问:
Does anyone know a no-nonsense intro to "logic for mathematics" that I can give to a Year 11 student?
I'm looking for material on propositional and first-order logic to give to a Year 11 student that explains how they're used "in practice." For example, I want to be able to write the null-factor law as $$\forall (a \in \mathbb{R},b \in \mathbb{R}) \,ab = 0 \rightarrow a = 0 \vee b = 0,$$
and have the student nod their head and say "yep, I get it" and be able to use this to deduce $$(x-1)(x-2) = 0 \rightarrow x-1 = 0 \vee x-2 = 0.$$ More generally, I want to be able to explain the laws necessary to solve Year 11 and 12 problems using precise logical notation, and have the student follow what I'm saying.

The material should:

Be phrased in straightforward language that any clever student in Year 11+ can follow.

Get straight to the point and not dawdle too much on the philosophy or analysis of language.

Emphasize the kind of reasoning needed in high-school problem solving, as opposed to "proofs." (Although in some sense, everything is just proofs.)

Not introduce too many words or phrases that aren't strictly necessary: e.g. the use of words like "proposition, statement, sentence, formula, expression, term, validity, and tautology" should be kept to a minimum.

Avoid truth trees. (Don't get me wrong, they're super cool. But inappropriate in this context.)

Any of the following would be fine: a short book, an online video series, or even a free online course (if it isn't too long).

答: A free resource is Hammack's "Book of Proof". It covers the proof techniques needed (not heavy on symbolic logic, though).</p>


问:
Conjecture: For almost all $n$, the sum of $n \bmod k$ for $k<n$ is unique.
It looks to me like $$f(n):= \sum_{k=1}^{n}{(n\bmod k)}$$ is almost unique to $n$, with the only systematic exceptions being $(2^m-1, 2^m)$ pairs, which share a value for any $m \in \mathbb N$. In other words, the set $\{f(n): n\in\mathbb N\land n\neq 2^m
\ \forall m\in\mathbb N\}$ shouldn't have many duplicate values.

As discussed in the answers, a handful of counterexamples have been found, but they seem more sparse than one would expect, and nobody has yet suggested why that might be, or what's special about the counterexamples found, apart from a moderate preponderance of small factors in the mean of the pair.

Although there's not much data to go on, it also looks like counterexamples may come in the form $(n,n+2i)$, or possibly even $(n,n+2^i)$, and I would be interested if anyone can find a case outside of those. The latter form would also include the $(2^m-1,2^m)$ exceptions, which seems elegant.

Also, if someone could set me straight on this: My understanding is that the term 'almost all' in this context means approaching $0$ density in the limit; so if one could show that the exceptions aren't necessarily finite, but they are less common than powers of two, would that be sufficient?

ContentLicense: CC BY-SA 4.0

Body:
I feel this is worth an answer, using Barry Cipra's technique, to report the first such equality over a gap of more than $2$:

$$f(113393278)=f(113393282)$$

with:

$\sigma(113393278)=171694080$ with $113393278 = 2\cdot 179\cdot 383\cdot 827$ (not needed)

$\sigma(113393279)=117765120$ with $113393279 = 43\cdot 67\cdot 39359$

$\sigma(113393280)=493516800$ with $113393280 = 2^7\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 59$

$\sigma(113393281)=123621120$ with $113393281  = 17\cdot 47\cdot 139\cdot 1021$
$\sigma(113393282)=172243200$ with $113393282 = 2\cdot 79\cdot 717679$


问:
Functions in Group Theory
I am reading a math book and in it, it says, "Let $V$ be the set of all functions $f: \mathbb{Z^n_2} \rightarrow \mathbb{R}.$ I know that $\mathbb{Z^n_2}$ is just the cyclic group of order $2$ taken to the $n$th power, but I don't get what the actual function means.

How would such a function, $f$, work? Could someone please give an example?

答:
The fact that $\mathbb{Z}_2^n$ is a group is completely irrelevant here. It is being used simply as a set. So let us discuss this construction without any reference to $\mathbb{Z}_2^n$.

Let $X$ be your favorite set. Define
$$\mathbb{R}^X=\{f\colon X\to \mathbb{R}\mid f\text{ is a function}\}.$$
No other conditions on $f$: just a function from the set $X$ to the real numbers.

We define addition of elements of $\mathbb{R}^X$ as follows: given $f,g\in\mathbb{R}^X$, I want to define a function from $X$ to $\mathbb{R}$ called “$f+g$”. To do so, I need to tell you how to evaluate this function “$f+g$” at each $x\in X$. The rule is:
$$(f+g)(x) = f(x) + g(x),$$
where the addition on the right hand side is the addition of real numbers. Note that since $f(x)$ and $g(x)$ are real numbers, this makes sense.

Now, the set $\mathbb{R}^X$ together with this operation $+$ on functions is a commutative group: verify that $+$ is associative, for example, by showing that the functions $(f+g)+h$ and $f+(g+h)$ take the same value at each $x\in X$. The identity of the operation $+$ on functions is the function $\mathbf{z}\colon X\to \mathbb{R}$ defined by $\mathbf{z}(x) = 0$ for all $x\in X$. And the additive inverse of a function $f\colon X\to \mathbb{R}$ is the function $(-f)\colon X\to \mathbb{R}$ defined by
$$(-f)(x) = -(f(x)),$$
where $-$ on the right hand side is the usual negative symbol of the real numbers.

We define a scalar multiplication on $\mathbb{R}^X$ as follows: given $\alpha\in\mathbb{R}$ and $f\in\mathbb{R}^X$, we define the function $(\alpha f)$ to be
$$(\alpha f)(x) = \alpha\cdot f(x)\text{ for all }x\in X,$$
where $\alpha\cdot f(x)$ means the product in $\mathbb{R}$ of $\alpha$ and $f(x)$, both real numbers.

These two operations, $+$ and the scalar multiplication, turn $\mathbb{R}^X$ into a vector space over $\mathbb{R}$.

Now, we define a distinguished set of elements of $\mathbb{R}^X$: for each $y\in X$, define $f_y\colon X\to\mathbb{R}$ by
$$f_y(x) = \delta_{yx} = \left\{\begin{array}{ll}
0 & \text{if }y\neq x,\\
1 & \text{if }y=x.
\end{array}\right.$$
So $f_y$ takes the value $1$ at $y$, and the value zero everywhere else.

If $X$ is finite, verify that the set $\{f_y\}_{y\in X}$ is a basis for the vector space $\mathbb{R}^X$.

This is the construction you are being presented with, except that the set $X$ is taken to be the underlying set of the group $\mathbb{Z}_2^n$. But the fact that it is a group doesn’t matter at all; what is being used is the fact that it is a (finite) set.

Title: Functions in Group Theory


问:
Spivak Calculus Chapter 5 (Limits) Problem 3 Part (i)
The question is: Find a $\delta$ such that $|f(x)-l|<\epsilon$ for all $x$ satisfying 0 < $|x-a|$<$\delta$.

(i) $f(x)=x^4; l=a^4$

This was my attempt, but I am lacking confidence in every single attempt of Spivak's exercises because I feel I not as rigorous as Spivak wants me to be.:

$|x^4-a^4|<\epsilon\\|x^4|-|a^4|<\epsilon\\|x|-|a|<\frac{\epsilon}{(|x|+|a|)\cdot (|x|^2+|a|^2)}$

So, $\delta=\frac{\epsilon}{(|x|+|a|)\cdot (|x|^2+|a|^2)}$

$\because |x-a|<\frac{\epsilon}{(|x|+|a|)\cdot (|x|^2+|a|^2)} \implies |x|-|a|<\frac{\epsilon}{(|x|+|a|)\cdot (|x|^2+|a|^2)}$

It feels correct to me but at the same time I have almost no confidence in my solution. (I'm trying to self-study the book after taking Calc I-III)

答:
An even easier approach is suggested by the proof of the lemma of Theorem 2, part 2, from earlier in the chapter:

If $|x-x_0| < \min(1, \frac{\epsilon}{2(|y_0|+1)})$ and $|y-y_0| < \frac{\epsilon}{2(|x_0|+1)}$, then $|xy - x_0y_0| < {\epsilon}$

In this case, you have:

$f(x)=x^4; l=a^4$

Substituting $x^2$ in for both $x$ and $y$ and $a^2$ in for both $x_0$ and $y_0$, we see that $|x^4 - a^4| = |(x^2)(x^2) - (a^2)(a^2)| < {\epsilon}$ as long as ${\delta} = \min(1, \frac{\epsilon}{2(|a^2|+1)})$

ContentLicense: CC BY-SA 3.0


问:
Proving the existence of minimal distance
Let $(X,d)$ be a metric space and $x \in X$. Let $Y$ be a compact subset of $X$. I want to show that there exists a point $\hat{y} \in Y$ such that:

$$d(x,\hat{y}) = \inf_{y\in Y}\{d(x,y)\}$$

not using the sequence characterization of the compactness of $Y$, but rather the topological defintion, that is every open subcover of $Y$ has a finite subcover.

My attempt was to derive a contradiction. Assume such a point $\hat{y}$ does not exits. Then consider the open cover of $Y$:

$$ U_n = \{z \in X| d(z,y) > \inf_{y \in Y} \{d(z,y)\}+\dfrac{1}{n}\}$$

Can we show that it has no finite subcover? Is there any other way to prove the statement using the topological defintion of compactness?

答: You've basically got it. As you noticed this is an open cover. Since $Y$ is compact, there is a finite subcover. Let it be $U_{n_1}, U_{n_2}, \cdots, U_{n_k}$. Then take the maximum of the $n_i$'s, call it $N$. So this means that every $z \in Y$ has the property that $d(z,x) > inf_{y \in Y}\{d(y,x)\} + 1/N$. But then this contradicts the definition of $inf_{y \in Y}\{d(x, y)\}$, i.e, it has to be at least $1/N$ bigger.


问:
Inequality $a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0$
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that:
$$a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0.$$
Now, I am supposed to solve this inequality by applying only the Rearrangement Inequality.

My Attempt:
W.L.O.G. let $a \geq b \geq c.$
Then, L.H.S. = $a \cdot ab(a-b) + b \cdot bc(b-c) + c \cdot ac(c-a)$.

There are $2$ cases to consider:

Case $1$:
$ab(a-b) \geq bc(b-c) \geq ac(c-a)$. Then, the L.H.S. is similarly sorted, so:

\begin{align}
\text{L.H.S.} \ & \geq c \cdot ab(a-b) + a \cdot bc(b-c) + b \cdot ac(c-a) \\
& = abc(a-b) + abc(b-c) +  abc(c-a) \\
& = a^2bc - ab^2c + ab^2c - abc^2 + abc^2 - a^2bc \\
& =0.
\end{align}

Case $2$: $bc(b-c) \geq ab(a-b) \geq ac(c-a)$. But Rearrangement doesn't seem to work for this case. Any hints on how to proceed?

答:
The standard substitution is $x=b+c-a , y=c+a-b , z=a+b-c $ ($x,y,z \geq 0$ by the triangle inequality.) These invert to give
\begin{eqnarray*}
2a&=&y+z \\ 2b&=&z+x \\ 2c&=&x+y.
\end{eqnarray*}
Subbing these gives
\begin{eqnarray*}
x^3z+y^3x+z^3x-xyz(x+y+z) \geq 0
\end{eqnarray*}
which you could $\color{red}{\text{not}}$ say is true by Muirhead's inequality ... but can be shown to be true because
\begin{eqnarray*}
xy(y-z)^2+yz(z-x)^2+zx(x-y)^2 \geq 0.
\end{eqnarray*}


问:
Sum of all $x$ satisfying $|x-6| = m$ where $m\in\mathbb{R}^+$
I have encountered this question in a book and it had me thinking about a different attempt to solve it, which I think isn't right but can't really comprehend.

Here is the basic way I solve it:

$$x-6 = m \implies x = m+6$$
$$x-6 = -m \implies x = 6-m$$
$$m+6+6-m = 12$$

Answer key also says $12$.

The other way is that we know that $|x-6|>0$ for all $x \in \mathbb{R}$ except for $x = 6$. Since $m>0$, we can say that the solution set of $x$ is $\mathbb{R}-\{6\}$ and therefore sum of all these values equals $-6$. (As far as I know, summing all $\mathbb{R}$ like this isn't valid. However and unfortunately, it's a commonly used way in solutions in this curriculum, so please ignore that part.)

Is the second way right in any sense? If yes or no please explain it because I couldn't come up with a good explanation other than saying $x$ depends on $m$, therefore it is not right.

Is $m$ a variable or a fixed number? I think it must be a fixed number or else we would end up with infinitely many solutions since it would state a function of $x$. Does $m$ being a variable or a fixed number have any impact on the first question?

Thank you in advance.

答:
Is $m$ a variable or a fixed number?

It's a fixed number.  And unknown fixed number.

I think it must be a fixed number or else we would end up with infinitely many solutions since it would state a function of x.

It's more that the question just wouldn't make sense.

You'd be expecten so solve for pairs of values of $m$ and for $x$.  And so long as $x = \begin{cases}m+6\\6-m\end{cases}$ then $\{(x,m)| x=m+6$ or $x = 6-m; m > 0\}$ would be the set of all pairs of solution.

But that wasn't what was asked.  You are asked to suppose $m\in \mathbb R^+$ and so solve for $x$ when $|x-6| = m$.  That is... $m$ is fixed.

Does m being a variable or a fixed number have any impact on the first question?

Has every impact.


问:
Let $R$ be a finite dimensional $F$-algebra, where $F$ is a field. Show that the following conditions are equivalent.
If $R$ is a finite ring, then we know that the following are equivalent:

i. $R$ is a division ring.

ii. $R$ is nontrivial and if $r$,$s \in R$, with $rs=0$, then either $r=0$ or $s=0$.

$\textbf{NOTE:}$ A commutative ring that satisfies (ii) is called an integral domain or sometimes just a domain. The problem tells us that finite domains are fields.

Let $R$ be a finite dimensional $F$-algebra, where $F$ is a field. Show that conditions (i) and (ii) from above are equivalent in this case, too.

I am not familiar with the definition of $F$-algebra. So that line is confusing me hence I cannot start this problem.

答:
An (associative unital) $F$-algebra is simultaneously a vector space over $F$ and a ring.

One formal definition is that $R$ is a ring equipped with a ring homomorphism $F\to R$.

(This induces multiplication by elements of $F$, and conversely, consider the map $c\mapsto c\cdot 1$ where $c\in F,\ 1\in R$.)

Classic examples are the complex numbers, split complex numbers and quaternions over $\Bbb R$.

The proof is similar to the finite case: for a nonzero $r\in R$, the map $\varphi_r:x\mapsto rx$ is injective by hypothesis.

Note also that $\varphi_r$ is an $F$-linear endomorphism of $R$, and since $R$ is finite dimensional, it must be an $F$-isomorphism, in particular it's surjective, so there's an $x\in R$ such that $rx=1$.


问:
Interesting Theorems on Finitely Generated Abelian Groups?
First time teaching algebraic topology, probably gonna be related to most of my questions on here for a while.

I was wondering if anyone knows of particularly interesting theorems or examples from the theory of finitely generated abelian groups, beyond the standard auxiliary stuff and the classification/primary decomposition.  By auxiliary I mean stuff like the basic results on subgroups, quotients, torsion/rank uniqueness and change of base.  Doesn't have to be something that can quickly be PROVEN from first principles, but as long as it's something digestible and believable (or even better, unbelievable).  Perhaps certain actions on manifolds or trees or the like?  Something for topology-minded folks rather than algebra-minded would be ideal, but whatever you think is cool I'd love to hear about!

It could also be surprising internal/structural features, or results about their automorphism groups.

Thanks, hope someone finds this topic interesting!

答:
For every finite generating set $S$ of $G=Z^n$, the generating function of its growth
$$
\sum_{n=0}^\infty \beta_S(n)t^n
$$
is rational. Here $\beta_S(n)$ is the number of group elements $g\in G$ of norm $|g|\le n$ and $|g|$ is the distance from $g$ to $0$ in the Cayley graph of $G$ associated with $S$.

This also holds for virtually abelian groups (i.e. groups containing $Z^n$ as a finite index subgroup) and may other groups. However, surprisingly, rationality becomes tricky for nilpotent groups: It depends on the generating set!

Benson, M., Growth series of finite extensions of $Z^n$ are rational, Invent. Math. 73, 251-269 (1983). ZBL0498.20022.

Stoll, M., Rational and transcendental growth series for the higher Heisenberg groups, Invent. Math. 126, No. 1, 85-109 (1996). ZBL0869.20018.


问:
Multiplying power formal series when reindexing is required?
Normally, when I multiply the two following series:

$$\sum_{n\ge0}a_nx^n$$

$$\sum_{n\ge0}b_nx^n$$

I have:

$$\sum_{n\ge0}(\sum_{i=0}^na_ib_{n-i})x^n$$

Which is nice and intuitive. However, what if the exponents of x, or the indices don't match up? For example, what if I wanted to multiply these two:

$$\sum_{n\ge0}a_nx^{100n}$$

$$\sum_{n\ge0}b_nx^n$$

If I were to try to do this I'd reindex the second series to:

$$\sum_{\frac{n}{100}\ge0}b_nx^{100n}$$

But then the indices wouldn't line up. Am I missing something?

答:
We have
\begin{eqnarray*}
\left( \sum_{n=1}^{ \infty} a_n x^{100n} \right)  \left(\sum_{  m=1}^{\infty }b_m x^m\right) = \sum \sum a_n b_m x^{ \color{red}{100n+m}}.
\end{eqnarray*}
So to collect "like" $x$ terms, we need to list solutions of $100n+m=N$ ... this will give
\begin{eqnarray*}
 \sum_{N=1}^{\infty}  \left(\sum_{m \mid m+100n=N} a_n b_m \right) x^{ \color{red}{N}}.
\end{eqnarray*}


问:
Does $f(x) = f(2x)$ for all real $x$, imply that $f(x)$ is a constant function?
If a Continuous function $f(x)$ satisfies $f(x) = f(2x)$, for all real $x$, then does $f(x)$ necessarily have to be constant function? If so, how do you prove it? If not any counter examples?

答: Counterexample: $f(x)=\sin(\log_a(x))$, select basis $a$ so that $\log_a(2)=2\pi$, that is, $a ^{2\pi}=2$.</p>


问:
What's the distance between two circumcenters
This problem was came to a Facebook post of mathigon ......the problem seems trickier than I expected...can you help me to find the distance??

答:
Let $\angle CAB = \angle ABC = \angle BAD = \angle DBA = \theta$

Area of the $\bigtriangleup CAD = \frac{1}{2} 1^2 \sin 2 \theta = \frac{1}{2} \sin 2 \theta$

Area of the sector $CAD$ (consider the right side of the circle on the left) $ = \frac{1}{2} 1^2 2\theta = \frac{1}{2} (2 \theta)$

So the common area between the circles $ = 2 (\frac{1}{2} (2 \theta) - \frac{1}{2} \sin 2 \theta)$

However, it is given that the common area between the circles $= \frac{\pi}{2}$

Hence we need to solve the following equation for $\theta$

$$2 \left(\frac{1}{2} \left(2 \theta \right) - \frac{1}{2} \sin 2 \theta \right) = \frac{\pi}{2}$$

Solving, $\theta = 1.15494$

Finally length of $AB$ is $2 \cos \theta = 0.80794684217$


问:
Show the graph commutes, diagonal function.
$\require{AMScd} \begin{CD}
                   X @>{f}>> Y\\
@V{\delta_X}VV @V{\delta_Y}VV \\
X\times X@>{f \times f}>> Y\times Y
\end{CD}$

Where, $\delta _{X,Y}$ is the diagonal of $X$ and $Y$ respectively.

I am trying to show the diagram commutes, and this is what I have so far:

$\delta _Y \circ f$, and $(f \times f)\circ \delta_X$ I am not sure I can see how the two functions are equal. Any Ideas?

答: You can use the UMP of $Y\times Y$, so if we have an arrow $X\to Y$, there is a unique arrow $X\to Y\times Y$, s.t. the induced triangles commute (so if $g: X\to Y$, and $h:X\to Y\times Y$, then this commutativity is where $g = h\pi_j$, where $\pi_j$ are the projections of $Y\times Y$ to the factors), but we have two such maps! Namely we have $f\times f\circ \delta_X$ and $<f,f>:X\to Y\times Y$ (the latter map is the one via the UMP of $Y\times Y$ from the arrow $X\to Y$), so they must be equal! We can see that $f\times f$ is compatible with the projections of $Y\times Y$ to the factors just by using the definition of $\delta_Y$ (induced by the map $1_Y:Y\to Y$) so you have the commutative triangle $\pi_j\delta_Y = 1_Y$, which makes $f\times f = \delta_Yf$. Your square commuting is a direct result of $<f,f> = f\times f\delta_y$ (I would draw all of these triangles out, namely the ones where $f\times f = \delta_y f$ which follows from connecting this with the triangle we know commutes right off the bat: where $\pi_j\delta_Y = 1_Y$ definitionally).


问:
Every ideal $J \supsetneq \sqrt{0}$ contains a non-zero-divisor
I'm looking for an example of a commutative ring with unity, that has elements in each of the following three classes:

Non-zero nilpotents

Non-nilpotent zero-divisors

Non-zero-divisors

and furthermore satisfies the following condition:

Any ideal $J$ which strictly contains the nilradical $\sqrt{0}$ also contains a non-zero-divisor.

Any comments or partial answers are appreciated.

答:
Such a ring doesn't exist because the second and last conditions are incompatible.

Let $x$ be a non-nilpotent element. By the last condition, the ideal $\big( x,\sqrt{0} \big) \supsetneq \sqrt{0}$ contains a non-zero-divisor, i.e. there exist elements $a, b$ with $ax + b$ a non-zero-divisor, and $b^n = 0$ for some positive integer $n$.

But then $(ax + b)^n$ is a non-zero-divisor, since if $(ax+b)^n k = 0$, then $(ax+b)(ax+b)^{n-1} k = 0$, contradicting that $ax+b$ is not a zero-divisor. Furthermore, expanding $(ax + b)^n$ shows it is of the form $cx$ for some element $c$, so $x$ is also a non-zero-divisor.


问:
Probability of network computer
If a computer blows up after turning it on with probability $0.05$. We have an order to create a network of $45$ computers, and we gathered $50$ computers for this purpose. What's the probability, that we can create a network with these computers?

We can calculate it by the simple way:

$\binom{50}{45}\cdot \left(0.95\right)^{45}\cdot \left(0.05\right)^2+...+0.95^{50}$

But my question is how we can calculate the probability by these 2 ways:

Central limit theorem and Normal distribution.

Approximation of the binomial distribution by Poisson distribution.

答:
Binomial Distribution

By the way, you computed the probability that you get 45 non-blown-up computers, but you can also create a network with 46 or more computers, so the actual exact answer is:

$\sum_{i=45}^{50}{50\choose i}(.95)^i(.05)^{50-i}$. This can be computed in R using the command

```<code>pbinom(44,50,.95,lower.tail=FALSE)
</code>```

to get $0.9622238$.

Normal Approximation

A binomial distribution can be approximated by a normal distribution like the following: $Binomial(n, p)\approx Normal(np, np(1-p))$. So $Bin(50,.95)\approx N(47.5,2.375)$. You need to calculate the area to the right of a standard normal distribution of the z-value $\frac{45-47.5}{2.375}\approx -1.05$.

You can do either of the following commands, one being the standard normal distribution and the other specifying the mean and standard deviation:

```<code>pnorm(-1.62, lower.tail=FALSE)
pnorm(45,47.5,sqrt(2.375),lower.tail=FALSE)
</code>```

to get $0.9473839$ and $0.9476213$.

Poisson Approximation

The poisson distribution can be a good approximation to the binomial distribution when n>20 and np<5, which both hold in our case. Then, $Bin(n,p)\approx Poisson(np)$. In this case, $Bin(50,.05)\approx Poisson(2.5)$. Now we are counting the number of defective computers, so we want this to run from 0 to 5. The answer is $e^{-2.5}\sum_{k=0}^5\frac{2.5^k}{k!}$

This can be calculated in R using the command

```<code>ppois(5,2.5,lower.tail=TRUE)
</code>```

to get $0.957979$.

Title: Probability of network computer

Body:
Binomial Distribution

By the way, you computed the probability that you get 45 non-blown-up computers, but you can also create a network with 46 or more computers, so the actual exact answer is:

$\sum_{i=45}^{50}{50\choose i}(.95)^i(.05)^{50-i}$. This can be computed in R using the command

pbinom(44,50,.95,lower.tail=FALSE)

to get $0.9622238$.

Normal Approximation

A binomial distribution can be approximated by a normal distribution like the following: $Binomial(n, p)\approx Normal(np, np(1-p))$. So $Bin(50,.95)\approx N(47.5,2.375)$. You need to calculate the area to the right of a standard normal distribution of the z-value $\frac{45-47.5}{2.375}\approx -1.05$.

You can do either of the following commands, one being the standard normal distribution and the other specifying the mean and standard deviation:

pnorm(-1.62, lower.tail=FALSE)
pnorm(45,47.5,sqrt(2.375),lower.tail=FALSE)

to get $0.9473839$ and $0.9476213$.

Poisson Approximation

The poisson distribution can be a good approximation to the binomial distribution when n>20 and np<5, which both hold in our case. Then, $Bin(n,p)\approx Poisson(np)$. In this case, $Bin(50,.05)\approx Poisson(2.5)$. Now we are counting the number of defective computers, so we want this to run from 0 to 5. The answer is $e^{-2.5}\sum_{k=0}^5\frac{2.5^k}{k!}$

This can be calculated in R using the command

ppois(5,2.5,lower.tail=TRUE)

to get $0.957979$.


问:
Using implicit differentiation to find equation of a tangent at a point
So my problem is that I have to use implicit differentiation to find the derivative of $F(x, y) = e^{xy} - x$ when $F(x, y) = 10$ and the equation of the tangent at the point $(1, \log(11))$. So I tried to solve this using two ways:

The first way I used was rearranging the equation to $10 + x = e^{xy}$ and then using $\ln$ to simply into $\ln(10 + x) = xy$ before using implicit differentiation. The result is $$\frac{dy}{dx} = \frac{1}{10x+x^2} - \frac{y}{x}$$ and after substituting the point, I get approximately $-0.950$ as the slope.

The second way I used was just differentiating it straight away rather than rearranging and I get $$\frac{dy}{dx} =\frac{\frac{1}{e^{xy}} - y}{x}.$$ But when I sub the point in I get approximately $-0.688$ as the slope. So I'm not sure if I've done something wrong when getting the derivatives or am I not allowed to rearrange the equation?

答:
Careful, I edited what I thought was a typo, but it turned out to be the source of your error, so I reversed the edit.

$F(x,y) = 10$ when $(x,y) = (1,\ln(11))$, not when $(x,y) = (1,\log(11))$, where I (and your calculator) are using $\log(11)$ to mean the base-$10$ logarithm, not the base-$e$ logarithm.  Both your derivatives are correct, and when you plug in $(1,\ln(11))$ they both give you $$\frac{dy}{dx}\approx -2.307.$$


问:
Is $\max$ a valid order for $\cup_{n\in\mathbb{N}}\Gamma_n$, each $\Gamma_n$ is finite.
The question is:

Consider the partitions $(\Gamma_n)_{n\in\mathbb{N}}$, where  $\Gamma_n=\{z,\cdots,z+\frac{1}{n} \}$ is a partition of $[z,z+\frac{1}{n}]$, for each $n\in\mathbb{N}$ (exclude $0$). Can we order decreasingly using $\max$ the set  $$ \bigcup_{n\in\mathbb{N}}\Gamma_n\ \ \ ?$$

Is it valid for example define: $$y_1=\max \left\lbrace\bigcup_{n\in\mathbb{N}}\Gamma_n\right\rbrace =z+1$$ and for $n\geq 2$ define $$y_n=\max \left\lbrace \left\lbrace\bigcup_{n\in\mathbb{N}}\Gamma_n\right\rbrace \backslash \left\lbrace \bigcup_{k\leq n-1 }\{y_k\} \right\rbrace \right\rbrace$$ so we should have $y_n<y_{n+1}$ for all $n$. You can think of this problem also as (not partitions but sets of finite elements), at infinite we have $\lim_{n\to\infty}y_n=z$.

答:
Yes, this will work. The crucial observation is that for each $\alpha\in(0,1)$ there are only finitely many elements of $\bigcup_{n\in\Bbb Z^+}\Gamma_n$ that are greater than $\alpha$. To see this, let $m\in\Bbb Z^+$ be large enough so that $\frac1m\le\alpha$; then $\Gamma_n\subseteq[0,\alpha]$ for all $n\ge m$, so

$$(\alpha,1]\cap\bigcup_{n\in\Bbb Z^+}\Gamma_n\subseteq\bigcup_{1\le n<m}\Gamma_n\,.$$

This means that the set $\bigcup_{n\in\Bbb Z^+}\Gamma_n$ is well-ordered by $\ge$: every non-empty subset of it has a largest element. Your procedure simply picks off the largest remaining element at each stage.

Note, though, that it is possible for some member of $\bigcup_{n\in\Bbb Z^+}\Gamma_n$ to belong to more than one of the sets $\Gamma_n$ (though only, as we’ve just seen, to a finite number). As long as you care only about the points themselves and not the specific sets $\Gamma_n$ to which they belong, you’ll be in good shape. If you need to distinguish a point $z$ in some $\Gamma_n$ from the same point in some different $\Gamma_m$, however, you’ll need to work with ordered pairs $\langle z,n\rangle$ and replace $\bigcup_{n\in\Bbb Z^+}\Gamma_n$ with $\bigcup_{n\in\Bbb Z^+}(\Gamma_n\times\{n\})$. You can then order the pairs by setting $\langle z_1,n_n\rangle\preceq\langle z_2,n_2\rangle$ if and only if $z_1<z_2$, or $z_1=z_2$ and $n_1\ge n_2$. You can still pick off the maximum element at each step, but it should be the maximum element with respect to the order $\preceq$.


问:
property of function
Let $f _n (x)=x ^n$  .

If I want to get $f_{n+1}'(x)$ ,

firstly I find $f _{n+1} (x)=x^ {n+1}$   and next differentiate $f _{n+1} (x)=x^ {n+1}$  ,

I obtain $f _ {n+1}' (x)=(n+1)x^ n $ .

But in other ways, firstly differentiate $f _n (x)$ , obtain $f  _n '(x)=nx^ {n−1} $ , and substitute $n+1$  for $n$  in $f  _n '(x)=nx^ {n−1}$  , I get $f_  {n+1}' (x)=(n+1)x ^n$   too.

I think second ways is not definition of $f ′ _{n+1} (x)$ .

Is there any function which satisfying first ways is not equal to second ways?

答: Many functions satisfy this, but only because you are viewing $n$ as a fixed parameter.  You really have a function $f(x,n)$ and are asking does $\frac {\partial f(x,n)}{\partial x}|_{n \to n+1}=\frac {\partial f(x,n+1)}{\partial x}$ or do partials and substitution commute?  So let me define $f_n(x)= \begin {cases} x&n=1\\5x&n=2 \end {cases}$ and it fails.  Good for you to be thinking about this.</p>

Title: property of function


问:
Are $D_{20}$ and $\mathbb{Z}_{2} \times \mathbb{Z}_{10}$ isomorphic?

Are  $D_{20}$ and $\mathbb{Z}_{2} \times \mathbb{Z}_{10}$ isomorphic?

For $D_{20}$ I am using the notation $D_{2n}$ and $\mathbb{Z}_{2}, \mathbb{Z}_{10}$ are additive groups.

I think not, and my justification is this: In group $\mathbb{Z}_{2} \times \mathbb{Z}_{10}$ we have 2 elements of order 2, namely, $(1,5) $ and $(0,5)$; while in $D_{20}$ there are 10 rotations and 10 flips, where each flip has order 2, therefore they are not isomorphic. Is that so?

答:
Since $\Bbb Z_{10}\times \Bbb Z_2$ is abelian
and  $D_{20}\cong \Bbb Z_{10}\rtimes \Bbb Z_2\cong\langle g,h\mid g^{10}, h^2, hg=g^{-1}h\rangle$ is not, the groups are not isomorphic.


答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\left.\begin{array}{r}
\ds{\sum_{n = 0}^{\infty}a_{n}x^{\mu n}}
\\[2mm]
\ds{\sum_{n = 0}^{\infty}b_{n}x^{\nu n}}
\end{array}\right\}
\implies
\left\{\begin{array}{rcl}
\ds{\sum_{j = 0}^{\infty}a_{j}\, x^{\mu j}\sum_{k = 0}^{\infty}b_{k}\,x^{\nu k}} & \ds{=} &
\ds{\sum_{j = 0}^{\infty}\sum_{k = 0}^{\infty}a_{j}\, b_{k}\sum_{n = 0}^{\infty}
\delta_{n,\mu j + \nu k}\,\,\,x^{n}}
\\[2mm] & \ds{=} &
\ds{\sum_{n = 0}^{\infty}\pars{\sum_{j = 0}^{\infty}\sum_{k = 0}^{\infty}a_{j}\,b_{k}\,
\delta_{n,\mu j + \nu k}}}x^{n}
\\[2mm] & \ds{=} &
\bbx{\ds{\sum_{n = 0}^{\infty}\pars{\sum_{j = 0}^{\infty}a_{j}
\,b_{\pars{n - \mu j}/\nu}\,\,\,
\bracks{{n - \mu j \over \nu} \in \mathbb{N}_{\geq 0}}}}x^{n}} \\ &&
\end{array}\right.
\end{align}

Title: Multiplying power formal series when reindexing is required?

Body:
Normally, when I multiply the two following series:

$$\sum_{n\ge0}a_nx^n$$

$$\sum_{n\ge0}b_nx^n$$

I have:

$$\sum_{n\ge0}(\sum_{i=0}^na_ib_{n-i})x^n$$

Which is nice and intuitive. However, what if the exponents of x, or the indices don't match up? For example, what if I wanted to multiply these two:

$$\sum_{n\ge0}a_nx^{100n}$$

$$\sum_{n\ge0}b_nx^n$$

If I were to try to do this I'd reindex the second series to:

$$\sum_{\frac{n}{100}\ge0}b_nx^{100n}$$

But then the indices wouldn't line up. Am I missing something?


问:
Harmonic functions $u$ and $v$ such that $\nabla u = a \nabla v$ for some function $a$.
Let $u,v\colon\Omega\to\mathbb R$ be harmonic functions on the open set $\Omega \subset\mathbb R^2$ such that:

Exists function $a\colon\Omega\to\mathbb R$ such that $\nabla u \left(x,y\right) = a \left(x,y\right) \nabla v\left(x,y\right)$ for every $\left(x,y\right)\in \Omega$.

$\nabla v \ne \mathbf 0$ for every $\left(x,y\right) \in\Omega$.

Prove that there exist $b,c\in\mathbb R$ such that $u\left(x,y\right)=b\,v\left(x,y\right)+c$ for every $\left(x,y\right)\in\Omega$.

答:
It seems that this works in any dimension.

Indeed, assume that $u,v$ are harmonic in a domain $\Omega\subseteq\mathbb R^n$, with $$\nabla u=a\nabla v$$ for a function $a$, and $\nabla v\neq 0$ everywhere in $\Omega$. Multiplying with $\nabla v$, we obtain that $a=\nabla u\nabla v|\nabla v|^{-2}$, which implies that $$|\nabla v|^2\nabla u=(\nabla u\nabla v)\nabla v.$$ Considering the divergence and using that $u,v$ are harmonic, we get $$\nabla\left(|\nabla v|^2\right)\nabla u=\nabla\left(\nabla u\nabla v\right)\nabla v.$$ On the other hand, the gradient of $a$ is equal to $$\nabla a=\frac{\nabla\left(\nabla u\nabla v\right)|\nabla v|^2-(\nabla u\nabla v)\nabla\left(|\nabla v|^2\right)}{|\nabla v|^4}=0,$$ so $a$ is constant.


问:
Does the existence of an onto mapping $f: \mathbb{N} \to A$ indicate the existence of a one-one mapping $g: A \to \mathbb{N}$?
This statement looks so intuitive, as the existence of an onto function from $\mathbb{N}$ to $A$ suggests that $A$'s size is not "smaller" than $\mathbb{N}$. But how should we construct a one-one mapping from $A$ to $\mathbb{N}$ without using the axiom of choice?

P.S. I am skeptical about whether we can escape the axiom of choice, as many other seemingly naive theorems, for instance, any relation can be reduced to a function of the same domain, has to be proved using the AC.

答:
Since $\mathbb{N}$ has an inherent well-order (viz., $\omega$), you don't need AC to construct
$$
g(a):=\min\{n\in\mathbb{N}\mid f(n)=a\}.
$$


问:
Probability assignment
There are two boxes. In the first one there are, 4 black and 3 white balls, and in the second one there are 3 black and 5 white balls. We take two balls from the first box, and one from the second one and we put then in a third box. If we take out one ball from the third box, what is the probability that the ball we took out is white?

Can someone help me with this?

答:
Let $A$ be the event that a white ball is taken from box 3. Also let $X_k$ be the event that $k$ white balls are in box 3.

Using the law of total probability we get,

$P(A) = \sum_{k=0}^3P(A|X_k)P(X_k)$.

$P(A|X_k)$ are easy to calculate. If there are NO white balls in box 3, that is $k=0$, then $P(A|X_0) = 0$. If there are 1 white balls in box 3 ($k=1$), then $P(A|X_1)=1/3$. Similar argument shows that $P(A|X_2) = 2/3$ and $P(A|X_3) = 1$.

Next we calculate $P(X_k)$.

$P(X_3)$ = $P$(2 white balls from box 1 AND 1 white ball from box 2). Since taking ball(s) from box 1 and box 2 are independent events we have

$P$(2 white balls from box 1 AND 1 white ball from box 2) = $P$(2 white balls from box 1)$P$(1 white ball from box 2)

$P$(2 white balls from box 1) =  $\frac{3\choose2}{7\choose2}$

$P$(1 white ball from box 2) = $5/8$.

Hence $P(X_3) = \frac{3\choose2}{7\choose2}\frac{5}{8} = \frac{5}{56}$

Similar arguement shows that

$P(X_2) = \frac{3\choose2}{7\choose2}\frac{3}{8} + \frac{{3\choose1}{4\choose1}}{7\choose2}\frac{5}{8} = \frac{23}{56}$.

$P(X_1)= \frac{4\choose2}{7\choose2}\frac{5}{8} + \frac{{3\choose1}{4\choose1}}{7\choose2}\frac{3}{8} = \frac{22}{56}$.

We do not have to calculate $P(X_0)$ since it will be cancelled out by $P(A|X_0)$ which equals $0$.

Finally, putting it all together,

$P(A) = 0 + \frac{1}{3}\frac{22}{56} + \frac{2}{3}\frac{23}{56} + \frac{5}{56}$

which is approximately $0.49404$

Title: Probability assignment


问:
Why do rectangular matrices have no inverse that can solve Ax = v?
We can reverse a linear transformation using an inverse matrix to find the input vector used in the original linear transformation:

$$A^{-1}A\vec{x} = A^{-1}\vec{v}$$

Let's say we have a 2D vector that we transform into a 1D "vector":
$$\begin{pmatrix}
a & b
\end{pmatrix}*\begin{pmatrix}
x  \\
y
\end{pmatrix} = \begin{pmatrix}
ax + by
\end{pmatrix}       $$

Isn't there some matrix that can then convert that 1D vector back into the original 2D vector? Something like this:

$$\begin{pmatrix}
c  \\
d
\end{pmatrix}*\begin{pmatrix}
ax + by
\end{pmatrix}  = \begin{pmatrix}
x  \\
y
\end{pmatrix}      $$

My TA says that the original transformation does not have an inverse because a single input vector would map to multiple  output vectors. But I'm not seeing those multiple outputs in my algebra.

答: The point is that there are infinitely many vectors $\pmatrix{x\cr y\cr}$ that would give the same $ax+by$.  For example, if $a=1$ and $b=1$, $\pmatrix{1 & 1} \pmatrix{t\cr -t\cr} = 0$ for all $t$.  So how is your inverse going to take that $0$ and find out what $t$ it came from?


答:
You can prove that both $\delta_Y \circ f$ and $(f \times f) \circ \delta_X$ are the product morphism associated with the pair of morphisms $f:X\rightarrow Y$ and $f:X\rightarrow Y$.

You can do this by drawing two diagrams.

$\require{AMScd} \begin{CD}
@.X\\
@.*@V{\delta_X}VV*\\
X @<{\pi_X}<<X\times X @>{\pi'_X}>> X\\
@V {f} VV@V{f\times f}VV @V{f}VV \\
Y @<{\pi_Y}<<Y\times Y@>{\pi'_Y}>> Y
\end{CD}$

(Please add the diagonal identity morphisms on $X$ to this diagram where there's a "$*$", AMScd doesn't let me draw those.)

This diagram is a composition of the definition diagram of the diagonal of $X\times X$ and the definition of the product $f\times f$. This shows that indeed $\delta_X\circ f\times f$ is a product morphism in $\text{Hom}(X, Y\times Y)$ of $f:X\rightarrow Y$ and $f:X\rightarrow Y$.

On the other hand also consider the diagram:

$\require{AMScd}\begin{CD}
@.X\\
@.@V{f}VV\\
@.Y\\
@.*@V{\delta_Y}VV *\\
Y@<{\pi_Y}<<Y\times Y @>{\pi'_Y}>> Y\\
\end{CD}$

(Again please draw in your mind the diagonal arrows that are identity morphisms from $Y$ to $Y$ at the "$*$".)

This diagram is a composition of the morphism $f$ and the definition of the diagonal on $Y$. It shows that $f\circ\delta_Y$ is also a product morphism in $\text{Hom}(X, Y\times Y)$ of $f:X\rightarrow Y$ and $f:X\rightarrow Y$. Showing through uniqueness that they must be the same.

Title: Show the graph commutes, diagonal function.

Body:
$\require{AMScd} \begin{CD}
                   X @>{f}>> Y\\
@V{\delta_X}VV @V{\delta_Y}VV \\
X\times X@>{f \times f}>> Y\times Y
\end{CD}$

Where, $\delta _{X,Y}$ is the diagonal of $X$ and $Y$ respectively.

I am trying to show the diagram commutes, and this is what I have so far:

$\delta _Y \circ f$, and $(f \times f)\circ \delta_X$ I am not sure I can see how the two functions are equal. Any Ideas?


问:
log power rule - how to manipulate polynomials
This is a question on manipulating polynomials from Udacity's Free Course in collaboration with Georgia Tech, Introduction to Graduate Algorithms, video number 190 (videos are downloadable for free), in the Divide and Conquer section.

$4^{\log_2n} = n^2$

$3^{\log_2n} = n^c$

$3 = 2^{log_23}$

How does one get $3 = 2^{\log_23}$?

I looked at the rules here: https://www.rapidtables.com/math/algebra/Logarithm.html

and tried to understand the rule by setting for example
$3 = \log_3(9^1) = 1 \cdot  \log_3(9)$ but cannot extrapolate to the original equation at the top.

答:
$3 = 2^{log_2^3}$

Let $\log_2 3= \lambda$

Thus, the equation becomes:
$$3=2^{log_2^3}=2^\lambda$$

But, from equation 2, we can see that $\log_2 3=\lambda$. Or, $2^\lambda= 3$
Hope this answers your question.


答:
For any base $a>0,a\ne 1$, we know that $a^x$ and $\log _ax$ are inverse functions. If we let $f(x)=a^x$ and $f^{-1}(x)=\log _ax$, we can easily say that $$f(f^{-1}(x))=a^{\log _ax}=x$$
In this case if $a=2$ and $x=3$, we have $$3=2^{\log _23}$$

Title: log power rule - how to manipulate polynomials

Body:
This is a question on manipulating polynomials from Udacity's Free Course in collaboration with Georgia Tech, Introduction to Graduate Algorithms, video number 190 (videos are downloadable for free), in the Divide and Conquer section.

$4^{\log_2n} = n^2$

$3^{\log_2n} = n^c$

$3 = 2^{log_23}$

How does one get $3 = 2^{\log_23}$?

I looked at the rules here: https://www.rapidtables.com/math/algebra/Logarithm.html

and tried to understand the rule by setting for example
$3 = \log_3(9^1) = 1 \cdot  \log_3(9)$ but cannot extrapolate to the original equation at the top.


答:
As long as you stick to real, continuous functions, no. Here's the reasoning:

In the first example you provided, you find the next term, then take its derivative. You end up with the derivative of the $(n+1)$th term.

In the second example you provided, you find the general derivative term, then plug in $n+1$. You end up with the derivative of the $(n+1)$th term.

To think about it another way:

$$f_n{x}\to f_{n+1}{x}\to f'_{n+1}{x}$$
$$f_n{x}\to f'_n{x}\to f'_{n+1}{x}$$

Both get you to the same place. There's no reason why this would change. However, for discontinuous functions, it could very well be different.

Title: property of function

Body:
Let $f _n (x)=x ^n$  .

If I want to get $f_{n+1}'(x)$ ,

firstly I find $f _{n+1} (x)=x^ {n+1}$   and next differentiate $f _{n+1} (x)=x^ {n+1}$  ,

I obtain $f _ {n+1}' (x)=(n+1)x^ n $ .

But in other ways, firstly differentiate $f _n (x)$ , obtain $f  _n '(x)=nx^ {n−1} $ , and substitute $n+1$  for $n$  in $f  _n '(x)=nx^ {n−1}$  , I get $f_  {n+1}' (x)=(n+1)x ^n$   too.

I think second ways is not definition of $f ′ _{n+1} (x)$ .

Is there any function which satisfying first ways is not equal to second ways?


问:
How to prove the given set is convex
I was given to prove the solution set of the following constraints is a convex set.

$x_1+2x_2+x_3 \leq 10$ , $x_1 - 2x_3 \leq 9$ , $x_1 \geq 2$ , $x_2 \geq 0$ , $x_3 \geq 0$

My Attempt:  That can be written as $AX \leq B$  where $A = \begin{bmatrix} 1 & 2 & 1 \\ 1 & 0 &  -2 \end{bmatrix}$  and $B = \begin{bmatrix} 10 \\ 9 \end{bmatrix}$

Let's suppose $Y  , Z$  are two solutions of $AX \leq B$ . It can be easily proved that  $c_1 Y + c_2 Z$  will also be solution of $AX \leq B$  and $c_1 Y + c_2 Z$ will satisfy other constraints.

答: You can also use the result that half-spaces are convex (which is not too difficult to prove), and the fact that the intersection of convex sets is also convex. Since each constraint of your solution set is a half-space (in $\mathbb{R}^3$), it follows that the solution set is convex.

Title: How to prove the given set is convex


问:
Solve with KKT min $x_1-4x_2+x_3$

Given min $x_1-4x_2+x_3$

s.t $x_1+2x_2+2x_3=-2$

$||x||^2\leq1$

(i)Given a KKT point of the problem above,must it be an optimal solution?

(ii) Find the optimal solution of the problem using KKT conditions.

for (i) taking the point $(0,-0.89,-0.11)$ we satisfy slater's condition and the objective function is convex and the first contrains is affine and second is convex with inequality. therefore a KKT point is optimal point.
for (ii) I know I just need to find 1 KKT point

$L(x,\lambda,\eta)=x_1-4x_2+x_3+\lambda(x_1+2x_2+2x_3+2)+\eta(x_1^2+x_2^2+x_3^2-1)$

(1) $\frac{\partial L}{\partial x_1}=1+\lambda+2\eta x_1 = 0$

(2) $\frac{\partial L}{\partial x_2}=-4+2\lambda+2\eta x_2 = 0$

(3) $\frac{\partial L}{\partial x_3}=1+2\lambda+2\eta x_3 = 0$

(4) $\lambda(x_1+2x_2+2x_3+2)=0$

(5) $\eta(x_1^2+x_2^2+x_3^2-1)=0$

(6) $(x_1+2x_2+2x_3+2)=0$

(7) $x_1^2+x_2^2+x_3^2-1\leq0$

if $\lambda=\eta=0$ we get from (1) that $1=0$ therefore not feasible

if $\lambda\neq0,\eta=0$ from (1) we get that $\lambda=-1$ and from (2) we get that $\lambda=2$ not feasible again.

if $\eta\neq0,\lambda=0$ we get from (1) and (3) that $x_1=x_3=\frac{-1}{2\eta}$ and $x_2=\frac{2}{\eta}$ plugging into (6) we get that $\eta=\frac{-5}{4}<0$ not possible because $\eta\geq0$ therefore not feasible again.

if $\lambda,\eta\neq0$ from (2)-(3) we get $\frac{-5}{2\eta}+x_2=x_3$ and (2)-2*(1) we get that $\frac{-6}{4\eta}+\frac{1}{2}x_2=x_1$ plugging into (6) we get that $\frac{1}{\eta}=\frac{9}{13}x_2+\frac{4}{13}$ plugging all that we got into (5) we know that $x_1^2+x_2^2+x_3^2=1$ because $\eta\neq0$ and I get that $x_2=0.0503$ or $x_2=-0.764$ the second options is not feasible because $\eta\geq0$ and from the first one I don't get the right answer(checked in matlab)

I wonder what am I doing wrong here, any help please?

Opt vector is $(-0.5194,0.1074,-0.8478)$ opt val $-1.7969$

答:
plugging all that we got into (5) (..) I get that $x_2=0.0503$ or $x_2=-0.764$

Something went wrong in this step. I put (5) into Wolfram Alpha and got $x_2 \approx 0.1074$ as one of the solutions.

Title: Solve with KKT min $x_1-4x_2+x_3$


问:
Set of natural numbers and Peano axioms
Under standard Peano axioms (below, from Wikipedia), what implies how the set of natural numbers actually looks like, e.g. that 1 = S(0), 2 = S(1), 3 = S(2), etc.?

Why not for example 2 = S(0), 4 = S(2), 6 = S(4), with no odd numbers or some other variation of it?

Or is the above also a part of the definition of natural numbers and I'm missing some axioms here?

0 is a natural number.

Every natural number has a successor which is also a natural number.

0 is not the successor of any natural number.

If the successor of x equals the successor of y, then x equals y.

The axiom of induction: If a statement is true of 0, and if the
truth of that statement for a number implies its truth for the
successor of that number, then the statement is true for every
natural number.

答:
I think you are missing the point of the Peano Axioms.
It postulates that there is a set $\,\mathbb{N}\,$ which
is by convention called the set of natural numbers. We
are given that zero is a natural number and is by convention
denoted by $\,0.\,$ In order to avoid confusion and emphasize
its nature, perhaps it would be better to use a distinct
notation such as "$0$". The successor of zero is a natural
number by convention denoted by "$1$". This conventional
defining property of "$1$" as the successor of "$0$" in
conjunction with other definitions and the Peano Axioms
leads to all of its properties. Similarly for all of the
other natural numbers. Each natural number is defined to
be the successor the the previous number. In other words,
Only the number zero is initially given and all of the rest
of the natural numbers are determined as the repeated
successors of zero. For example, "$1$" $=S($"$0$"$),\,$
"$2$"$:=S(S($"$0$"$)),\,\dots .\,$ The actual identity of
the other natural numbers is not important. What is of
great importance is that they are all definite successors
of zero. You are allowed to use any given set as the
set of natural numbers as long as one element is singled
out as the zero element and all the rest of the elements
are the successors of zero.

Thus, if you wish, you can use the set of even numbers as
a model of the natural numbers and then define
$\,2=S(0),\,$ $4=S(2),\,$ $6=S(4),\,$ $\dots.\,$
In this model the number denoted by $\,2\,$ is the successor of
zero but this does not change its properties in the model
of the natural numbers. For example, in this model, we
have $\,2\times 2=2\,$ using the Peano definition of $\,\times\,$ (multiplication) of natural numbers. This is because $\,2\,$
models the natural number "$1$" and has the same properties in the model as "$1$" has.

This is an example of the abstract nature of modern axiomatic mathematics. The natural numbers are not defined by what they
are, but by what they do. All models of the natural
numbers relative to the Peano Axioms are equivalent in
the sense that they all have the same properties in the model.
The Wikipedia article Peano axioms
has a lot of details, but the fundamental idea is that it is an axiomatic model of the natural numbers. Previously generations of mathematicians took them as given somewhat as in the quote
"God made the integers, all else is the work of man". The natural
numbers have not changed, but our view of their nature has.

Title: Set of natural numbers and Peano axioms


问:
Conditional Probability of drawing balls
A bag has $10$ red balls and $12$ blue balls. You draw $6$ times from the bag, without replacement. What is the probability that the first two draws were red given that you drew EXACTLY $2$ red balls?

My attempt:

I am assuming that the problem is equivalent to finding the probability of drawing $4$ blue balls in the last $4$ draws. By this logic, we get

$P($4 blues in 4 draws$) = \frac{12}{20} \cdot \frac{11}{19} \cdot \frac{10}{18} \cdot \frac{9}{17}$

Is this correct or did I make a mistake somewhere? Any assistance is appreciated.

答:
Let $\Gamma$ be the event $RRBBBB$ and $\Delta$ be the event $\{2R,4B\}$.

$\Gamma \cap \Delta = \Gamma$,

So $$P(\Gamma|\Delta)=\frac{P(\Gamma \cap \Delta)}{P(\Delta)}=\frac{P(\Gamma)}{P(\Delta)}$$

$$P(\Gamma) = \frac{10}{22} \frac{9}{21} \frac{12}{20} \frac{11}{19} \frac{10}{18} \frac{9}{17}$$

$$P(\Delta) =\frac{ \pmatrix{10\\2} \pmatrix{12\\4}}{\pmatrix{22\\6}}$$

Simplifying, this gives $$P(\Gamma|\Delta)= \frac{2!4!}{6!}=\frac{1}{15}$$
which is consistent with Graham's answer which is more elegant.


答:
I am not sure if you are misreading the expression, or just mistyped it, but the exponent on the $2$ is not $\log_2^3$ (which doesn't mean anything) but rather $\log_2 3$.  As for why $2 = 2^{\log_2 3}$, this is a case of a general identity:

$$b^{\log_b x} = x$$
for any $x>0$ and any base $b>0$.

As for why this identity is true:  The function $f(x) = \log_b x$ and $g(x) = b^x$ are inverse functions.  (Indeed, that's usually how the logarithm function is defined, as the inverse of the exponential function.)  That means that $f(g(x)) = x$ and $g(f(x))=x$.  The first one of these equations says exactly that $b^{\log_b x} = x$.

Here's an informal way of explaining it that might help you understand better:

The expression $\log_2 3$ can be thought of as the answer to the question "What power do you raise $2$ to, if you want to get $3$?"  The expression $2^{\log_2 3}$ means "raise 2 to the power that, if you raised $2$ to it, would produce $3$."  When you say it that way, it should be obvious that the result is $3$.

If it's still confusing, try it again with different numbers, like $10$ and $1000$.  Then $\log_{10} 1000$ means "find the exponent which, when $10$ is raised to it, produces $1000$."  The answer to this is $3$.  Now $10^{\log_{10} 1000}$ means "Raise $10$ to the $3$rd power", which produces $1000$, of course.  So $10^{\log_{10} 1000} = 1000$.


问:
Equality of triple integrals over unit sphere $ \iiint_{\text{unit ball}} x e^{ax + by + cz} dV$
I have to calculate
$$ \iiint_{\text{unit ball}} x e^{ax + by + cz} \,dV,$$

where by "unit ball" I mean the region $x^2 + y^2 + z^2 \leq 1$.

I know how to calculate this (rotation matrix that takes $(a,b,c)$ to $(0,0,\sqrt{a^2 + b^2 + c^2})$ and then spherical coordinates). The answer gives
$$\frac{4\pi a}{r^5}((3 + r^2)\sinh(r) - 3r \cosh(r)), \,\,\,\,\,r = \sqrt{a^2 + b^2 + c^2}$$

Now this would suggest that we have the following equality:

$$ \iiint_{\text{unit ball}} \frac{x}{a} e^{ax + by + cz} \,dV = \iiint_{\text{unit ball}} \frac{y}{b} e^{ax + by + cz} \,dV = \iiint_{\text{unit ball}} \frac{z}{c} e^{ax + by + cz} \,dV$$

Is it true? How would one quickly prove it if it is? It would be relevant because if one could quickly spot and prove this, then by naming the value of the above integrals $I$, we would have
$$(a^2 + b^2 + c^2)I = \iiint_{\text{unit ball}} (ax+by+cz) e^{ax + by + cz} \,dV$$

This integral would be much easier to calculate than the first since it wouldn't require to calculate the rotation explicitly (just considering an arbitrary rotation that orients $(a,b,c)$ along the $z$-axis suffices.)

Any ideas?

答:
In the Cartesian coordinates $(u,v,w)$ with the unit vector $\hat w=\frac{1}{\sqrt{a^2+b^2+c^2}}(a,b,c)$,
$$\hat x= \cos\theta_u \hat u + \cos\theta_v \hat v + \cos\theta_w \hat w$$
where $\cos \theta_w =\frac{a}{\sqrt{a^2+b^2+c^2}}$. Then
\begin{align}
\iiint_{{r<1}} \frac xa  e^{ax + by + cz} \,dV
& =  \frac1a\iiint_{{r<1}}( \cos\theta_u u + \cos\theta_v v + \cos\theta_w w) e^{\sqrt{a^2+b^2+c^2}w }\,dV\\
&= \frac1a\iiint_{r<1} \cos\theta_w  w e^{\sqrt{a^2+b^2+c^2}w } \,dV\\
&= \frac{1}{\sqrt{a^2+b^2+c^2}} \iiint_{r<1} w e^{\sqrt{a^2+b^2+c^2}w } \,dV=I
\end{align}
where the integrations over $u$ and $v$ vanish due to symmetry of the unit ball. Likewise,
\begin{align}
\iiint_{r<1} \frac yb  e^{ax + by + cz} \,dV
= \iiint_{r<1} \frac zc e^{ax + by + cz} \,dV
= I
\end{align}


问:
Check if $\lim_{x\to\infty}{\log x\over x^{1/2}}=\infty$, $\lim_{x\to\infty}{\log x\over x}=0$
Could anyone tell me which of the following is/are true?

$\lim_{x\to\infty}{\log x\over x^{1/2}}=0$, $\lim_{x\to\infty}{\log x\over x}=\infty$

$\lim_{x\to\infty}{\log x\over x^{1/2}}=\infty$, $\lim_{x\to\infty}{\log x\over x}=0$

$\lim_{x\to\infty}{\log x\over x^{1/2}}=0$, $\lim_{x\to\infty}{\log x\over x}=0$

$\lim_{x\to\infty}{\log x\over x^{1/2}}=0$, but $\lim_{x\to\infty}{\log x\over x}$ does not exists.

For I know $\lim_{x\to\infty}{1\over x}$  exist and so is for $1\over x^{1/2}$.

答: $3$ is correct as $\log x$ grows slower than any $x^n$. So $x^{-1}$ and $x^{-\frac{1}{2}}$ will manage to pull it down to $0$. And $\displaystyle\lim_{x\to\infty}{1\over x}=\lim_{x\to\infty}{1\over \sqrt{x}}=0$.</p>


问:
Finding a consistent Estimator for $\mathbb{E}(X^2)$
Given $X_1,\dots,X_n\stackrel{iid}{f}(x;\mu,\sigma^2)$ with population mean $\mu$ and population standard deviation $\sigma$. I want to find a consistent estimator for $\mathbb{E}(X^2)$.

By the variance formula we know $\mathbb{E}(X^2)=\mathrm{Var}(X_i)+[\mathbb{E}(X_i)]^2=\sigma^2+\mu^2$

So if I am not mistaken $\sigma^2+\mu^2$ is our estimand and we want to find an estimator that will converge to this estimand as $n\rightarrow\infty$. $\color{red}{\text{correct me if im wrong!}}$

We know $\bar{X}\stackrel{p}{\longrightarrow}\mu$ , I am assuming squaring will preserve continuity so $\bar{X}^2\stackrel{p}{\longrightarrow}\mu^2$.  $\color{red}{\text{correct me if im wrong!}}$

But I do not know how to proceed from here. It does not feel right to say $\mathbb{E}(\bar{X}^2)\stackrel{p}{\longrightarrow}\mu^2+\sigma^2$. I am assuming I will have to show it converges using chebyshev's.

I know in general $\mathbb{E}(\bar{X}^2)=\frac{\sigma^2}{n}+\mu^2$, so I was considering $n\bar{X}$   $\color{red}{\text{ (is this a good idea?)}}$

Any help or guidance would be appreciated!

答:
If you want to estimate $\mathbb{E}[X^2]$, a natural estimator would be to simply take the sample mean of $X_i^2$. Then by the weak law of large numbers, $\frac{1}{n} \sum^n_{i=1} X_i^2 \stackrel{p}{\rightarrow} \mathbb{E}[X^2]$.

If you want to use your approach (which seems to also work when pushed a bit), a useful tool would be the continuous mapping theorem. Thus we know $\overline{X} \stackrel{p}{\rightarrow}
\mu$, and so since squaring is continuous, indeed $\overline{X}^2 \stackrel{p}{\rightarrow} \mu^2$. For the variance, we can use the usual sample variance estimator, i.e. $\frac{1}{n} \sum^n_{i=1} (X_i - \overline{X})^2$, which converges to $\sigma^2$ in probability. Then the sum of these two estimators will also converge to $\mu^2 + \sigma^2$ in probability (again, by the continuous mapping theorem).


答:
My attempt:

I am assuming that the problem is equivalent to finding the probability of drawing $4$ blue balls in the last $4$ draws. By this logic, we get

$P($4 blues in 4 draws$) = \frac{12}{20} \cdot \frac{11}{19} \cdot \frac{10}{18} \cdot \frac{9}{17}$

This is the probability that the third to sixth draws are blue balls given that the first two balls are red. That is not what you seek.

You are given that a specific drawing has exactly two red balls among its six balls.

Each of those balls has an equal chance of being in any particular place in this draw.

So you seek the probability that those red balls were in the first two among the six positions of that drawing.

Ie: that you have obtained 2 from 2 positions when selecting 2 from 6 positions without bias.

Title: Conditional Probability of drawing balls

Body:
A bag has $10$ red balls and $12$ blue balls. You draw $6$ times from the bag, without replacement. What is the probability that the first two draws were red given that you drew EXACTLY $2$ red balls?

My attempt:

I am assuming that the problem is equivalent to finding the probability of drawing $4$ blue balls in the last $4$ draws. By this logic, we get

$P($4 blues in 4 draws$) = \frac{12}{20} \cdot \frac{11}{19} \cdot \frac{10}{18} \cdot \frac{9}{17}$

Is this correct or did I make a mistake somewhere? Any assistance is appreciated.


问:
Significant figures: Is the answer on this question correct?
A google form question

So I got this question in a test and options 2, 3 and 4 were wrong, in my view. So I chose a random option and I got it wrong. Upon further enquiry, the teacher explained that the second and fourth option was actually correct because it could be ROUNDED to 4 significant figures and 3 significant figures respectively. However, the third option was impossible to be rounded to 2 significant figures. So my question is, was option three really the right answer, or options 2, 3 and 4 were all the right answer?

It may not really be math but its just the wording of the questions and answers are ambiguous.

答: All the proposed answers are wrong except the first.  $97.040$ has five significant figures because the trailing zero is not needed to set the decimal point, so it is just there as a significant figure.  $21000$ has only two significant figures.  All the zeros are there to set the scale.  $0.03$ has only one and $970$ has two.  All the zeros in the answers except the first are there to set the decimal point.  The question does not have a unique answer.  I suspect the writer lost track of the not in the question when proposing the answers.  I don't think the wording is ambiguous, the question and answers do not match.


问:
Why is the direction of angle theta in circular motion towards and inwards?
Why is the direction of angle $\theta$ in circular motion towards and inwards of plane x-y axis?

I am not getting this concept at all.As the angel theta is changing ; the arc length (s in diagram ) is also changing;

We know about vector cross product of two vectors a and b whose direction is perpendicular to the plane of paper( or perpendicular xy axis on my paper i.e xz axis).So my sir tells me that to find direction of angle θ (it is on a 2D plane, X-Y axis), then its direction is on X-Z plane. (You can interpret the direction of vector cross product by right hand thumb rule ).

Also, that z axis is drawn from the centre of circle. Only from there only.

答:
This answer is based on the comment that the statements in the question occurred during instruction in physics.

There is a convention in physics of using a vector to represent rotation around a fixed line in space (the "axis of rotation", which may or may not also be a coordinate axis). The length of the vector is proportional to the amount of rotation. This convention has some advantages when representing rotational speed,
but in introductory physics it might also be used to represent the total angle through which the rotation was performed. If you rotate twice as much, you make the vector twice as long.

If you have an object traveling in a circle around the origin in the $x,y$ plane, this is a rotation around the $z$ axis.
To represent a rotation through a certain angle, which moves the object a certain distance around the circle, you make a vector in the same direction as the $z$ axis and make its length equal to the rotation measured in radians
(equal to the distance traveled around the circle if the circle has radius $1.$)

To decide whether the vector should point in the positive $z$ direction or the negative $z$ direction, you use another convention, usually a "right hand" rule.
The idea that the vector has to "start" in the center of the circle is your instructor's convention, perhaps to help you remember graphically where the axis of rotation is. For a mathematician, a three-dimensional vector in three-dimensional space does not start anywhere: it merely has a direction and a length.

This particular use of a rotation vector has limited usefulness. In special cases you can represent the result of doing one rotation and then another rotation by adding the rotation vectors, but only if the rotations are around the exact same axis (or at least around parallel axes). The triangle rule for adding vectors gives wrong answers to this kind of problem.


答:
The question is flawed in my opinion.

Let's think of the options as being "rounded off" figures and try to see if we can decide how many significant figures the original number (usually the result of a measurement or calculation incorporating measured quantities) had to begin with. But first let me express these using standard form (also known as scientific notation) as I often find it easier to decide on significant figures this way.

When you express these choices in standard form, you get:

$9.7040 \times 10^4$

$2.1 \times 10^4$ (but this is ambiguous, explained below)

$3 \times 10^{-2}$

$9.7 \times 10^2$ (but this is ambiguous, explained below)

Now, of those choices, only with the first and the third can you explicitly state the number of significant figures the number has. This is because with the first choice, there is a trailing zero - if this were not significant, it would have been clipped off (truncated). You must represent this trailing zero in the standard form as well. The fact that it's left intact means it is significant. So the first option is definitely true.

The third option definitely has just the one significant figure because you only have the single digit in the standard form, without any trailing zeroes. Remember, zeroes that "just" indicate the magnitude of a number do not count as significant digits. They disappear in standard form.

OK, now on to the second option. Now, what I wrote is one perfectly reasonable standard form for the result (with two significant figures), which will give $21000$ in "longhand" form. The problem is that I could also have written $2.10 \times 10^4$ or $2.100 \times 10^4$ or $2.1000 \times 10^4$. These are all numerically equivalent to $21000$ but all have different numbers of significant figures. Note that the representation $2.100 \times 10^4$ does have $4$ significant figures, so $21000$ could have $4$ significant figures. But there are other representations that have different numbers of significant figures. To me it's impossible to decide simply based on the representation $21000$, so I would leave it as ambiguous (but $4$ s.f. is possible).

Similarly, the last one has two significant figures in the representation I chose but it could also have been represented as $9.70 \times 10^2$, which would mean it had $3$ significant figures. Again, there is ambiguity here. But $3$ s.f. is a possible answer.

So, I would have gone with choice number three based on it being the only unequivocally incorrect choice. But it is a poor question as options two and four are ambiguous.

Some might think to take the "longhand" representations as exact numbers, and ask how many significant figures they possess based on the nonzero digits they had (excluding trailing zeroes). On this basis, they might state that the options had (respectively)  $5, 2, 1, 2$. In my opinion, this is an incorrect way of thinking about it. Fundamentally, significant figures are based on measurement or derived from measured values in some way. They are much more intimately linked to the physical sciences than to pure mathematics. So if those values are taken as "exact", that implies that some measuring instrument returned a value that was exactly equal to that, in which case, you would actually have $5, 5, 1, 3$ s.f. as every "non-magnitude determining" digit in a value returned by a measurement method is to be considered significant. We are basically positing here that the putative measuring instrument in the second option could have returned a value like $21002$  but instead returned "exactly" $21000$ (in which case the value $21000$ has exactly five signficant figures) and that the putative measuring instrument in the last option could have returned $973$, but instead returned "exactly"  $970$. It is not sensible in my view to talk about significant figures in a context outside measurement (or calculated values derived from measured values).

Title: Significant figures: Is the answer on this question correct?

Body:
A google form question

So I got this question in a test and options 2, 3 and 4 were wrong, in my view. So I chose a random option and I got it wrong. Upon further enquiry, the teacher explained that the second and fourth option was actually correct because it could be ROUNDED to 4 significant figures and 3 significant figures respectively. However, the third option was impossible to be rounded to 2 significant figures. So my question is, was option three really the right answer, or options 2, 3 and 4 were all the right answer?

It may not really be math but its just the wording of the questions and answers are ambiguous.


问:
Continuity of Complex Function of Two Functions Dependent of Re(z)

I believe this function is continuous at $0$ and hence is a continuous function.

It's continuous if $\lim_{z \rightarrow 0^-}\bar{z} = 0$, is that right?

If so, how do I prove that that's the limit? It seems really intuitive for me. It sounds as intuitive as, for example, $y=-x$ and hence $\lim{y}_{x \rightarrow 0^-} = 0$. Does it need proof?

答:
No. To show continuity of $f$ at $0$ you need to show that $\lim_{|z|\to 0} f(z)=0$ instead of just showing that the left side limit converges to $0$. To show continuity see that $$ |f(z)-f(0)|=|f(z)|=|z|$$
Since $|z|=|\overline{z}|$. So for any $\epsilon > 0$, take $\delta =\epsilon$. Then $|z-0|=|z|<\delta$ implies that $|f(z)-f(0)|<\epsilon$. Hence $f$ is continuous at $z=0$.


问:
Prove Function G Is Linear
First of all $v, w, z$ are in $\mathbb R^3$ as vectors, and there are functions $G(v, w, z) = (v \times  w) \times z$, this is cross product. So I know this function is linear in the 3 arguments, so for example $G(x+ y)= G(x)+G(y)$ and $G(ax)=aG(x)$, and $a$ is a constant. First of all what is means to be linear in second argument? And, also, prove this $G$ function is linear in the second argument.

答:
A function of multiple vectors can be said to be linear or not with respect to each of its arguments (whereas a function of one vector is said to be linear or not).

G is linear in its second argument if $G(v,aw_1+bw_2,z)=aG(v,w_1,z)+bG(v,w_2,z)$.

The left hand side is $(v \times(aw_1+bw_2))\times z=(v\times aw_1+v\times bw_2)\times z=(v\times aw_1)\times z+(v\times bw_2) \times z$, which equals the right hand side.

Title: Prove Function G Is Linear


答:
A logarithm is just an exponent. It’s just the exponent that the base needs to carry in order to produce the given value.

So if you think about $3^?=9$, you know $2$ is the exponent that $3$ needs to carry to produce $9$. That means $\log_3 9= 2$.

So now if you consider $2^?=3$, you are asking what exponent $2$ needs to carry to produce $3$. This is exactly what $\log_2 3$ is, so of course it is true that $2^{\log_2 3}=3$.

Likewise $5^{\log_5 3}=3$ because $\log_5 3$ is the exponent that $5$ must carry to produce $3$.

And $17^{\log_{17}91}=91$ because $\log_{17} 91$ is the exponent that $17$ must carry to produce $91$.

And so on.


问:
If $x^5=2$, find $\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$

If $x^5=2$, find
$$\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$$

My attempt-

Since $x^5=2,x^6=2x,x^7=2x^2..$and so on

The equation is equivalent to
$$x^5\left(\frac{1}{2x+x^4}+\frac{1}{2x^2+x^3}+\frac{1}{x^3+x^2}+\frac{1}{x^4+x}\right)$$

Which simplifies to
$$2\left[\frac{3x^4+7x^3+5x^2+8x+12}{2x^4+6x^3+4x^2+6x+12}\right]$$
Now I am struck here.

Any help will be appreciated.

Note: Please don't use root tables and complex numbers too!

Body:
$$
y=\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$$
$$
=\frac{x (x^4 + x^3 + x^2 + 1) (x^6 - x^5 + 2 x^4 + x + 1)}{(x + 1) (x^2 + 1) (x^2 - x + 1) (x^4 + 1)}$$
$$
=\frac{8 + 5 2^{1/5} + 7 2^{2/5} + 3 2^{3/5} + 6 2^{4/5}}{3 + 2 2^{1/5} + 3 2^{2/5} + 2^{3/5} + 3 2^{4/5}}$$

$$x=\sqrt[\large 5]{2}\approx 1.1487  \implies y\approx 2.37411
$$


问:
$x^3 - 6 = 25y^2 + 35y$ has no non-zero integer solutions
I need to show that $x^3 - 6 = 25y^2 + 35y$ has no non-zero integer solutions.
I tried moving things around and factor. Or use modulo 5 and 7 reductions for something shady to become a quadratic residue of a prime. I also couldn't really come up with something to use infinite descent. Hints would be appreciated.

This was a problem for non-linear diophantine equations (Pythagorean triples, rational points on curves, Fermat's last theorem, Pell's equations, etc.) in last year's elementary number theory exam.

答: We have $$x^3=25y^2+35y+6=(5y+6)(5y+1)$$Now, not both $5y+6$ and $5y+1$ are cubes, since there are no two positive cubes that differ by less than $7$, so there is a prime $p$ that divides both factors.  But then $p$ divides their difference, and $p=5$.  But this is absurd, since $p\nmid 5y+1$.


问:
Prove corollary of Darboux's Theorem
I'm currently going through some analysis problems in preparation for an upcoming test, and I was struggling with the following problem:

Let ${f:[a,b]} \rightarrow \mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$ and let $x \in (a,b)$. Suppose $lim_{t \rightarrow x^{-}}f'(t)=L_{1}$ and $lim_{t \rightarrow x^{+}}f'(t)=L_{2}$.

Prove $L_1=L_2.$

Now, I can see that this likely follows from a contradiction, if we can apply Darboux's Theorem after assuming $L_1<L_2$ (without loss of generality), but I'm a little confused by how to do such. Any help is appreciated!

答: Suppose $f'(x)=L$. Find $m\in \mathbb{N}$ so that $x+\frac{1}{n}\in (a,b)$ for all $n\geq m$. From MVT we have the existence of $c_n\in \Big(x,x+\frac{1}{n}\Big)$ such that $$n\Bigg(f\Big(x+\frac{1}{n}\Big)-f(x)\Bigg)=f'(c_n)$$ Since $\lim_{t \rightarrow x^{+}}f'(t)=L_2$ and $c_n \rightarrow x^+$ we have $f'(c_n)\rightarrow L_2$. Moreover, since we know that $f'(x)$ exists and equals $L$, we have $n\Bigg(f\Big(x+\frac{1}{n}\Big)-f(x)\Bigg) \rightarrow L$ telling us that $L=L_2$. An identical argument considering a sequence of the form $x-\frac{1}{n}$ will tell us $L=L_1$.

Title: Prove corollary of Darboux's Theorem


问:
Example of a continuous function which is bounded and not contained in any $L_p$-space ($p\gt 0$)
I'm struggling to find an example of a continuous function $f:(0,\infty)\to \mathbb R$ which is bounded, not contained in any $L_p$-space ($p\gt 0$) and goes to zero when x goes to infinity.

I need help.

Thanks a lot!

答: Another example is $f:(1,\infty)\to\mathbb R$ defined by $$f(x)=\sum\limits_{n=1}^\infty2^{-n}x^{-1/n}.$$</p>


问:
Is a recursively given series monotone?
The series is given as $a_{n+1}=\frac{8a_{n}-3}{2a_{n}+1}$ and $a_{1}=1$

I have no idea how to prove if this series is monotone or not.

I have tried rearranging the equation to form $a_{n+1}>a_{n}$ but nothing seems to work.

Please if you know how to solve this it would help a ton!

答:
Just for your curiosity.

In my answer to this question, I detailed the steps for the solution of  first-order rational difference equation (have a look here) $$a_{n+1} = \frac{m\,a_n + x}{a_n + y} $$

With your parameters $m=4$, $x=-\frac 32$ and $y=\frac 12$, the results for $a_n$ are extremely simple.

On the other side, do not forget that, if the limit $L$ exists, it is the solution of $L=\frac{8 L-3}{2 L+1}$.

Title: Is a recursively given series monotone?


问:
Strange use of Fubini for infinite sum
Let $k(n)$ be a sequence of natural numbers and $y_m$ a sequence of non-negative numbers. I encountered this line I don't understand:

$$ \sum_{n=1}^{\infty} \frac{1}{k(n)^2} \sum_{m=1}^{k(n)} y_m = \sum_{m=1}^{\infty} y_m \sum_{k(n)\geq m} \frac{1}{k(n)^2}$$

justified as an application of Fubini's theorem. I am aware of this analogue of Fubini but I don't see how this equality could be true because the "integration interval" changes too. Can someone help me see how this would work out?

答:
A convenient way to think through this (whether its exchanging sums or exchanging integrals) is to replace bounds of integration by characteristic/indicator functions.

In your case, let's replace the upper bound by an indicator function and use (standard) Fubini:
\begin{align*}
\sum_{n = 1}^{\infty} \frac{1}{k(n)^{2}} \sum_{m = 1}^{k(n)} y_{m} = \sum_{n = 1}^{\infty} \frac{1}{k(n)^{2}} \sum_{m = 1}^{\infty} y_{m} 1_{\{m \leq k(n)\}} = \sum_{n = 1}^{\infty} \sum_{m = 1}^{\infty} \frac{1}{k(n)^{2}} y_{m} 1_{\{m \leq k(n)\}}.
\end{align*}
Now that there are no $m$- or $n$-dependent bounds in the sums, we can interchange them at will and then impose a bound on $n$ instead of $m$:
\begin{align*}
\sum_{n = 1}^{\infty} \frac{1}{k(n)^{2}} \sum_{m = 1}^{k(n)} y_{m} = \sum_{n = 1}^{\infty} \sum_{m = 1}^{\infty} \frac{1}{k(n)^{2}} y_{m} 1_{\{m \leq k(n)\}} = \sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} \frac{1}{k(n)^{2}} y_{m} 1_{\{m \leq k(n)\}} = \sum_{m = 1}^{\infty} y_{m} \sum_{k(n) \geq m} \frac{1}{k(n)^{2}}.
\end{align*}

The same principle applies to integrals.  For instance,
\begin{equation*}
\int_{0}^{1} \int_{x^{2}}^{1} f(x,y) \, dy \, dx = \int_{0}^{1} \int_{0}^{1} f(x,y) 1_{\{x^{2} \leq y\}} \, dx \, dy = \int_{0}^{1} \int_{0}^{1} f(x,y) 1_{\{x \leq \sqrt{y}\}} \, dx \, dy.
\end{equation*}
Thus, instead of integrating first in $y$ and then in $x$, we can write
\begin{equation*}
\int_{0}^{1} \int_{x^{2}}^{1} f(x,y) \, dy \, dx = \int_{0}^{1} \int_{0}^{\sqrt{y}} f(x,y) \, dx \, dy.
\end{equation*}

Title: Strange use of Fubini for infinite sum


答:
Prove by contradiction. Suppose the contrary that $L_{1}\neq L_{2}$.
Consider the case that $L_{1}<L_{2}.$ (The case $L_1>L_2$ can be treated similarly.) Choose $c,d\in\mathbb{R}$
such that $L_{1}<c<d<L_{2}$. By property of limit, there exists $\delta>0$
such that $f'(t)<c$ whenever $t\in(x-\delta,x)$ and $f'(t)>d$ whenever
$t\in(x,x+\delta)$. Choose $x_{1}\in(x-\delta,x)$ and $x_{2}\in(x,x+\delta)$.
Choose $y\in(c,d)\setminus\{f'(x)\}$. Observe that $f'(x_{1})<y<f'(x_{2})$,
so, by Darboux's Theorem applied on $[x_1,x_2]$, there exists $\xi\in(x_{1},x_{2})$ such
that $f'(\xi)=y.$ Clearly $\xi\neq x$ because $y\neq f'(x)$. If
$\xi\in(x_{1},x),$ then $f'(\xi)<c$. If $\xi\in(x,x_{2})$, then
$f'(\xi)>d$. Either case contradicts to $f'(\xi)=y$.

Title: Prove corollary of Darboux's Theorem

Body:
I'm currently going through some analysis problems in preparation for an upcoming test, and I was struggling with the following problem:

Let ${f:[a,b]} \rightarrow \mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$ and let $x \in (a,b)$. Suppose $lim_{t \rightarrow x^{-}}f'(t)=L_{1}$ and $lim_{t \rightarrow x^{+}}f'(t)=L_{2}$.

Prove $L_1=L_2.$

Now, I can see that this likely follows from a contradiction, if we can apply Darboux's Theorem after assuming $L_1<L_2$ (without loss of generality), but I'm a little confused by how to do such. Any help is appreciated!


问:
For any $x, x' \in \mathbb{R}^n$ of dimension $n > 0$, if for any $y \in \mathbb{R}^n$ of dimension $n > 0$ one has $x . y = x' . y$, then $x = x'$.
Since $x \cdot y = x' \cdot y$,
$x_1y_1 + x_2y_2 + x_3y_3 = x_1'y_1 + x_2'y_2 + x_3'y_3$.

I'm not sure how to go from here.
Thank you.

I'm also curious about the variation $x \times y = x' \times y$, then $x = x'$, but in the  $x, x' \in \mathbb{R}^3$ .

答:
Consider $y= e_1$. What does $xe_1 = x'e_1$ tell you in that case? Can you conclude $x = x'$ from that?

As for the dot product, what if $x' = \lambda x, y = \mu x$ for some nonzero $\lambda, \mu$?


问:
What is $\lim_{x \to \infty} \frac{\sin x}{\sin x}?$
As the title suggests, I am interested in computing $$\lim_{x \to \infty} \frac{\sin x}{\sin x}.$$ Some of my friends say that the limit is $1$ because the domain of function is $\mathbb{R} \setminus \{n \pi : n \text{ is an integer}\},$ whereas others say that this limit is undefined.

Which opinion is correct?

答:
By the familiar definition (i.e., the "Wikipedia" definition), for a real-valued function $f(x),$ we have that $\lim_{x \to \infty} f(x)$ is a real number $L$ (if it exists) such that for every real number $\varepsilon > 0,$ there exists a real number $M$ such that for all $x > M,$ we have that $|f(x) - L| < \varepsilon.$

Like you have mentioned, if $x$ is not an integer multiple of $\pi,$ we have that $f(x) = \frac{\sin x}{\sin x} = 1$; however, for any integer $n,$ we have that $f(n \pi)$ is not defined. What can you conclude?


问:
Can a countable union of sets of measure zero have nonzero measure?
I'm fairly certain the answer is no, since if $E$ is a countable union of sets $E_k$, then countable subadditivity gives $m(E)\leq \sum_{k=1}^{\infty} m(E_k)$. But I can't figure out why this "counterexample" isn't actually a counterexample:

If we let $F_n$ be the set of real numbers in $[0,1]$ with decimal expansion of length $n$, each $F_n$ is a subset of $\mathbb{Q}$ so they all have measure zero. But every $x\in[0,1]$ has a countably infinite decimal expansion (or at least has a representation $x=\sum_{k=1}^{\infty}a_k 10^{-k}$), so the (countable) union $\bigcup_{n=1}^{\infty}F_n$ is equal to a set of nonzero measure, namely $[0,1]$. Can anyone help me figure out where this reasoning breaks down?

答: All of the partial sums, or in other words all of the rationals in $[0,1]$ with terminating decimal expansions are in the union, but not necessarily their limit points. You are proceeding as though the union is closed, but it isn't. Keep in mind that this union would be a countable set with full measure, and by countable sub-additivity that is impossible since its measure is equal to 0 after re-indexing and comparing it to the countable sum of singletons.


问:
How to tell which of these two relational-algebra expression is more effective?
I have this table :

And these two expressions :

I have to tell which one of these two expression is more effictive to find "nom".

I have choice between A or B or both are equally effective.

My teacher said in class that a join is a really demanding operation but it is the most used. Other than that he didn't address this topic so i am kinda lost to answer this question.

They have the same amount of join operations so i am not sure what else i am supposed to look at..

Thank you for your help.

答:
As a general principle, filtering should happen as early as practical: joins tend to multiply the number of rows in each table, which means more work for the filters and for later joins.

In this case:

$S1 ⋈ C1$ creates 14 rows; $E1 ⋈ (S1 ⋈ C1)$ creates another 14 rows; $\sigma_{sujet=BD}(E1 ⋈ (S1 ⋈ C1))$ filters those 14 rows down to 5.  The filter is run 14 times, and 28 rows are calculated.

$\sigma_{sujet=BD} (C1)$ filters 6 rows down to 2; $S1 ⋈ \sigma_{sujet=BD} (C1)$ creates 5 rows;  $E1 ⋈ (S1 ⋈ \sigma_{sujet=BD} (C1))$ creates another 5 rows.  The filter is run 6 times, and 10 rows are calculated.

Both of these calculations give the same result, but the second - filtering as early as possible - has reduced the number of filter attempts and row calculations, because it doesn't need to come up with all the rows for course data that it doesn't care about.


问:
Interesting calculus problems of medium difficulty?
I would like to know sources, and examples of good "challenge" problems for students who have studied pre-calculus and some calculus. (differentiation and the very basics of integration.) Topics could be related to things such as:

Taylor Series.

Product Rule, Quotient Rule, Chain Rule.

Simple limits.

Delta Epsilon Proofs.

Induction proofs for the sum of the first n, integer, squares, etc.

Integration by substitution.

Other topics...

What I have found so far are too many problems that are just a bit too difficult. The problem can have a "trick" but it needs to be something a freshman could do.

Here is one problem that I thought was just at the right level:

If $f(x) = \frac{x}{x+\frac{x}{x+ \frac{x}{x+ \vdots}}}$, find $f'(x)$*

*To be honest this problem makes me a little nervous. Still, I like it.

Body: Evaluate $\displaystyle \lim_{x\to0}\frac{x^2\tan{\left(\sin{x^2}-2x\sin{x}\right)}}{\sin{x^2}}$.


问:
Prove the sequence ${|(−1)^n|}n∈N$ has limit = 1.
Hello I am working on a problem where I have to prove that a sequence ${a_n}, n \in \mathbb{N}$ may not converge but $\lim_{n\rightarrow\infty} |a_n|$ may exist. I am using an example where $a_n = (-1)^n$. I know how to prove that ${a_n},  n \in \mathbb{N} $ does not converge but I am having trouble proving that $\lim_{n\rightarrow\infty} |(-1)^n| = 1$. I was wondering if anyone had any idea where I should start. I was thinking proof by induction.

答: Let $x_n=|(-1)^n|=1$, For given $\epsilon>0$, there exist $M\in \Bbb{N}$ such that $|x_n-1|=0<\epsilon, ~\text{provided} ~n\geq M$.


答:
It depends on exactly what your definition of limit is.

In calculus, limits of real-valued functions are often defined in such a way that in order for the limit as $x\to a$ to exist, the function must be defined on some interval of the form $(a-\delta,a+\delta)$, except perhaps at $a$. For limits as $x\to\infty$, you require the function to be defined on some interval of the form $(M,\infty)$. The definition using $\epsilon$s would look something like:

$\lim_{x\to\infty} f(x)=L$ if and only if for all $\epsilon\gt 0$ there exists $M$ such that if $x\gt M$, then $|f(x)-L|\lt\epsilon$.

However, sometimes limits are defined differently (slightly more generally, applicable to more functions), and the prerequisite for the limit as $x\to a$ to exist is that $a$ be an “accumulation point” of the domain of $f$. That is, that for all $\delta\gt 0$, there exist points $b$ in the domain of $f$ such that $0\lt |b-a|\lt \delta$. For limits as $x\to\infty$, one requires that for all $M$ there exist $b$ in the domain of $f$ such that $b\gt M$.

Here, the definition using $\epsilon$s would be:

$\lim_{x\to\infty} f(x)=L$ if and only if (i) for every $M$ there exists $x\gt M$ such that $x\in\mathrm{dom}(f)$; and (ii) for every $\epsilon\gt 0$ there exists $N$ such that if $x\gt N$ and $x\in\mathrm{dom}(f)$, then $|f(x)-L|\lt\epsilon$.

If your definition of limits is of the first type, then the limit you write does not exist, because the function does not meet the requirement to be able to even talk about a potential limit as $x\to\infty$: it is not defined on any interval of the form $(M,\infty)$, so you can always find an $x$ with $x\gt M$, and where $|f(x)-1|\lt\epsilon$ fails because the expression doesn’t even make sense. On the other hand, if your definition of limit is of the second type, then the limit is $1$, since the satisfies both conditions of that definition.

In short, neither opinion is right and neither is wrong: it depends on exactly what the symbol $\lim_{x\to\infty}f(x)=L$ means.

Title: What is $\lim_{x \to \infty} \frac{\sin x}{\sin x}?$

Body:
As the title suggests, I am interested in computing $$\lim_{x \to \infty} \frac{\sin x}{\sin x}.$$ Some of my friends say that the limit is $1$ because the domain of function is $\mathbb{R} \setminus \{n \pi : n \text{ is an integer}\},$ whereas others say that this limit is undefined.

Which opinion is correct?


问:
Maximum of continuous functions.
Let f be a continuous function and non decreasing , can we write
$$f(\max(a,b))=\max(f(a),f(b))$$.

答:
If a function is non-decreasing then since $a\leq max(a,b)$ and $b\leq max(a,b)$, we have $f(a) \leq f(max(a,b))$ and $f(b)\leq f(max(a,b))$. So $max(f(a) ,f(b))\leq f(max(a,b))$.

The other side of the inequality always holds for any function you don't even need the assumption of non-decreasing since $max(a,b)\in ${a,b} so $f(max(a,b))\in ${f(a),f(b)}, and therefore $f(max(a,b)) \leq max(f(a),f(b))$ for any function.

With both inequalities we get $max(f(a) ,f(b))= f(max(a,b))$.

Title: Maximum of continuous functions.


答:
Given that $f(x)$ is non-decreasing, then for $c_1\le c_2$, we have $f(c_1)\le f(c_2)$. Hence $$f(\max(a,b))=\begin{cases}
f(a), & a\ge b\\
f(b), & a<b
\end{cases}$$
Note that $$\begin{cases}
f(a)\ge f(b), & a\ge b\\
f(b)\ge f(a), & a<b
\end{cases}$$ So we conclude that
$$f(\max(a,b))=\begin{cases}
\max(f(a),f(b)), & a\ge b\\
\max(f(a),f(b)), & a<b
\end{cases}$$
and hence $f(\max(a,b))=\max(f(a),f(b))$.

Title: Maximum of continuous functions.

Body:
Let f be a continuous function and non decreasing , can we write
$$f(\max(a,b))=\max(f(a),f(b))$$.


问:
$U\subset\mathbb R$ is open, then $\forall x\in U(\inf\{a:(a,x]\subset U\},\sup\{b:[x,b)\subset U\}) \subset U$

If $U\subset\mathbb R$ is open, $x\in U$, then $$I_x = (\inf\{a:(a,x]\subset U\},\sup\{b:[x,b)\subset U\}) \in U$$

My work (1):

Let $a_x = \inf\{a:(a,x]\subset U\}$ and $b_x = \sup\{b:[x,b)\subset U\}$. Let $y\in I_x$, so $a_x<y<b_x$. If $y=x$, we are done, so let $y\neq x$. Assume $y < x$. In this case, $x > y > a_x$ proves useful. Now if $a_x \in \{a:(a,x]\subset U\}$ then $(a_x,x] \subset U$ and so $y\in U$. What happens if $a_x \notin \{a:(a,x]\subset U\}$?

Similarly, assume $x<y$. We have $x<y<b_x$. If $b_x\in\{b:[x,b)\subset U\}$ then $[x,b_x)\subset U$ and $y\in U$. What happens if $b_x\notin\{b:[x,b)\subset U\}$?

My work (2):

Let $a_x = \inf\{a:(a,x]\subset U\}$ and $b_x = \sup\{b:[x,b)\subset U\}$. Let $y\in I_x$, so $a_x<y<b_x$. If $y=x$, we are done, so let $y\neq x$. Assume $y < x$. In this case, $x > y > a_x$ proves useful. There exists some $a$ such that $a_x<a<y<x$ such that $a\in \{a:(a,x]\subset U\}$. This is because if such an $a$ doesn't exist, then $y$ will be a lower bound for $\{a:(a,x]\subset U\}$, which contradicts the fact that $a_x$ is its infimum. Now $(a,x]\subset U \implies y\in U$.

Similarly for the case $x<y$.

While (2) seems fine, I am unable to complete my thoughts in (1).

Thank you.

答: Pick your favorite $x \in U$. Since we know that open sets in $\mathbb{R}$ are a countable union of disjoin open intervals, this $x$ must fall in some interval $(m,n)$. Remains to show that $inf\{...\} = m$. Let $\epsilon>0$. Suppose $inf\{...\} = m - \epsilon$, then contradiction because that means $m\in U$. Suppose $inf\{...\} = m + \epsilon$, then contradiction because there exists an interval starting from $m+\epsilon/2$ that is the candidate for being the inf. Therefore, $inf\{...\} = m$ and a symmetric argument shows $sup\{...\} = n$. And clearly, $(m,n)\subset U$.


问:
Integral of $\sqrt{4-x^2}$ by Riemann sum
I was trying to compute $\int_{0}^{2}\sqrt{4-x^2}\,dx$ by Riemann sum like this

$$\sum_{i=1}^{n}\sqrt{4-x_i^2}\,\Delta x_i$$

where $x_i = \frac{2i}{n}$ and $\Delta x_i = \frac{2}{n}$, but I couldn't get something useful for the answer. I think it should be with another kind of partition, but I don't know which kind of partition to use.

答:
The right Riemann sum using the partition with points $x_k = 2 \sin \frac{\pi k}{2n}\,\,(k = 0,1,2,\ldots, n)$ is

$$\tag{*}S_n = \sum_{k=1}^n \sqrt{4 - 4  \sin^2 \frac{\pi k}{2n}}\left(2  \sin \frac{\pi k}{2n}- 2\sin \frac{\pi (k-1)}{2n}\right)\\ = 4\sum_{k=1}^n \cos \frac{\pi k}{2n}\left(  \sin \frac{\pi k}{2n}- \sin \frac{\pi k}{2n}\cos \frac{\pi}{2n}+ \sin \frac{\pi }{2n}\cos \frac{\pi k}{2n}\right)\\ = 4\left(1 -\cos\frac{\pi}{2n} \right)\sum_{k=1}^n \cos \frac{\pi k}{2n}\sin \frac{\pi k}{2n} + 4 \sin \frac{\pi}{2n}\sum_{k=1}^n \cos^2 \frac{\pi k}{2n}$$

Note that

$$\sum_{k=1}^n \cos \frac{\pi k}{2n}\sin \frac{\pi k}{2n}= \frac{1}{2}\sum_{k=1}^n\sin \frac{2\pi k}{2n} =\frac{1}{2}\sum_{k=1}^n\sin \frac{\pi k}{n},\\\sum_{k=1}^n \cos^2 \frac{\pi k}{2n}= \sum_{k=1}^n \left(\frac{1}{2} - \frac{1}{2} \cos \frac{2\pi k}{2n} \right) = \frac{n}{2}- \frac{1}{2}\sum_{k=1}^n\cos \frac{\pi k}{n}$$

Substituting into (*), we get

$$S_n = \underbrace{2\left(1 -\cos\frac{\pi}{2n} \right)\sum_{k=1}^n \sin \frac{\pi k}{n}}_{A_n} + \underbrace{2n \sin \frac{\pi}{2n}}_{B_n}- \underbrace{\sin \frac{\pi}{2n}\sum_{k=1}^n \cos\frac{\pi k}{n}}_{C_n} $$

It is shown below that $A_n - C_n \to 0$ as $n \to \infty$, and, therefore,

$$\int_0^2 \sqrt{4-x^2}\, dx= \lim_{n \to \infty}S_n = \lim_{n \to \infty}2n \sin \frac{\pi}{2n} = \lim_{n \to \infty}\pi \frac{\sin \frac{\pi}{2n}}{\frac{\pi}{2n}} = \pi$$

Limits of $A_n$ and $C_n$:

We have, since $\frac{1-\cos x}{\sin x} \to 0$ as $x \to 0$,
$$A_n = 2\left(1 -\cos\frac{\pi}{2n} \right)\sum_{k=1}^n \sin \frac{\pi k}{n}= 2 \left(1 -\cos\frac{\pi}{2n} \right) \frac{\sin \frac{n}{2}\frac{\pi}{n}\sin \frac{n+1}{2} \frac{\pi}{n}}{\sin \frac{\pi}{2n}}\\ = 2 \frac{1 -\cos\frac{\pi}{2n}  }{\sin \frac{\pi}{2n}}\sin \left[\left(1 + \frac{1}{n}\right)\frac{\pi}{2}\right]\underset{n\to \infty}\longrightarrow 0$$

Also,

$$C_n = \sin\frac{\pi}{2n} \sum_{k=1}^n \cos \frac{\pi k}{n} =  \sin\frac{\pi}{2n}\frac{\sin \frac{n}{2}\frac{\pi}{n}\cos \frac{n+1}{2} \frac{\pi}{n}}{\sin \frac{\pi}{2n}}\\ = \cos \left[\left(1+\frac{1}{n} \right)\frac{\pi}{2}\right]\underset{n\to \infty}\longrightarrow 0$$


问:
A surjective ring homomorphism which is not an isomorphism

Give an example of a ring $R$ and a surjective homomorphism $R \to R$ that is not an isomorphism.

Here is my approach. I know a lemma that says that if $R$ is a Noetherian ring, and $f: R \to R$ is a surjective homomorphism, then $f$ is an isomorphism. So the ring $R$ above must be non-Noetherian, which is hard(?) to find, as 'most' rings are Noetherian (or so I've been told in lectures). One example of a non-Noetherian ring that I know is $K[X,Y]/\langle X,XY, XY^2, XY^3, \dots \rangle$, and from this, I was trying to construct such a homomorphism. But I'm stuck; I can see no intuition, and no systematic way to do this except for guessing. Does anyone have a nice solution to this?

Note: I've done some searching, and it seems that upon learning higher mathematics, this is a standard problem. For example, When is a ring homomorphism $A \to A$ surjective but not injective. has a solution which uses a lot of mathematics that I am unfamiliar with. This problem is from a basic undergraduate algebra course, and I'm trying to work out the answer with only what I know.

Edit: The example of a 'non-noetherian' ring I give above is wrong.

答: Let $R=\mathbb{Z}[x_1,x_2,\ldots,x_n,\ldots]$. Consider the map induced by sending $x_1$ to $0\in R$, and sending $x_{k+1}$ to $x_k$ for $k=2,3,4,\ldots$. Verify this gives a surjective ring homomorphism, but that it is not injective (hence not an isomorphism).


问:
Theorem about the Moore Plane
I'm going through Topology Through Inquiry by Francis Su and Michael Starbird, and ran into the following exercise:

3.10.3: In $\mathbb{H}_{bub}$, let $A$ be a countable set on the $x$-axis and let $z$ be a point on the $x$-axis not in $A$. Show that there exist disjoint open sets $U$ and $V$ such that $A \subseteq U$ and $z \in V$.

Where $\mathbb{H}_{bub}$ is the "Upper Half-Plane with the Sticky Bubble Topology", aka the Moore Plane.

I'm confused why this is true because if we take $A = \mathbb{Q}$ and $z = \pi$, what possible open sets could satisfy 3.10.3? Any radius chosen for the open set containing $\pi$ will be too large to not intersect with a rational arbitrarily close to $\pi$ given some other radius.

The exercise also asks whether or not the countability requirement on $A$ is necessary, but I feel like this is false with any dense set $A \in \mathbb{R}$ where $A \neq \mathbb{R}$.

I'd love some guidance on this exercise but above all I'd like to understand why my counterexample is wrong.

Edit: as pointed out in the comments, it's clear to me now why my counterexample is correct. We can pick some $r_\pi$ for the tangent open ball containing $\pi$, and for any rational $q$, we can find a radius that doesn't intersect with that open ball. Now the question is why wouldn't this process work for uncountable sets?

答:
Let $A$ be any subset of the $x$-axis, and let $z$ be any point of the $x$-axis not in $A$. You can indeed separate $A$ and $z$ by disjoint open sets by first choosing a basic tangent disk nbhd of $z$ and then choosing for each $x\in A$ a sufficiently small tangent disk nbhd; there is no need to restrict $A$ to countable sets.

The result also follows from the fact that the space is Tikhonov: every subset of the $x$-axis is closed, so in particular $A$ is closed, and there is a continuous $f:\Bbb H_{\text{bub}}\to[0,1]$ such that $f(z)=0$ and $f[A]=\{1\}$.

Title: Theorem about the Moore Plane


问:
Prove that X is compact if and only if X is a finite set.
Let X be a set of real numbers such that for each a∈X, a is an isolated point in X. Prove that X is  compact if and only if X is a finite set. (Hint: see the  proof of theBolzano-Weierstrass Theorem)</p>

答: Each $a$ is isolated which means that there is a neighborhood $N_{a}$ so that $N_{a}\cap X$ = {a}. If X is compact these $N_{a}$ form a cover of $X$ with no proper subcover, therefore there must be finitely many such $N_{a}$ and hence finitely many $a$, so $X$ is finite. If X is finite the set of distinct sets over X is finite call the totality of such sets J, any open cover of $X$ can find a finite subcover by discarding any repeated sets so that all sets are distinct, then there can be no more than J sets in this subcover, so every open cover has a finite subcover and X is compact.


问:
For which hypergraphs $H=(V,E)$ does there exist a linear order $<$ of $V$ such that $\small\forall X,Y\in E[X\subset Y\implies \max(X)<\max(Y)]$?
I'm trying to characterise hypergraphs $H=(V,E)$ such that there exists a linear order $<$ of $V$ which satisfies $\small\forall X,Y\in E[X\subset Y\implies \max(X)<\max(Y)]$?. Are there a few 'elementary' criteria on $H$ that I can use to determine when such an order exists?

I'm thinking maybe I could transform the problem into determining when a certain poset is ranked.

答:
Assuming that $E$ is a family of nonempty finite subsets of $V$, a necessary and sufficient condition for the existence of such a linear order is that there is no nonempty finite set $W\subseteq V$ such that, for each vertex $w\in W$, there are hyperedges $X,Y\in E$ such that $w\in X\subsetneqq Y\subseteq W$.

The necessity is obvious; if $W$ is such a set, then for any linear order there are hyperedges $X,Y\in E$ with $X\subsetneqq Y$ and $\max(X)=\max(Y)=\max(W)$.

In proving the sufficiency, we may assume that $V$ is finite; the general case will then follow by the usual sort of compactness argument.

Suppose that $V$ is an $n$-element set, and that every nonempty subset $W\subseteq V$ contains an element $w$ such that $\{X\in E:w\in X\subseteq W\}$ is an antichain. If $i\in\{1,\dots,n\}$ and if $w_1,\dots,w_{i-1}$ have already been chosen, let $W_i=V\setminus\{w_1,\dots,w_{i-1}\}$, and choose $w_i\in W_i$ so that $\{X\in E:w\in X\subseteq W_i\}$ is an antichain. Finally, define the linear order $\lt$ so that $w_1\gt\cdots\gt w_n$, i.e., $w_i=\max(W_i)$.


问:
What's the distance between two circumcenters
This problem was came to a Facebook post of mathigon ......the problem seems trickier than I expected...can you help me to find the distance??

答:
The area of the lens formed by two overlapping circle of radius $r$ and at distance $d$ apart is
given by https://mathworld.wolfram.com/Lens.html

$$Area= \pi r^2-2r^2 \tan^{-1}\frac d{\sqrt{4r^2 -d^2}}-\frac12d \sqrt{4r^2 -d^2}
$$
Set $Area =\frac\pi2$ and $r=1$ to obtain numerically $d= 0.8079$.


问:
How do you solve a question like
Suppose that a, b, and c are real numbers, and $P(x) = ax^2 + bx + c$. If $P(x + 2) + P(x + 1) − P(x) = 2x^2 + 15x + 20$, what is the value of $P(10)$?

This is a grade 8 Canadian Math Challenger's question (Regional, 2014 Round 1, Q24). Students get about 90 seconds to solve each question in this round. I tried different values of X but it seems that there has to be a trick to solve something like this in 90 or so seconds. This question can be found on page #525 of this PDF egbc.ca/getmedia/5cfa019a-f50b-4a12-97b0-e8ebf020ce71/

Title: How do you solve a question like

Body:
$$P(x) = ax^2 + bx + c$$
$$P(x+1) = a(x+1)^2 + b(x+1) + c=ax^2+(2a+b)x+(a+b+c)$$
$$P(x+2) = a(x+2)^2 + b(x+2) + c=ax^2+(4a+b)x+(4a+2b+c)$$
$$P(x+2)+P(x+1)-P(x)=ax^2+(6a+b)x+(5a+3b+c)=2x^2 +15x+20$$

Thus, comparing coefficients, we see that:
$$a=2$$
$$12+b=15; b=3$$
$$10+9+c=20; c=1$$

Thus, $P(x)=2x^2+3x+1$

$P(10)=231$


答: It is simply the inclusion-exclusion principle... Choosing $1$ from each of $k$ groups of $n$ is equivalent to choosing $k$ from all $k$ groups (combined) minus choosing $k$ from only $(k-1)$ groups plus choosing $k$ from only $(k-2)$ groups ...

Title: Find $^{505}C_5-5\cdot{^{404}C_5}+10\cdot{^{303}C_5}-10\cdot{^{202}C_5}+5\cdot{^{101}C_5}-1$

Body:
Find the sum

$$^{505}C_5-5\cdot{^{404}C_5}+10\cdot{^{303}C_5}-10\cdot{^{202}C_5}+5\cdot{^{101}C_5}-1$$

Critiques and other answers are always welcome!


问:
Exhaustive events and the law of total probability
Suppose $X$, $Y$, and $Z$ are exhaustive events with $Y$ and $Z$ being mutually exclusive. Can $Y$ and $Z$ be treated as a partition? Also, as a follow-up, can the law of total probability be applied to $X$ this way: $P(^cX) = P(^cX \cap Y) + P(^cX\cap Z)$, where $^cX$ is the complement of $X$?

答:
No. Consider P(X) = .4, P(Y) = .2, P(Z) = .2, with X,Y,Z disjoint and P(X) + P(Y) + P(Z) = 1 (exhaustive). Then Y and Z do not partition our probability space.

For the second part, since Y and Z do not partition our probability space, no, we cannot say that. If Y and Z did partition our probability space, then yes, we can say that.


问:
Is there any function that satisfies the limit exists?
Is there any function that satisfies the limit below exists ?

$$
\lim_{x \to \infty}\frac{f(x+1)}{f(x)}=0
$$

Body:
Try $f(x) = x^{-x}$

$$\lim_{x\to\infty}\frac{(x+1)^{-(x+1)}}{x^{-x}} = \lim_{x\to\infty}\frac{1}{x+1}\left(1+\frac{1}{x}\right)^{-x} = 0\cdot e^{-1} = 0$$


问:
Example of sequence converging in $d_{l^\infty}$ but not in $d_{l^1}$.
I'll denote by $X$ the space of real sequences $(a_n)$ such that $\sum |a_n|$ converges. Let $d_{l^1}$ be the metric
$$
d_{l^1}((a_n),(b_n))=\sum|a_n-b_n|
$$
and $d_{l^\infty}$ be the metric
$$
d_{l^\infty}((a_n),(b_n))=\sup\{|a_n-b_n|\}.
$$

It seems clear to me that if a sequence (of sequences) $(x^{(n)})$ in $X$ converges with respect to $d_{l^1}$, then it must also converge in $d_{l^\infty}$.

However, I believe I recall that $d_{l^1}$ and $d_{l^\infty}$ are not equivalent, so I'm trying to find an example of a sequence in $X$ which converges in $d_{l^\infty}$ but does not converge in $d_{l^1}$. Does anyone have an example? Cheers.

Body: $(1/n,1/n,\ldots,1/n,0,0,\ldots)$ with $1/n$ in $n$ spots.


答:
Plugging numbers which have a P(0) somewhere in the equation may help you go a little faster to get your linear equations. Like so:

Plugging in $x=-1$:

$$2b+c = 7$$

Plugging in $x=0$:

$$5a+3b+c=20$$

Plugging in $x=-2$:

$$ -3a + b +c = -2$$

a=2, b=3, c=1

Then $P(10) = 231.$

Title: How do you solve a question like

Body:
Suppose that a, b, and c are real numbers, and $P(x) = ax^2 + bx + c$. If $P(x + 2) + P(x + 1) − P(x) = 2x^2 + 15x + 20$, what is the value of $P(10)$?

This is a grade 8 Canadian Math Challenger's question (Regional, 2014 Round 1, Q24). Students get about 90 seconds to solve each question in this round. I tried different values of X but it seems that there has to be a trick to solve something like this in 90 or so seconds. This question can be found on page #525 of this PDF egbc.ca/getmedia/5cfa019a-f50b-4a12-97b0-e8ebf020ce71/