = \frac{1}{\sqrt {2 \pi \sigma^2}} \exp-\frac{(x-\mu)^2}{2 \sigma^2}$$

and

$$ p(\mu | x_o) = \frac{1}{\sqrt {2 \pi \sigma^2}} \exp-\frac{(\mu - x_0)^2}{2 \sigma^2}$$

Therefore:

\begin{align}
      p(x) & = \int p(x|\mu) \, \,p(\mu|x_0) \, d\mu \\\\
      & = \frac{1}{2\pi\sigma^2} \int \exp\big[-\frac{1}{2\sigma^2}\big( (x-\mu)^2 + (\mu-x_0)^2 \big)\big]d\mu
\end{align}

Expanding now the demoninator inside the exponential function we get
\begin{align}
(x-\mu)^2 + (\mu-x_0)^2 & = x^2 + \mu^2 - 2 x \mu + \mu^2 + x_0^2 - 2 x_0 \mu \\\\
& = 2\mu^2 -2\mu(x + x_0) +x^2+x_0^2 \\\\
& = 2\mu^2 -2\mu(x + x_0) +x^2+x_0^2 + 2\, x\,x_0 - 2\, x\,x_0 \\\\
& = 2\mu^2 -2\mu(x + x_0) +(x+x_0)^2 - 2\, x\,x_0 \\\\
& = \mu + (\mu-(x+x_0))^2 -  2\, x\,x_0
\end{align}

Can I work this any further please and get the distribution of $x$ (given $x_0$) or am I stuck? Is there another more (mathematically) convenient distribution for $\mu|x_0$ that is typically used in these cases?

答:
A simpler way to look at this, without needing to use the density, is to recognise that $\mathcal{N}(\mu,\sigma^2)\sim\mathcal{N}(0,\sigma^2)+\mu$. From that, you can say

\begin{align}
x\vert x_0 &\sim \mathcal{N}(0,\sigma^2)+\mu\vert x_0\\
&\sim \mathcal{N}(0,\sigma^2)+\mathcal{N}(0,\sigma^2)+x_0\\
&\sim\mathcal{N}(0,2\sigma^2)+x_0\\
&\sim\mathcal{N}(x_0,2\sigma^2)
\end{align}


问:
limit on incomplete gamma functions
I am trying to prove that
$$
\frac{x^{k}}{{\rm e}^{xk}\int_{x}^{\infty}
{\rm e}^{-tk}\,\,t^{k}\,{\rm d}t} =
ak + b + o\left(1\right)
$$
when $k\to \infty$ for any $x> 1$, or equivalently
$$
\int_{0}^{}\left({\rm e}^{-s}\,\,
\frac{x + s}{x}\right)^{k}\,{\rm d}s =
ck^{-1} + dk^{-2} + o\left(k^{-2}\,\right)
$$
where I can derive what should be the correct $a,b,c,d$.

Wolfram seems unable to expand or take limits when functions like $\Gamma(k + 1,kx)$ are involved.

Any idea $?$.

答:
For $k \in \mathbb{N}$ define $f_k \colon (1,\infty) \to \mathbb{R}$ by
$$ f_k (x) = \frac{\mathrm{e}^{k x}}{x^k} \int \limits_x^\infty t^k \mathrm{e}^{-k t} \, \mathrm{d} t \stackrel{t \,= \,x \,+ \,\frac{u}{k}}{=} \frac{1}{k} \int \limits_0^\infty \left(1 + \frac{u}{k x}\right)^k \mathrm{e}^{-u} \, \mathrm{d} u \, .$$
Using the series definition of the exponential function and the monotone convergence theorem we find
$$ \lim_{k \to \infty} k f_k(x) = \int \limits_0^\infty \lim_{k \to \infty} \left(1 + \frac{u}{k x}\right)^k \mathrm{e}^{-u} \, \mathrm{d} u = \int \limits_0^\infty \mathrm{e}^{\frac{u}{x}} \mathrm{e}^{- u} \, \mathrm{d} u = \int \limits_0^\infty \mathrm{e}^{- \left(1 - \frac{1}{x}\right)u} \, \mathrm{d} u  = \frac{x}{x-1}$$
for $x > 1$ (so $c = \frac{x}{x-1}$ in your notation).

In order to compute the next term(s) of the expansion we write
$$ f_k(x) = \frac{1}{k} \int \limits_0^\infty \mathrm{e}^{k \log \left(1 + \frac{u}{k x}\right)} \mathrm{e}^{-u} \, \mathrm{d} u = \frac{1}{k} \int \limits_0^\infty \mathrm{e}^{- \left(1 - \frac{1}{x}\right)u} \exp \left[-\frac{u}{x} \left(1 - \frac{\log\left(1 + \frac{u}{k x}\right)}{\frac{u}{k x}} \right)\right] \, \mathrm{d} u $$
for $k \in \mathbb{N}$ and $x > 1$. This implies
$$ k^2 \left[\frac{x}{(x-1) k} - f_k(x)\right] = \int \limits_0^\infty \mathrm{e}^{- \left(1 - \frac{1}{x}\right)u} k \left\{1 - \exp \left[-\frac{u}{x} \left(1 - \frac{\log\left(1 + \frac{u}{k x}\right)}{\frac{u}{k x}} \right)\right]\right\} \, \mathrm{d} u$$
for $k \in \mathbb{N}$ and $x > 1$. Using elementary inequalities for the logarithm and the exponential function we can show that the expression in curly brackets is smaller than $\frac{u^2}{2 k x^2}$, so the dominated convergence theorem allows us to compute
\begin{align}
\lim_{k \to \infty} k^2 \left[\frac{x}{(x-1) k} - f_k(x)\right] &= \int \limits_0^\infty \mathrm{e}^{- \left(1 - \frac{1}{x}\right)u} \lim_{k \to \infty} k \left\{1 - \exp \left[-\frac{u}{x} \left(1 - \frac{\log\left(1 + \frac{u}{k x}\right)}{\frac{u}{k x}} \right)\right]\right\} \, \mathrm{d} u \\
&= \int \limits_0^\infty \mathrm{e}^{- \left(1 - \frac{1}{x}\right)u} \frac{u^2}{2 x^2} \, \mathrm{d} u = \frac{x}{(x-1)^3}
\end{align}
for $x > 1$ (so $d = - \frac{x}{(x-1)^3}$ in your notation).

Therefore, for $x > 1$ we have
$$f_k (x) \sim \frac{x}{(x-1)k} \left[1 - \frac{1}{(x-1)^2 k} + \mathcal{o} \left(\frac{1}{k}\right)\right]$$
as $k \to \infty$ (the order of the next term is (at least) $\frac{1}{k^3}$, as can be seen by expanding the logarithm and the exponential function further). We see that the condition $x > 1$ is required. In fact, the asymptotic behaviour at $x = 1$ is
$$ \lim_{x \to 1^+} f_k (x) = \frac{1}{k} \int \limits_0^\infty \left(1 + \frac{u}{k}\right)^k \mathrm{e}^{-u} \, \mathrm{d} u \stackrel{k \to \infty}{\sim} \sqrt{\frac{\pi}{2 k}} \, , $$
as is shown in this answer to another question.

Title: limit on incomplete gamma functions


问:
Projections of standard basis
Suppose that we are given the standard basis $e_1,\ldots, e_n$ is $\mathbb{R}^n$, where $n$ is odd.
Is there an orthogonal projection $P$ to two dimensional subspace and a positive integer $k$ such that $Pe_i=u$ for $1\leq i\leq k$ and $Pe_j=v$ for $k+1\leq j\leq n$ where $u,v$ are orthogonal and have the same length?

答:
When $k\in\{1,\dots,n-1\}$, the image of the proposed linear map would be $\operatorname{span}(u,v)$. Let's find the kernel. For $x=\lambda_1 e_1+\dots+\lambda_n e_n$ we have
$$
Px = (\lambda_1+\dots+\lambda_k) u + (\lambda_{k+1}+\dots+\lambda_n)v,
$$
so that $Px=0$ is equivalent to $\lambda_1+\dots+\lambda_k=0$ and $\lambda_{k+1}+\dots+\lambda_n=0$, since $u$ and $v$ are linearly independent. We conclude that
$$
\ker(P) = \operatorname{span}(e_i-e_{i+1} : i=1,\dots,k-1,k+1,\dots,n-1).
$$
The orthogonal complement of $\ker(P)$ consists of all vectors of the form
$$
\lambda(e_1+\dots+e_k)+\mu(e_{k+1}+\dots+e_n),
$$
so in order for kernel and image to be orthgonal, $u$ and $v$ have to be of this form. Let $u$ be given by choosing $\lambda_u, \mu_u$ and $v$ by choosing $\lambda_v,\mu_v$ in the expression above. In order for $P$ to be a projection, we need to have
\begin{align*}
u &= Pu = k\lambda_u u + (n-k)\mu_u v, \\
v &= Pv = k\lambda_v u + (n-k)\mu_v v.
\end{align*}
Since $u,v$ are linearly independent, this uniquely determines $u$ and $v$ as
\begin{align*}
u &=\frac 1 k (e_1+\dots+e_k), \\
v &=\frac 1 {n-k} (e_{k+1}+\dots+e_k).
\end{align*}
Those vectors are in fact orthogonal, but their square norms are $\frac 1 k$ and $\frac 1 {n-k}$ respectively, so they can only have the same length when $n=2k$.

Hence, there is no such projection when $n$ is odd.

Title: Projections of standard basis


问:
Does ZF prove that there are more cardinals than elements of any set?
In ZFC, it is easy to prove there are more cardinals than elements of any set.  Specifically, given a set $X$, pick a well-ordering of $X$, and then you can inject $X$ into the cardinals by mapping its $\alpha$th element to $\aleph_\alpha$.

My question is whether this can be proved in ZF (where "cardinal" is interpreted in the appropriate way in the absence of choice, not restricted to well-ordered cardinalities).  To be precise, can ZF prove the following?

For any set $X$, there exists a function $f$ on $X$ such that for any distinct $x,y\in X$, there is no bijection between $f(x)$ and $f(y)$.

I imagine the answer is no but don't really know anything about how you could prove such a global statement about the cardinals in a model of ZF in order to get a counterexample.

答:
At least ZFA cannot prove this assertion. The model constructed here should be the standard model of ZFA where choice for families of sets of size 2 fails.

Start with a model of ZFA with a countable set $A$ of atoms. Label the elements of $A$ as $a_n^i$ with $n<\omega$ and $i<2$. Think of $a_n^0, a_n^1$ of pairs and the set $A$ as an infinite set of these pairs. Let $G$ be the group of permutations $\pi$ of $A$ that maybe swap pairs, but do not do anything else, i.e. $\pi(a_n^i)\in\{a_n^i, a_n^{1-i}\}$ for all $n<\omega, i<2$. Let $\mathcal F$ be the filter generated by $\mathrm{fix}_G(E)$, where $E$ ranges over the finite subsets of $A$. This induces a permutation model $\mathcal V$ of hereditarily symmetric sets.
Note that the group $G$ is commutative. This has the following consequence: As usual, any $\pi\in G$ exptends to an automorphism $\pi^+$ of $\mathcal V$. If $X\in\mathcal V$ then $\pi^+\upharpoonright X:X\rightarrow\pi^+(X)$ is a bijection that lies in $\mathcal V$.  The reason being that $\pi^+$ commutes with any other $\mu^+$, $\mu\in G$, which is enough to see that $\pi^+\upharpoonright X$ is (hereditarily) symmetric.

If $f:A\rightarrow\mathcal V$ is a function in $\mathcal V$, then for cofinitely many $n$, $f(a_n^0)$ and $f(a_n^1)$ cannot be significantly different (as that would violate symmetry). In particular they cannot have differnt size:
Let $\mathrm{fix}_G(E)$, $E\subseteq A$ finite, be a support for $f$. Find any $n<\omega$ such that $a_n^0, a_n^1\notin E$ and let $\pi$ be the permutation of $A$ that swaps $a_n^0, a_n^1$ and is the identity everywhere else (observe that $\pi\in\mathrm{fix}_G(E)$). Then
$$f(a_n^1)=f(\pi(a_n^0))=\pi(f)(\pi(a_n^0))=\pi(f(a_n^0))$$
Now $\pi^+\upharpoonright f(a_n^0)$ is a bijection between $f(a_n^0)$ and $\pi(f(a_n^0))=f(a_n^1)$ in $\mathcal V$.

I am not quite sure at the moment how to turn this into a model of ZF with the same property...


问:
What function gives this sequence?
Recently, I have been trying to find what function $f(n)$, for natural $n$, gives the following sequence:

$$1, 5, 105, 4725, 363825, 42567525, \cdots $$

It turns out that $\frac{d^{2n}}{dy^{2n}} (1-y^2)^{-5/2}|_{y=0}$, generates it. However, is there any function $f(n)$ that gives the same?

So far, it has been kind of hard to spot a pattern:

$$1, 5, \underbrace{3 \cdot 5 \cdot 7}_{105}, \underbrace{3^3 \cdot 5^2 \cdot 7}_{4725}, \underbrace{3^3 \cdot 5^2 \cdot 7^2 \cdot 11}_{363825}, \underbrace{3^5 \cdot 5^2 \cdot 7^2 \cdot 11 \cdot 13}_{42567525}$$

Any suggestions would be very much appreciated! Thanks

答:
hint

Observe that

$$(1-y^2)^\frac{-5}{2}=(1+y)^\frac{-5}{2}(1-y)^\frac{-5}{2}$$
$$=f(y).g(y)$$

and use Leibnitz formula which gives the $ n^{\text{th}} $ derivative of a product.

$$(f.g)^{(2n)}(y)=$$
$$\sum_{k=0}^{2n}\binom{n}{k}f^{(k)}(y)g^{(2n-k)}(y)$$

Title: What function gives this sequence?


问:
Prove that a number is the product of two primes under certain conditions.
Suppose $p$ be the smallest prime dividing $n \in \mathbb{Z}^+$. Prove that if $n$ is not a perfect square and that $p<n<p^3$, then $n$ must be the product of two primes.

Clearly, the smallest $p$ can be is $2$ and $n$ must be an integer that is greater than $1$ in order to be divisible by a prime. I know that the Fundamental Theorem of Arithmetic (FTA) guarantees that every positive integer greater than $1$ is the product of two or more primes. So is it enough to argue that by the FTA, $n$ is the product of two primes?

答:
Here is yet one more way to see that your proposition is true:

$n\ne p$ because $p<n$

$n\ne p^2$ because $n$ is not a perfect square

$n\ne p^3$ because $n<p^3$

Hence, $n$ has one or more other prime factors.

Assume $n$ has one additional (larger) prime factor, $q=p+a$. Then $n=pq=p^2+ap$, which is less than $p^3$ whenever $a<p^2-p$

Assume $n$ has two additional (larger) prime factors, $q=p+a$ and $r=p+b$. Then $n=pqr=p^3+(a+b)p^2+abp>p^3$, which necessarily contradicts the assumption $n<p^3$

Plainly, even more prime factors of $n$ only makes the issue in point 5 worse. So the only possibility not ruled out is 4, which is what you set out to prove.


问:
Show $60 \mid (a^4+59)$ if $\gcd(a,30)=1$

If $\gcd(a,30)=1$ then $60 \mid (a^4+59)$.

If $\gcd(a,30)=1$ then we would be trying to show $a^4\equiv 1 \mod{60}$ or $(a^2+1)(a+1)(a-1)\equiv 0 \mod{60}$. We know $a$ must be odd and so $(a+1)$ and $(a-1)$ are even so we at least have a factor of $4$ in $a^4-1$. Was thinking I could maybe try to show that there is also a factor of $3$ and $5$ necessarily giving that $a^4-1\equiv0 \mod{60}$.

Other things I was thinking was that as $Ord_n(a) \mid \phi(n)=\phi(60)=16$ that we just need to show that $Ord_n(a) \in \{1,2,4 \}$.

Any hints? I have the exam soon =/

答:
The question is answered successfully, I would like find the number for which $a^4\equiv1\pmod n$ for all $a$ such that $(a,n)=1$

Using Carmichael Function, If $\lambda(n)=2^m$ where integer $m\ge 1$

$n$ can not have any odd prime factor with exponent $>1$ and for each prime factor $p_i,p_i-1$ must be of the form $2^{2^{r_i}}$ where integer $r_i\ge0$

$$\text{Then if }n=2^s\cdot\prod p_i=2^s\cdot (2^{2^{r_i}}+1)$$

If $s\ge 3, \lambda(n)=$lcm $(2^{2^{r_i}},2^{s-2})$ as $\lambda(2^s)=2^{s-2}$ for $s\ge3$

$\implies\lambda(n)=$lcm $(2^{s-2},2^{2^{r_1}},2^{2^{r_2}},\cdots)=$lcm$(2^{s-2},2^{2^{\text{max}(r_i)}})$

If  $\lambda(n)$ divides $4=2^2,$  $$2^{\text{max}(r_i)}\le 2 \text{ and }s-2\le 2 \implies 0\le max(r_i)\le1 \text{ and } 3\le s\le 4 $$

For the  maximum value of $n,r_1=0,r_2=1$ and $s=4$

So, the maximum value of $n$ will be $3^1\cdot5^1\cdot2^4=240$

$$\implies x^4\equiv1\pmod {240}\implies x^4\equiv1\pmod {60}\equiv1-60=-59$$


问:
Short exact sequence with center right splits
Suppose that $G= \mathbb{Z}^n \times H$ for some group $H$. Then, there is a short exact sequence $$ 1\rightarrow \mathbb{Z}^n \rightarrow G \rightarrow H \rightarrow 1$$ which right splits (clearly).

Let $K< G$ be a subgroup. Then, the previous short exact sequence descends to another SES $$ 1 \rightarrow \mathbb{Z}^n \cap K \rightarrow K \rightarrow p(K) \rightarrow 1.$$

Does this short exact sequence right split? If $x\in p(K)$, then there is an element $z x \in K$ for some $z\in \mathbb{Z}^n$. So I could define $\iota \colon p(K) \mapsto K$ such that $\iota(x)= zx$. However, is $\iota$ a homomorphism?

答:
It does not necessarily split, because $K$ can be a nontrivial subdirect product.

As an example, take $n=1$, $H = {\mathbb Z}$, where the normal subgroup and $H$ are generated by $x$ and $y$, respectively, and let $K$ be the subgroup $\langle x^2,y^2,xy \rangle$.


问:
Computing cross product using norm and angle
Sorry for the weird title, if someone finds a better title for my problem be my guest to edit it ;)

For $\mathbf{v,w} $ in R³ with $\mathbf{||v||=1 ;||w||=4; \theta
 =\frac{2\pi}{3}}$

Solve the following:

$\lVert (3v+w)\times(v-2w)\rVert$

$\langle w\times 5v, v-3w\rangle$

I've already calculated

$$\langle v,w\rangle = \frac{-1}{2}||v|| \cdot||w|| =-2$$

Also, I've transformed 1. into

$$\lVert-7(v\times w)\rVert$$

But I'm a bit lost here.

Next I masked $w\times 5v $ as $x$ so as 2. we now have

$$\langle x,v-3w\rangle$$

This all seems to bring me not a bit closer to the solution, can someone suggest a valid approach?

答: Just as $v \cdot w = |w| |v| \cos \theta$, there is a corresponding formula $|v \times w| = |v| |w| \sin \theta$.  Since the angle and the lengths are known, you should be able to compute this magnitude.</p>


问:
cauchy-preserving metric spaces

Let $d, p$ be two metrics on a nonempty set $X$. Say $d\sim p$ if for any $(x_n)\subseteq X$ and $x\in X, \lim_{n} x_n = x$ in $(X,d)$ iff $\lim_n x_n = x$ in $(X,p)$. Say $d\approx p$ if $\exists m, M > 0$ so that $m d(x,y) \leq p(x,y) \leq Md(x,y)$ for $x,y \in X.$

Give an example of an $X$ and two metrics $d,p$ on $X$ with $d\not \approx p$ but such that a sequence $(x_n)_{n=1}^\infty \subseteq X$ is Cauchy in $(X,d)$ iff it is Cauchy in $(X,p).$

Give an example of an $X$ and two metrics $d,p $ on $X$ with $d\sim p$ and so that there is a sequence $(x_n)_{n=1}^\infty \subseteq X$ that is Cauchy in one of $(X,d)$ or $(X,p)$ but not the other.

I know that on $\ell_1, \lVert \cdot \rVert_r \not \approx \lVert \cdot \rVert_p,$ where $1\leq p < r \leq \infty$ and $\lVert x\rVert_r := (\sum_{i\in\mathbb{N}} |x_i|^r)^{1/r}$ for $x= (x_i)_{i\in\mathbb{N}}\in \ell_1,$ but it seems that there are Cauchy sequences in $(\ell_1, \lVert \cdot \rVert_r)$ that aren't Cauchy in $(\ell_1, \lVert \cdot \rVert_p).$ For example, take the sequence $(x_n)_{n\in\mathbb{N}}, x_n = (1, \frac{1}{2^{1/p}}, \frac{1}{3^{1/p}},\cdots, \frac{1}{n^{1/p}}, 0,0, \cdots)$ for each $n.$ It's not Cauchy in $(X, p)$ because if one fixes $n\in\mathbb{N}$, then as $m\to \infty, \lVert x_n - x_m\rVert_p =(\sum_{i=n+1}^\infty \frac{1}i)^{1/p},$ which diverges. It is Cauchy in $(X,r)$ because for any $\epsilon > 0,$ there exists $N\in\mathbb{N}$ so that for $n\geq N, (\sum_{i=n+1}^\infty \frac{1}{i^{r/p}})^{1/r} < \epsilon$ (as $ \sum_{i=1}^\infty \frac{1}{i^{r/p}})^{1/r} < \infty$ by the $p$-series test as $r/p > 1$). However, this may not be useful for the second part as I'm not sure if $\lVert \cdot \rVert_p\sim \lVert \cdot \rVert_r$ in $\ell_1.$

答:
Let $X=\Bbb Z^+$. Let $d$ be the discrete metric on $X$:

$$d(m,n)=\begin{cases}
1,&\text{if }m\ne n\\
0,&\text{if }m=n\,.
\end{cases}$$

Let $p$ be the usual Euclidean metric: $p(m,n)=|m-n|$. These metrics both generate the discrete topology, so the Cauchy sequences in both are precisely the sequence that are eventually constant, but clearly $d\not\approx p$.

For the second problem let $X=\Bbb R$, let $d(x,y)=|\tan^{-1}x-\tan^{-1}y|$ for all $x,y\in\Bbb R$, and let $p$ be the usual Euclidean metric. Let $\langle x_n:n\in\Bbb N\rangle$ be a sequence in $\Bbb R$. Then

$$\lim_{n\to\infty}|x_n-x|=0$$

iff

$$\lim_{n\to\infty}|\tan^{-1}x_n-\tan^{-1}x|=0\,,$$

so $d\sim p$, but the sequence $\langle n:n\in\Bbb N\rangle$ is $d$-Cauchy and not $p$-Cauchy.

Title: cauchy-preserving metric spaces


问:
Given ring with identity $R$ and ideals $I, J$, with $I+J=R$ and $IJ=0$, prove $I\cap J=0$
Give a ring with identity $R$ and its ideals $I, J$, with $I+J=R$ and $IJ=0$, I was required to prove $R\simeq R/I \times R/J$.

My understanding is $R/I\cap J \simeq R/I \times R/J$ so $I\cap J=0$ must hold if I'm able to do so.

Work done:

Since $1$ belongs to $R$, $1=u+(1-u)$ with $u$ in $I$ and $1-u$ in $J$, with $u(1-u)=0$.

For arbitary $v\in I\cap J$ we have $uv = 0$ and $v(1-u)=0$ i.e. $v=vu$.

If the ring is commutative or just $JI=0$ (we have $vu=0$ and $v=uv$) will force $v=0$. Neither commutativity nor $JI=0$ is given. I am aware of this conterexample but this example uses left ideals $(x)=Rx$, while the ideals defined in my book is always $(x)=Rx+xR+RxR$.

Edit: It seems $R=\mathbb{Z}\langle x,y|2x+2y-1=xy=0\rangle,I=(x),J=(y)$ have $I+J=R$, $IJ=0$ and $I\cap J=JI=(yx)$ (note x and y does not commute). Is this a counterexample or something is wrong?

答: Here’s maybe an easier counterexample. Take $R$ to be the ring of lower triangular 2x2 matrices over a field, with basis $e,f,x$ such that $e,f$ are orthogonal idempotents and $x=fxe$. Set $I=ReR$ having basis $e,x$, and $J=RfR$ having basis $f,x$. Then $I+J=R$ and $IJ=0$ but $JI$ has basis $x$.


问:
Is the indefinite integral of a piecewise continuous function a continuous function?
I had looked around on the web and can't find much information related to the integration of piecewise continuous functions.

Let's say we have a simple function

$$f(x)=
\begin{cases}
      0 & x\leq 0 \\
      x & 0\leq x\leq 1 \\
      0 & x>1
\end{cases}$$

and we are looking to find the integral $\int f(x) dx$

WolframAlpha gives me the following result:

$$\int f(x) dx=
\begin{cases}
      c & x\leq 0 \\
      \frac{x^2}{2}+c & 0\leq x\leq 1 \\
      \frac{1}{2}+c & x>1
\end{cases}$$

The cases $x\leq 0$ and $0\leq x\leq 1$ are clear. But why do we need the $\frac{1}{2}$ for the $x>1$ case? I see that this makes the integral continuous but is it necessary, i.e. is it wrong to have simply $c$ for the $x>1$ case for the integral?

答:
Actually, the indefinite integral of a continuous function is a family of continuous functions. If you want to find exactly one of these functions, you have to fix an initial condition $F(x_0) = C_0$.

If you want to obtain the same function obtained by WolframAlpha, solve for the following equalities: lateral limits


问:
Showing weak convergence of a sequence in $L^p(R)$
I have a sequence of functions $f_k$ in $L_p(R)$, with $1<p<\infty$ and I'd like to show that it weakly converges to $0$.
This is the sequence, where $k\in N$
$$f_k = 1_{[k,k+1]}$$

What I've tried:
If $f_k$ converges to $0$, then we should have
$$\lim_{k\to \infty}\int_R(f_k-0)\phi dx =0$$
where $\phi \in L_{p'}(R)$ (the dual, or here, $L_q(R)$.
Putting in the function, one gets
$$\lim_{k\to \infty}\int_k^{k+1}1.\phi dx$$
Now I need to show this goes to $0$ for all functions $\phi \in L_q(R)$$, but that isn't necessarily true right?

答:
With $q = p'$ you can use Hölder's inequality to say that
$$\int_{\mathbb{R}}1_{[k,k+1]}|\phi(x)|dx\leq \left(\int_{\mathbb{R}}1_{[k,k+1]}|\phi(x)|^qdx\right)^{1/q}$$
The question is therefore if
$$\lim_{k\rightarrow \infty}\int_{\mathbb{R}}1_{[k,k+1]}|\phi(x)|^qdx = 0$$
which you can show using the Dominated Convergence Theorem.


答:
as mentioned in the comments
$(1-y^2)^{-5/2}= \sum \limits_{n=0}^\infty \binom{-5/2}{n}(-1)^n y^{2n}$

differentiating $2n$ times the general form would look like $\binom{-5/2}{n}(-1)^n (2n)!$

Edit

I tried the first $5$ values with Maple, they match your sequence.

you can also simplify the generalized binomial coefficient to get:

$f(n) = \frac{1}{3} \cdot \frac{1}{2^{2n+1}} \cdot \frac{(2n)! (2n+3)!}{n!(n+1)!}$ for $n=0,1,...$

Edit #2

Also note how the prime factorizations of the numbers in your sequence contain no powers of $2$, that is because of the denominator $2^{2n+1}\cdot n! \cdot (n+1)!$

Title: What function gives this sequence?

Body:
Recently, I have been trying to find what function $f(n)$, for natural $n$, gives the following sequence:

$$1, 5, 105, 4725, 363825, 42567525, \cdots $$

It turns out that $\frac{d^{2n}}{dy^{2n}} (1-y^2)^{-5/2}|_{y=0}$, generates it. However, is there any function $f(n)$ that gives the same?

So far, it has been kind of hard to spot a pattern:

$$1, 5, \underbrace{3 \cdot 5 \cdot 7}_{105}, \underbrace{3^3 \cdot 5^2 \cdot 7}_{4725}, \underbrace{3^3 \cdot 5^2 \cdot 7^2 \cdot 11}_{363825}, \underbrace{3^5 \cdot 5^2 \cdot 7^2 \cdot 11 \cdot 13}_{42567525}$$

Any suggestions would be very much appreciated! Thanks