问:
Parentheses in the set $\Omega \subseteq\mathbf R\times \mathbf R^n$?

Consider a system of differential equations of first order with $n$ equations
$$
\mathbf x'(t)=\mathbf f(t,\mathbf x)
$$
and the initial condition $\mathbf x(t_0)=\mathbf x_0$.
Here $\mathbf f(t,\mathbf x)$ is a continuous function from some open, connected set $\Omega \subseteq\mathbf R\times \mathbf R^n$ to $\mathbf R^n$.

Does  $\Omega \subseteq\mathbf R\times \mathbf R^n$ mean

$$
(\Omega \subseteq\mathbf R )\times \mathbf R^n  \tag 1
$$
Or
$$
\Omega \subseteq(\mathbf R\times \mathbf R^n) \tag 2
$$
?

Update:

If $\Omega \subseteq(\mathbf R\times \mathbf R^n)$ is correct, isn't $\Omega \subseteq\mathbf R\times \mathbf R^n$ ambiguous?

答:
I see no reason why it shouldn't mean $\Omega\subseteq(\Bbb R\times \Bbb R^n)$, seeing as:

this is the standard way to read $X\subseteq A\times B$;

it makes sense in the context;

the other candidate is highly non-standard (not to say unheard of), and I daresay any decent writer would spend his time articulating "$\Omega\times \Bbb R^n$ with $\Omega\subseteq \Bbb R$".

For the record, keep in mind that modern mathematical language is modelled over (present-day) Indo-European languages. To the best of my knowledge these languages organise their proposition around the system subject-predicate-complement. This means that grammatically there is a very strict separation between the verb, which is a word that can only be used as the part of the sentence ("predicate") expressing the action, and the sustantives, which are the objects that may do ("subject") or receive / partake of ("complements") the action. In $X\subseteq A\times B$ this machinery comes into play because "$\subseteq$" replaces the predicate "being subset of", whereas "$X$" and "$A\times B$" indicate objects which are respectively subject and complement of the sentence. Therefore "$\subseteq$" is a verb, and "$X$" and "$A\times B$" are sustantives: this is effectively their indivisible nature as grammatical entities. An interpretation such as your $(X\subseteq A)\times B$ is not sound in this context because it mixes those things: in fact it cannot be interpreted as a standalone sentence or group of sentences, but I don't want to go into further detail.


问:
Bound for integral-sum estimate using the total variation
Let $f:[0,2\pi]\rightarrow \mathbb{R}$ be a continuous function with total variation $V$. Let $\{\vartheta_k\}$ denote the equidistant points $\vartheta_k = \frac{2\pi}{2(n+1)}(2k-1)$ where $n$ is some positive integer. I want to show that
$$\left|\frac{1}{n+1}\sum_{k=1}^{n+1}f(\vartheta_k)-\frac{1}{2\pi}\int_{0}^{2\pi}f(\theta)d\theta\right|\leq \frac{V}{2(n+1)}$$
but I can only get $\leq \frac{V}{n+1}$. The book that I'm reading is suggesting that the above inequality should be true so I am wondering if this is so.

My estimate: Let $h = \frac{2\pi}{2(n+1)}$. Then since $2h(n+1) = 2\pi$
$$\left|\frac{1}{n+1}\sum_{k=1}^{n+1}f(\vartheta_k)-\int_{0}^{2\pi}f(\theta)d\theta\right| = \left|\frac{1}{n+1}\sum_{k=1}^{n+1}\frac{1}{2h}\int_{\vartheta_k-h}^{\vartheta_k+h}f(\vartheta_k)d\theta-\frac{1}{2\pi}\int_{0}^{2\pi}f(\theta)d\theta\right| \\ = \left|\frac{1}{2\pi}\sum_{k=1}^{n+1}\int_{\vartheta_k-h}^{\vartheta_k+h}f(\vartheta_k)-f(\theta)d\theta\right| \leq \frac{1}{2\pi}\sum_{k=1}^{n+1}\int_{\vartheta_k-h}^{\vartheta_k+h}|f(\vartheta_k)-f(\theta)|d\theta \\ \leq \frac{1}{2\pi}\sum_{k=1}^{n+1}2h|f(\vartheta_k)-f(\tilde\vartheta_k)| = \frac{1}{n+1}\sum_{k=1}^{n+1}|f(\vartheta_k)-f(\tilde\vartheta_k)|\leq \frac{V}{n+1}$$
where I applied the mean value theorem of integral calculus to get the points $\tilde\vartheta_k$. Now I know that the last inequality is probably strict but I don't see another method which would give the $\frac{1}{2}$ in front.

答: Look up Koksma's inequality. See if it gives the factor of 2. If so, look at its proof.</p>

ContentLicense: CC BY-SA 4.0


问:
Proving two sets are equal if the size of the intersection and union are equal
I want to prove the following:
if $|X \cap Y| = |X \cup Y|$, then $X=Y$. But not sure how to do this mathematically. I am using this prove the property of indiscernibles. But not sure where to start with this when the size of the intersection/union is thrown into the mix. Would it be the same as proving $X \cap Y = X \cup Y$ or are there edge cases I'm missing?

Would the right step be to prove the inverse using DeMorgan's law?

答:
Hint:
$$X \cup Y = (X \cap Y) \sqcup (X \setminus Y) \sqcup (Y \setminus X),$$
where $\sqcup$ indicates disjoint union.
So
$$|X \cup Y| = |X \cap Y| + |X \setminus Y| + |Y \setminus X|.$$
Now what can you conclude about the last two terms?


问:
Proving $D(G) ≤ w(1 + (t − 1) \log w) $
Given,$$D(G) ≤ \exp(G)(1 + \log(|G|/\exp(G))),$$ where $D(G)$ is Davenport's constant, for any finite Abelian group $G$. Let the cyclic group of the order $w$  is denoted by  $C_w$.

If $$  g_1, \dots, g_k \in  C_w^t,    k ≥ tw \log(w), G = C_w^t,$$ and $$\exp(G) = w,$$ then how the below inequality holds?

$$D(G) ≤ w(1 + (t − 1) \log w) ≤ tw \log w$$

Because, $\exp(G)(1 + \log(|G|/\exp(G))) = w (1+\log (k/w)) > w (1+\log (tw \log(w)/w)) = w (1+\log (t \log(w))) =  w (1+\log\log(w^t))$

Source:

答:
We have $G$ is $t$-th Cartesian power of $C_w$, so $|G|=w^t$, $\exp (G)=w$, therefore

$$\exp(G)(1 + \log(|G|/\exp(G))) = w (1+\log (w^{t-1}))= w (1+(t-1)\log w)\le tw\log w,$$

for $w>2$, because in this case $t\log w\ge\log w\ge 1$.

Title: Proving $D(G) ≤ w(1 + (t − 1) \log w) $


问:
Use a calculator to approximate $\int_0^{7\pi} \sqrt{5-4\cos(t)}dt$ to four decimal places
Use a calculator to approximate $\int_0^{7\pi} \sqrt{5-4\cos(t)}dt$ to four decimal places

Can somebody explain what this means? Do they want me to use a series expansion or something?? What are some ways this question could be interpreted and could I have a bit help getting through it??

Maybe I'm suppose to use Simpsons rule? I feel like if that were so they would have explicitly said so..

Thank you for your time.

答:
Actually you don't need hundreds of intervals to compute
$$I=\int_0^{7 \pi } \sqrt{5-4 \cos (t)} \, dt$$
Define $f(t)= \sqrt{5-4 \cos (t)}$
$$I_n=\frac{7 \pi  \left(\sum \limits_{k=1}^{n-1} f\left(\frac{\pi  k}{n}\right)+\frac{1}{2} (f(0)+f(\pi ))\right)}{n}$$
for $n=7$ I got $|I-I_7|<\frac{1}{10^4}$


问:
Would adding a symmetric positive semi-definite matrix to a non-symmetric positive definite matrix increase the spectral radius?
Let $A$ be a non-symmetric positive definite matrix with the spectral radius $\rho(A) = \max_i|\lambda_i|$ (note $\lambda_i$ can be complex). By positive definite, I mean $x^\top A x > 0$ for all possible $x$.

Let $B$ be a symmetric PSD matrix with the same shape as $A$.

My question is: would the spectral radius $\rho(A+B)$ be larger than (or equal to) $\rho(A)$? This looks intuitive but I have difficulty in proving it, any hints?

For general case, I have found a counter-example. But what if I know some structures of A. Say A has the following form:
$$A = \begin{bmatrix} A_{11} & A_{12} \\ -A_{12}^\top & A_{22}\end{bmatrix}$$
where $A_{11}$ and $A_{22}$ are symmetric positive definite.

答:
I didn't expect this to be true because to increase $\rho$ you need to rule out rotations, but your special structure doesn't rule them out (ie, skew-symmetric matrix + identity is a rotation)

It's easy to generate counter-examples programmatically. For instance the following two matrices form a counter-example:

$$
A=
\left(
\begin{array}{cccc}
 2.01 & 0.01 & -1.00421 & -0.475834 \\
 0.01 & 2.01 & 1.84741 & -1.73169 \\
 1.00421 & -1.84741 & 2.01 & 0.01 \\
 0.475834 & 1.73169 & 0.01 & 2.01 \\
\end{array}
\right)
$$

$$
B=
\left(
\begin{array}{cccc}
 \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
 \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
 \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
 \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
\end{array}
\right)
$$

Generated using following Mathematica code

```<code>rho[mat_] := Max[Abs[Eigenvalues[mat]]];
try[d_] := (
   B = ConstantArray[1, {d, d}]/2;
   ii = IdentityMatrix[d/2];
   AB = RandomReal[{-2, 2}, {d/2, d/2}];
   A = ArrayFlatten[{{2 ii + .01, AB}, {-AB\[Transpose], 2 ii + .01}}];
   result = rho[A + B] - rho[A];
   If[result < 0,
    Print[A];
    Print[B];
    ];
   result
   );
Table[try[4], {20}]
</code>```

Body:
I didn't expect this to be true because to increase $\rho$ you need to rule out rotations, but your special structure doesn't rule them out (ie, skew-symmetric matrix + identity is a rotation)

It's easy to generate counter-examples programmatically. For instance the following two matrices form a counter-example:

$$
A=
\left(
\begin{array}{cccc}
 2.01 & 0.01 & -1.00421 & -0.475834 \\
 0.01 & 2.01 & 1.84741 & -1.73169 \\
 1.00421 & -1.84741 & 2.01 & 0.01 \\
 0.475834 & 1.73169 & 0.01 & 2.01 \\
\end{array}
\right)
$$

$$
B=
\left(
\begin{array}{cccc}
 \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
 \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
 \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
 \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
\end{array}
\right)
$$

Generated using following Mathematica code

rho[mat_] := Max[Abs[Eigenvalues[mat]]];
try[d_] := (
   B = ConstantArray[1, {d, d}]/2;
   ii = IdentityMatrix[d/2];
   AB = RandomReal[{-2, 2}, {d/2, d/2}];
   A = ArrayFlatten[{{2 ii + .01, AB}, {-AB\[Transpose], 2 ii + .01}}];
   result = rho[A + B] - rho[A];
   If[result < 0,
    Print[A];
    Print[B];
    ];
   result
   );
Table[try[4], {20}]


问:
Factorize the polynomial $f(x)=4x^2-4x+8$ over $\mathbb{Z}[x]$, $\mathbb{Q}[x]$ and $\mathbb{Z}_{11}[x]$

Factorize the polynomial $f(x)=4x^2-4x+8$ in $\mathbb{Z}[x]$, $\mathbb{Q}[x]$ and $\mathbb{Z}_{11}[x]$

My approach

First we consider in $\mathbb{Z}[x]$.
Let $$f(x)=4x^2-4x+8=2*2*(x^2-x+2)$$ I claim that $2$ and $(x^2-x+2)$ are ireducible,the fisrt is irreduzible since $\mathbb{Z}$ is a UFD and PID and then the primes numbers are the irreducible factors, the second comes from , since  the discriminant is $\Delta<0$ and $(x^2-x+2)$ have degree $2$ then have two complex roots and therefore are irreducible in $\mathbb{R}[x]$ therefore since $\mathbb{Z}[x]\subset \mathbb{R}[x]$ then $(x^2-x+2)$ is irreducible and therefore:
$$f(x)=2*2*(x^2-x+2)$$

For the second case in $\mathbb{Q}[x]$ is the same argue since  $\mathbb{Q}[x] \subset  \mathbb{R}[x]$ and $\Delta<0$ implies that $(x^2-x+2)$ is irreducible in  $\mathbb{Q}[x]$  and therefore
$$f(x)=2*2*(x^2-x+2)$$

In the thrid case in $\mathbb{Z}_{11}[x]$ notice that there exists $c_1=\overline{5}$ and $c_2=\overline{7}$ such that $f(c_1)=f(c_2)=0$ and that implies that $c_1,c_2$ are roots of the polynomial and therefore
$$(x)=2*2*(x^2-x+2)=2*2*(x-5)(x-7)$$ and note that
$$(x-5)(x-7)=x^2-12x+35=x^2+10x+2=x^2-x+2$$
Is my  approach right?

Someone can give me an other prove of the fact that $x^2-x+1$ are irreducible in $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$

答: Your tittle question has the following answer. First, $f(x)=4(x+6)(x+4)$ is the factorisation over the finite field $\Bbb F_{11}$, as you said. Secondly, $x^2-x+2$ is irreducible over $\Bbb Z$ by the rational root theorem, and as a monic polynomial then also over $\Bbb Q$.


问:
$xy= z^2$ and $x, y$ are individual squares
We know that concatenation of $xy = z^2$ for $x = 4^2 = 16$ and $y = 3^2 = 9$. Here $169 = 13^2 \implies z = 13$. Now my question is how to prove this is the first such set of positive integers $(x, y, z) = (16, 9, 13)$ for concatenation of $xy = z^2$? Or disprove if it is not such first set of pairs? Thanks in advance.</p>

答:
I assume that your idea of "first" is based on the last numbre, i.e. if $(a,b,c)$ is a triple of positive integers where the concatenation of $a^2$ and $b^2$ gives $c^2$, and $(d,e,f) $ is another such triple, then $(a,b,c)$ comes before $(d,e,f)$ if $c<f$. Note that $c=f$ mmight be possible even with different triples.

We see by inspection that $(2,3,7)$ is a triple satisfying your criteria.
Any triple $(a,b,c)$ coming before that must have $c\le 6$. As one readily checks, $36,25,16,9,4,1$ cannot be written as concatenation of positive squares.

A quick computer search reveals as first solutions:
$$ (2,3,7),
(4,3,13),
(6,1,19),
(1,15,35),
(12,2,38),
(4,9,41),
(18,3,57),
(2,15,65),
(2,30,70),\\
(1,75,125),
(4,30,130),
(6,10,190),
(2,45,205),
(7,27,223),
(80,3,253),
(9,15,285),\ldots$$
Note that each solution $(a,b,c)$ gives rise to $(a,10b,10c)$.

ContentLicense: CC BY-SA 3.0


问:
Evaluating $ \lim_{n\to\infty}\sum_{i=0}^n\frac{n}{n^2+i^2} $
A few days ago I confronted this limit question, and all my attempts to solve it failed somehow.
$$
\lim_{n\to\infty}\sum_{i=0}^n\frac{1}{n^2+i^2}
$$
I feel I should relate it to $\arctan(n)$. I tried integration and derivation to reach my point, it went well at first but eventually nowhere.

Any kind of help and enlightenment would be appreciated.

答:
$$
\sum_{i=0}^n\frac{1}{n^2+i^2} = \frac 1 n \sum_{i=0}^n \frac 1 {1 + \left( \frac i n \right)^2} \cdot \frac 1 n = \frac 1 n \sum_{i=0}^n \arctan'\left( \tfrac i n \right)\cdot \tfrac 1 n
$$

That last sum is a Riemann sum that approximates $\displaystyle\int_0^1 \arctan'(x)\,dx = \arctan 1 - \arctan 0 = \frac \pi 4,$ but outside the sum there is a factor of $1/n$ that approaches $0,$ so ultimately the limit is $0.$


问:
Relation \ Equivalence relation
I have a problem with this task. If anyone had a similar problem it would help me.

The task is:

In the set $S={(-2,-1,0,1,2,3)}$ the relation is defined $\rho$ with
$(\forall t,m \in S) t \rho m \iff 2tm\ge t+m$. Examine the properties of the relation $\rho$.
If an equivalence relation is given find all classes.

I tried this:

Reflexivity:

$(\forall t\in S) t\rho t$

$2t^{2}\ge2t$
Reflexivity is satisfied

Symmetric:

$(\forall t,m\in S) t\rho m \iff m\rho t\\2tm\ge t+m \iff t+m \le2tm$

The symmetry is satisfied

And now I don’t know how to prove transitivity ?

Thanks in advance !

答:
To show transitivity, you need to show that if $t,m,n\in S$, $t\rho m$, and $m\rho n$, then $t\rho n$. In other words, you must show that if $2tm\ge t+m$ and $2mn\ge m+n$, then $2tn\ge t+n$. That looks rather messy, so you might first try finding a counterexample to transitivity, but if you do try, you’ll not find one. At that point you could go back and tinker with the algebra to try to prove transitivity, but the small size of $S$ makes a much simpler approach feasible: you can simply go through and check all $6^2=36$ pairs of elements of $S$ to see which ones are related by $\rho$. Better yet, since you’ve already shown that $\rho$ is reflexive and symmetric, you need check only the $15$ pairs $\langle t,m\rangle$ with $t<m$. If you do this, you should get the following table, where the numbers in the first column are possible values of $t$, the numbers across the top are possible values of $m$, and a $+$ in row $t$, column $m$ indicates that $t\rho m$.

$$\begin{array}{r|cc}
&-2&-1&0&1&2&3\\\hline
-2&+&+&+&-&-&-\\
-1&+&+&+&-&-&-\\
0&+&+&+&-&-&-\\
1&-&-&-&+&+&+\\
2&-&-&-&+&+&+\\
3&-&-&-&+&+&+
\end{array}$$

This table shows that $t\rho m$ if and only if either $t\le 0$ and $m\le 0$, or $t>0$ and $m>0$. Now use that information either to show that $\rho$ is transitive or to find a counterexample to transitivity.

Title: Relation \ Equivalence relation


答:
Note that $X\cap Y\subset X\subset X\cup Y$ so if these two sets have the same size, and $X,Y$ are finite, then you must have equality instead of inclusion. Likewise for $Y$.
If they are infinite, by the way, you could have $|X\cap Y|=|X\cup Y|$ while $X\neq Y$, for example choosing $X=Z$ and $Y=N$ would do the trick. However, if $X\cap Y=X\cup Y$, this always works.

Title: Proving two sets are equal if the size of the intersection and union are equal

Body:
I want to prove the following:
if $|X \cap Y| = |X \cup Y|$, then $X=Y$. But not sure how to do this mathematically. I am using this prove the property of indiscernibles. But not sure where to start with this when the size of the intersection/union is thrown into the mix. Would it be the same as proving $X \cap Y = X \cup Y$ or are there edge cases I'm missing?

Would the right step be to prove the inverse using DeMorgan's law?


问:
Interpretation of dP in Radon-Nikodym Theorem
[Radon-Nikodym Theorem]

Let $(\Omega, \Sigma, P)$  be a probability space. Suppose that $(\Omega, \Sigma, \mu)$ be a measure space with $\mu(A)=0$ implies $P(A)=0$, then there exist a function $f:X \rightarrow \mathbb{R}$ such that

\begin{equation}
P(A)=\int_A f d\mu.
\end{equation}

Further, function $f$ is called Radon-Nikodym derivative of $P$ with respect to $\mu$.
(So far I know, why this definition makes sense, because it's similar with derivative in calculus, that is, $g'$ is derivative of $g$ because $g(x)=\int g'(x) dx$).

In calculus, we also know differential, that is, if $y=f(x)$ be a differentiable function, then differential is defined by $dy=f'(x)dx$. Could we interpret $dP$ as a differential like in calculus? How could we interpret $dP$ in Radon-Nikodym context?

答:
The $dP$ symbol appears first in probability in the context of integration, not the Radon-Nikodym theorem. In that setting it is analogous to the $dx$ in Riemann integration: $dx$ is a notational artifact that remains when passing from a Riemann sum to integration and refers to the Lebesgue measure of an interval of length $\Delta x$ as $\Delta x \to 0$. So,
$$
  \sum_{i=1}^n f(x_i)\Delta x_i \ \mbox{ is to } \ \sum_{i=1}^n \sup_{x_i \in A_i}f(x_i) P(A_i)
$$
as
$$
  \int_a^b f(x) dx \ \mbox{ is to } \ \int_{\cup A_i} f dP.
$$
Put another way, a differential like $dx$ can be interpreted as an infinitesimally small length, and the $dP$ can be interpreted as infinitesimally small measure.

In the Radon-Nikodym context, if $\Omega$ has a topological structure and if you have
$$
  P(A)
= \int_A f(x) d\mu(x)
$$
for some $f:\Omega \to \mathbb{R}$ satisfying: $\forall \epsilon > 0$, $\exists$ $\delta > 0$ such that $x,y \in U$ and $P(U) < \delta$ implies $|f(x) - f(y)| < \epsilon$, then for a set $\Delta A$ of small $P$-measure
$$
        P(\Delta A)
=       \int_{\Delta A} f(x) d\mu(x)
\approx \sup_{x \in \Delta A} f(x) \mu(\Delta A)
$$
which, if you take the limit as $P(\Delta A) \to 0$, you can think of as being analogous to $dy = f'(x) dx$.


问:
Is there a way to meaningfully express $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}(x-1)^n}{n^k}$ for $k\neq 1$?
For the following series:
$$\sum_{n=1}^{\infty}\frac{\left(-1\right)^{\left(n+1\right)}\left(x-1\right)^{n}}{n^{k}}$$
other than $k=1$, is there any way to express this series as a function for other values of $k$?

答:
Your series is related to the polylogarithm:

$$\sum_{n=1}^\infty \frac {(-1)^{n+1}(x-1)^n}{n^k}=-\sum_{n=1}^\infty\frac {(1-x)^n}{n^k}=-\operatorname{Li}_k(1-x)$$


问:
Probability and the first uncountable ordinal number
Let's suppose we can put a probability measure on the set of countable ordinals. (which is the same as the first uncountable ordinal). Let's now play a game. I pick a countable ordinal, say $\alpha$. Now you pick one. (Clearly our choices are independent). But $\alpha$ has countably many ordinals less than it and for you the number of choices greater than $\alpha$ are uncountable. So with a large probability, possibly one, your choice is bigger than mine. But the choices were independent!  How to explain? Of course, you could take this as a proof that no such measure exists but intuitively it seems to make sense that such a game could exist.

答: A probability measure is on the set of measurable subsets of the set of outcomes, so which subsets do you want to consider measurable? If it's a discrete probability measure (i.e. all point masses), then only countably many outcomes can have positive measure (that is well known and easy to prove). But let's say you have a one-to-one correspondence between $\mathbb R$ and the set of countable ordinals (that can be done if the continuum hypothesis and the axiom of choice hold). Then say you consider all sets corresponding to Lebesgue-measurable sets to be measurable. In that case, the set of all ordinals less than any particular outcome has measure $0.$


问:
Find the length of the loop of the given parameterized curve: $x=12t-4t^3$ and $y=12t^2$
Find the length of the loop of the given parameterized curve: $x=12t-4t^3$, $y=12t^2$

I'd like to get some views on different ways of solving this problem. Is there an easy way to know how to sketch this curve? Using a computer I saw what the graph looks like and that indeed there is a loop, so I could find the values of $t$ that parameterize that loop, but I'd like to be able to do this without computer aide.

Would eliminating the parameter be a good idea here? I couldn't find a nice way to do that.

Thank you very much for your time.

答:
You want to find the length of the loop in the curve. As the curve is symmetric wrt $y$-axis we can limit to the interval $[0,\sqrt 3]$
$$L=2\int_0^{\sqrt{3}} \sqrt{x'(t)^2+y'(t)^2} \, dt$$
$$x'(t)=12-12 t^2,y'(t)=24 t$$
$$L=2\int_0^{\sqrt{3}} \sqrt{576 t^2+\left(12-12 t^2\right)^2} \, dt=\\=
2\int_0^{\sqrt{3}} \sqrt{144 \left(t^2+1\right)^2}\,dt =24\int_0^{\sqrt{3}} \left(t^2+1\right)dt=24\left[ {\frac{{t^3 }}{3} + t} \right]_0^{\sqrt 3 }=48 \sqrt{3}$$


问:
Evaluating $\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})} x$
I don't know how can this be solved:
$$
\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})} x
$$
I've tried to multiply by $\pi\sqrt{\cos x}$:
$$
\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})}{x\pi\sqrt{\cos x}}
$$

but we have here a hard problem $\lim_{x\to0}(1/x)$.

You think about the limit on the right and on the left but the exercise supposed to be solved by the limit in $0$ and not on the right or the left of it.

答:
Note that for $x\to 0$, $\dfrac {\sin(\pi\sqrt{\cos x})}{\pi\sqrt{\cos x}} \to \dfrac {\sin \pi}{\pi}=0$. The limit for $\dfrac {\sin x}x$ is $1$ for $x \to 0$ only.

Hence we transform the argument of the sine so it tends to $0$, by seeing

$$\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})} x = \lim_{x\to0} \frac{\sin(\pi - \pi\sqrt{\cos x})} x = \lim_{x\to0} \frac{\sin(\pi - \pi\sqrt{\cos x})} {\pi-\pi\sqrt{\cos x}}\cdot \frac {\pi-\pi\sqrt{\cos x}}x$$

Now the first part of the product has limit $1$. Do you know how to calculate the limit of the second part?


问:
Is this equation sum setup right passing the N an i values?
So I have been working on this equation which sums up two summations from i = 1 to N.

This is the equation.

This is the latex equation for it for help in editing if its wrong.

```<code>AL =  \sum_{i=1}^{N}\left ( \sum_{i=1}^{N}max(0, m + d(r_{a},r_{p}) - d(r_{a}, r_{n})) + (-\frac{1}{N}(\sum_{i=1}^{N}L_{i}\cdot log(s_{i}))) \right  )
</code>```

This looks right to me but do I need the N and i on every sum or only the first one?
Thank you

答: I doubt that any compiler would let that by without an error message!  You are using the same variable name, i, for two different variables.  Interpreted rigorously it would mean that the first sum would set each occurance of i 1 to N, then add them.  There would be no "i" for the outer sum.  if you were to write it as [tex]AL =  \sum_{i=1}^{N}\left ( \sum_{j=1}^{N}max(0, m + d(r_{a},r_{p}) - d(r_{a}, r_{n})) + (-\frac{1}{N}(\sum_{j=1}^{N}L_{i}\cdot log(s_{i}))) \right)[/tex] where I have change the "i" for the sums inside the parentheses to "j".  Notice that I can use the same "j" for both sums inside the parentheses because those two sums are done independently of each other.</p>


答:
$$\frac{\sin(\pi\sqrt{\cos x})}x=\frac{\sin(\pi(1-\sqrt{\cos x}))}x.$$

As the argument of the sine tends to $0$, we can bypass the sine and evaluate

$$\pi\lim_{x\to 0}\frac{1-\sqrt{\cos x}}x.$$

Now

$$\frac{1-\sqrt{\cos x}}x=\frac{1-\cos x}{(1+\sqrt{\cos x})x}=\frac{2\sin^2\dfrac x2}{(1+\sqrt{\cos x})x}\to0.$$

Title: Evaluating $\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})} x$

Body:
I don't know how can this be solved:
$$
\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})} x
$$
I've tried to multiply by $\pi\sqrt{\cos x}$:
$$
\lim_{x\to0} \frac{\sin(\pi\sqrt{\cos x})}{x\pi\sqrt{\cos x}}
$$

but we have here a hard problem $\lim_{x\to0}(1/x)$.

You think about the limit on the right and on the left but the exercise supposed to be solved by the limit in $0$ and not on the right or the left of it.


问:
On a passage from 'Graphs and Homomorphisms' concerning Reconstruction Conjecture
For context, I'm reading Graphs and Homomorphisms by Hell and Nesetril. I have some questions / confusions about a passage concerning the reconstruction conjecture (page 16-17 for anyone with the book). The book presents the conjecture:

Conjecture 1.28  Graphs $G$ and $H$ are isomorphic if and only if all their edge-removed subgraphs are isomorphic, i.e., if and only if there exists a bijection $\beta: E(G)\rightarrow E(H)$ such that $G-e \simeq H-\beta(e)$ for all $e\in E(G)$.

and then the passage after that I'm struggling to understand:

Since we are only making statements about isomorphic graphs, we may assume that both $G$ and $H$ have the same vertex set $V$. All graphs in this proof will have vertex set $V$, and we describe them by just giving their edge sets. If the desired bijection $\beta: E(G)\rightarrow E(H)$ exists, we may extend it to associate subsets of $E(G)$ with subsets of $E(H)$ so that for any nonempty set of edges $A$ we have $G-A \simeq H - \beta(A)$. Equivalently, for any proper subset $A\subset E(G)$ we have the graph formed by $A$ isomorphic to the graph formed by $\beta(A)$. This is possible, as can easily be seen by counting. Assume that $A$ consists of $k < |E(G)| = |E(H)| = m$ edges, let $N(G)$ denote the number of copies[1] of $A$ in $G$, and $N_e(G)$ the number of copies of $A$ in $G-e$. Then counting in two ways[2] we obtain $$(m-k)N(G) = \sum_{e}N_e(G) = \sum_e N_e(H) = (m-k)N(H),$$ and hence the number of copies of any particular $A$ is the same in $G$ and $H$ [3].

Here are my questions:

[1] It's not clear to me what 'number of copies' means here. Does it mean the number of subgraphs of $G$ that are isomorphic to $A$? Or is $N(G)$ just an 'indicator' function that says $A$ appears as a subgraph of $G$, with exactly the same labels on the edges (So $N(G)$ would always be $0$ or $1$)? Normally I'd assume the former isomorphic subgraphs interpretation, but that makes no sense for the counting argument that follows.

[2] What is being counted on the left hand side to obtain $(m-k)N(G)$? My first assumption was that for each edge $e\notin A$ (of which there are $m-k$), we track the (one singular??) appearance of $A$ in $G-e$, but this only makes sense if we assume $N(G)$ is the 'indicator that $G$ contains $A$'. But if that is what the notation means, then it's easy to get $N(G)\neq N(H)$ in general. My guess for this is that the notation hasn't been fully explained, and that $N(H)$ actually means the number of copies of $\beta(A)$ in $H$, but I'm not sure?

[3] In the book, the argument ends here. Why is the fact that the number of copies of $A$ (whatever that means in this context) is the same in $G$ and $H$ enough to conclude that $A\simeq \beta(A)$?

Short Version: What's going on in the counting argument here, and why does it suffice to show $A\simeq \beta(A)$?

答:
I agree that this is written in a confusing way. At first sight, I thought that the authors are extending $\beta$ to subsets simply by defining $\beta(A) = \{\beta(e) : e \in A\}$ for each $A \subseteq E(G)$, and then showing that this satisfies $G - A \simeq H - \beta(A)$. But indeed, this is not what they seem to prove.

At second sight, I think that what they want show is that for each $A$, the number of copies of $A$ (more on that in a minute) in $G$ and in $H$ are equal. Then we can define $\beta$ simply by choosing a one-to-one mapping between these subgraphs in $G$ and $H$, and then doing this for each (isomorphism class of) $A \subsetneq E(G)$. This ensures that $A \simeq \beta(A)$. But in this case I can't see how $A \simeq \beta(A)$ implies $G - A \simeq H - \beta(A)$. On the other hand, later on they only seem to use the fact that the subgraphs of $G$ and $H$ are in one-to-one correspondence.

Under this second interpretation, the answers to your questions (1) and (2) are as follows:

(1): Your first interpretation is correct - $N(G)$ is the number of subsets $A' \subseteq E(G)$ such that the $A'$ and $A$ induce isomorphic subgraphs. Analogously, $N(H)$ is the number of subsets $A' \subseteq E(H)$ such that $A'$ and $A$ induce isomorphic subgraphs in $H$ and $G$, respectively. Note that $\beta$ does not appear in this definition.

(2): We count the set $X(G) = \{(e,A'): e \notin A' \subseteq E(G), A' \text{ and } A \text{ induce isomorphic subgraphs}\}$. For each $A'$, we have $m-k$ edges not in $A'$, so $|X(G)| = (m-k)N(G)$. On the other hand, for each edge $e$, we have $N_e(A)$ choices of $A'$ that do not contain $e$, so $|X(G)| = \sum_e N_e(G)$. We may define $X(H)$ analogously, and since $G - e \simeq H-e$ implies $N_e(G) = N_e(H)$, we have $|X(G)| = |X(H)|$ and consequently $N(G) = N(H)$, as desired.


答:
We know that:
$\sin(\alpha)=\sin(\pi-\alpha)$ and

$$\lim_{x\to 0}\frac{\sin(x)}x=1\quad\&\quad\lim_{x\to 0}\frac{1-\cos(x)}{x^2}=\frac12$$

$$\begin{aligned}\lim_{x\to 0}\frac{\sin(\pi\sqrt{\cos(x)})}x&=\lim_{x\to 0}\frac{\sin(\pi(1-\sqrt{\cos(x)}))}{\pi(1-\sqrt{\cos(x)})}\frac{(1-\sqrt{\cos(x)})}{x^2}\frac{1+\sqrt{\cos(x)}}{1+\sqrt{\cos(x)}}\pi x\\&=\lim_{x\to 0}\frac{\sin(\pi(1-\cos(x)))}{\pi(1-\sqrt{\cos(x)})}\frac{1-\cos(x)}{x^2}\frac{\pi x}{1+\sqrt{\cos(x)}}=0\end{aligned}$$


问:
Is the sequence $(f_n)_n $ a convergent sequence in $(L,N) $?
We denote by $L$ the real vector space of the functions $g: [0,1] \to\mathbb {R}$ such that $\displaystyle\sup_{\substack{x, y\in [0,1]\\x\neq y}}{\frac {|g (x) - g (y)|}{| x - y | }}$ exists and is finite. We denote
$$
\sup_{\substack{x, y\in [0,1]\\x\neq y}}{\frac {|g (x) - g (y)|}{| x - y | }}=k (g)
$$
The nonnegative-valued function $N:g\to k (g)+\|g\|_{\infty}$ defines a norm on $L$.

Let the sequence of function $f_n\colon[0,1]\to \mathbb{R}$, defined by
$$f_n(x)\rightarrow \begin{cases}x\,, & x\in[0,\frac {1}{n}]\\ \frac {1}{n}\,, & x\in \mathopen]\frac {1}{n},1]\end{cases}$$

For all $n\geq 1$ we have $N (f_n)=1+\frac {1}{n} $. Is the sequence $(f_n)_n $ a convergent sequence in $(L,N) $?

An idea please.

答:
If $f_n$ is convergent there is some function $f$ in your space such that $N(f_n-f)\xrightarrow{n\to\infty}0$. Note that $N(f_n-f)=k(f_n-f)+\|f_n-f\|_\infty$ and both $k(\cdot)$ and $\|\cdot\|_\infty$ are nonnegative, so if $f_n$ is convergent then

$k(f_n-f)\xrightarrow{n\to\infty}0$ and $\|f_n-f\|\xrightarrow{n\to\infty}0$. Of course, the converse is also true.

If $f_n$ is convergent, we're specially interested in $\|f_n-f\|\xrightarrow{n\to\infty}0$, since we know that means $f_n$ converges pointwise to $f$, so select some $x\in[0,1]$ and let's see what $f(x)$ looks like.

If $x=0$ then for every $n$ we have $f_n(0)=0$, so $f(0)=0$. If $x\in(0,1]$ then there is some $n_0$ such that $\frac{1}{n_0}<x$, so $f_n(x)=\frac{1}{n}$ for every $n\ge n_0$,

which implies $f_n(x)\xrightarrow{n\to\infty}0$, so $f(x)=0$. Therefore, if $f_n$ is convergent, it converges to the zero function (let's call it $f$).

Now, if $f_n$ is convergent, let's see what happens to $N(f_n-f)=N(f_n)$.

As you pointed out $N(f_n)=1+\frac{1}{n}$, so $N(f_n-f)=N(f_n)\xrightarrow{n\to\infty}1\neq0$, which is not what should happen. Therefore $f_n$ is not convergent.


问:
Prove that the odd polynomial equation $p(x)=0$ has at least one real solution (use intermediate value theorem)
Let $a_0,a_1,....a_n$ be real numbers and $p(x) = a_nx^n+a_{n-1}x^{n-1}+...a_1x+a_0$ for $x \in R$

I am trying to prove that the odd polynomial equation $p(x)=0$ has at least one real solution by using the intermediate value theorem.

Here's how I started it.

We can assume without loss of generality that $a_n=1$ and can write $p(x)= x^n(1+\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+...\frac{a_0}{x_n})$

How can I show that there exists $M>0$ such that $x\leq -M$ implies $p(x) <0$

and
$M>0$ such that $x\geq M$ implies $p(x) >0$

答: Hint: consider the limits $\lim_{x\to \infty}p(x)$ and $\lim_{x\to -\infty}p(x)$. If these are large and small enough you might be able to find such $M$’s quite easily.


问:
Possible typo in Shoenfeld's Mathematical Logic?
The following is written in the first chapter of Shoenfeld's text on mathematical logic.

"...suppose that a collection C is defined by a generalized inductive definition. Then in order to prove that every object in C has property P, it suffices to prove that the objects having property P satisfy the laws of the definition. Such a proof is called a proof by induction on objects in C".

Am I wrong to think that this is mistaken? Wouldn't this be merely showing that objects with property P belong to C, rather than showing that objects in C have property P?

答:
The phrase "laws of the definition" is a bit slippery. What it means is that:

$C$ is defined as the smallest set of objects with some closure properties (including containment of some "basic objects" analogous to the base case of an induction proof).

If we then show that the set of things having property $P$ satisfies the same closure properties (including the containment of the same "basic objects"), then we have $C\subseteq P$.

For a very silly example, the set $C$ of even natural numbers is the smallest set of natural numbers containing $0$ and closed under the operation $x\mapsto 2+x$. Let $P$ be the property "Is a multiple of $2$ or is $>17$ (or both)." Then we have $P(0)$, and $P(x)\rightarrow P(2+x)$ for all $x$; so $C\subseteq P$.


问:
If $X \times Y = X \times Z$ and $X \neq \emptyset$, then $Y = Z$.
I got a pretty good idea for the proof but it feels like it's missing some details.

Proof:
$$X \times Y \Leftrightarrow a\in X \land b \in Y \Leftrightarrow a\in X \land b\in Z$$

Since $b\in Y \Leftrightarrow b\in Z$, then $Y = Z \blacksquare$

I feel like the jump to $b\in Y \Leftrightarrow b\in Z$ is flawed is some way, missing some details. Can anyone help me justify it more or tell me that it's good?

答: Let $y \in Y$. Since $X \neq \emptyset $ , let $x \in X$. Consider $(x,y) \in X \times Y = X \times Z$ which implies $y \in Z$. The other way is proven similarly.


问:
Principal Component Analysis of variables $x$, $x^2$, $x^3$?
I know the working of PCA, but was baffled by question asked in the interview that:

Principal Component Analysis of variables $x$, $x^2$, $x^3$. How can
we derive it mathematically?

Do we need to create matrix of it and then calculate covariance matrix?

答: Yes, you would just use your vector $x$ to make a matrix $[x, x^2, x^3]$ that can then be used for the standard PCA steps.

OwnerDisplayName: MAC

OwnerDisplayName: Cameron Chandler


答:
You do not clearly state where and how you use $X\ne \emptyset$, and you should start from $Y$ or $Z$ and not the products.

Let $y\in Y$ be arbitrary. As $X\ne\emptyset$, there exists some $x\in X$.
Then $(x,y)\in X\times Y=X\times Z$, i.e., $(x,y)=(u,v)$ for some $u\in X$, $v\in Z$. By definition of ordered pairs, this means ($x=u$ and) $y=v$. In particular, $y\in Z$. We conclude $Y\subseteq Z$.
The other direction, $Z\subseteq Y$ , follows the same way,

Title: If $X \times Y = X \times Z$ and $X \neq \emptyset$, then $Y = Z$.

Body:
I got a pretty good idea for the proof but it feels like it's missing some details.

Proof:
$$X \times Y \Leftrightarrow a\in X \land b \in Y \Leftrightarrow a\in X \land b\in Z$$

Since $b\in Y \Leftrightarrow b\in Z$, then $Y = Z \blacksquare$

I feel like the jump to $b\in Y \Leftrightarrow b\in Z$ is flawed is some way, missing some details. Can anyone help me justify it more or tell me that it's good?


答:
Since $\cos x=1-x^2/2+o(x^2)$, we also have
$$
\sqrt{\cos x}=1-\frac{x^2}{4}+o(x^2)
$$
and so your limit can be rewritten as
$$
\lim_{x\to0}\frac{\sin(\pi-\pi x^2/4+o(x^2))}{x}=\lim_{x\to0}\frac{\sin(\pi x^2/4+o(x^2))}{x}=\lim_{x\to0}\frac{\pi x^2/4+o(x^2)}{x}=0
$$
With $x^2$ at the denominator you'd get $\pi/4$.

A different way to solve the problem is to notice that the given limit is the derivative of $f(x)=\sin(\pi\sqrt{\cos x})$ at $0$; since
$$
f'(x)=\cos(\pi\sqrt{\cos x})\frac{-\pi\sin x}{2\sqrt{\cos x}}
$$
we get $f'(x)=0$


问:
Folks theorem for hypercube: In high dimension, everyone is alone (the curse of dimensionality)
I'm reading a slide on dimensionality curse

I've searched on Google, but can not find any reference for this Folks theorem. Could you please suggest me some of them?

答: My professor has provided with the paper containing the theorem. It's On the Surprising Behavior of Distance Metrics in High Dimensional Space by Charu C. Aggarwal, Alexander Hinneburg, and Daniel A. Keim.</p>


问:
Show that the following inequality is true (about power sum)
Given $a$ and $b$ are non negative real number and $m\in\mathbb{R}_{\ge1}$

Can it be shown that, If $b\ge a$ then

$$\left|\sum_{i=0}^{n-1}(-1)^i(a+ib)^m\right |\le\left|\sum_{i=0}^{n}(-1)^i(a+ib)^m\right |$$

Example let $a=0$,$b=1$ so $|1^m-2^m\pm...+(-1)^{n-1}(n-1)^m|\le|1^m-2^m\pm...+(-1)^{n}(n)^m|$

I know about triangular inequality but not get exact proof. thanks.

答:
Hint: square both sides and try mathematical induction (treat $n$ even and odd separately), then you'll find some famous inequality useful.

==================================================

Let $c_n=(-1)^i(a+ib)^m$ and $S_n \equiv \sum_{i=0}^n c_n$.

$$
\left| S_{n-1} \right| \leqslant \left| S_n \right| \iff S_n^2-S_{n-1}^2 \geqslant 0 \iff c_n^2+2c_n S_{n-1}\geqslant 0.
$$

Case 1: $n$ is even, $n=2k$, $c_n\geqslant 0$.
$$
2 S_{n-1} +c_n = 2\sum_{i=0}^{2k-1}c_i + c_n = c_0 + \sum_{i=0}^{k-1} (c_{2i}+2c_{2i+1}+c_{2i+2}).
$$

Apply Jensen's inequality,
$$
c_{2i} + 2c_{2i+1} + c_{2i+2} = (a+2ib)^m-2[a+2(i+1)b]^m + [a+2(i+2)b]^m \geqslant 0.
$$

Therefore $c_n^2+2c_n S_{n-1}=c_n (2S_{n-1} + c_n) \geqslant 0.$

Case 2: $n$ is odd, $n=2k+1, c_n \leqslant 0.$
$$
2 S_{n-1} +c_n = 2\sum_{i=0}^{2k}c_i + c_n = 2c_0 +c_1 + \sum_{i=1}^k (c_{2i-1}+2c_{2i}+c_{2i+1}).
$$

Again, Jensen's inequality gives
$$
c_{2i-1} + 2c_{2i} + c_{2i+1} = -[a+(2i-1)b]^m+2(a+2ib)^m - [a+(2i+1)b]^m \leqslant 0.
$$

And, $2c_0+c_1=2a^m-(a+b)^m \leqslant (a+b)^m-(a+b)^m = 0$.

Therefore $c_n^2+2c_n S_{n-1}=c_n (2S_{n-1} + c_n) \geqslant 0$ also holds for odd $n. \blacksquare$


问:
How do you find the distance between two points on a parabola
So I've been wanting to figure out a formula for an odd pattern I found... but to write a proof, I need to know one thing...

How do I find the distance between two points on a parabola?

Like, if I have y=x^2, how do I find out how much actual line is between say, (4,16) and (2,4)? If you use the normal line distance thing, it tells you how much space is between each point, but not how much of the actual parabola is between the points.

So can you help me out?

Thanks.

答:
Rectification (that's what this is called) of the curve given by $f(x)$ is obtained by the integral
$$\int_a^b \sqrt{1+f'(x)^2}\,\mathrm dx.$$
Hre you need to evaluate
$$\int_2^4\sqrt{1+4x^2}\,\mathrm dx.$$</p>


问:
Find the sum of the series $\displaystyle \sum_{n \ge 0} \frac{2^{3n}}{5^{n-1}}$ and specify if it is convergent or divergent.
I have to find the sum of the following series and specify if it is convergent or divergent:

$$\sum_{n \ge 0} \frac{2^{3n}}{5^{n-1}}$$

This is what I have so far:

$$\sum_{n \ge 0} \frac{2^{3n}}{5^{n-1}}
= \sum_{n \ge 0} \frac{5 \cdot 2^{3n}}{5^{n}}
= \sum_{n \ge 0} \frac{5 \cdot (2^{3})^{n}}{5^{n}}
= 5 \cdot \sum_{n \ge 0} \bigg (\frac{8}{5} \bigg )^n$$

From here I said that:

$$\frac{8}{5} > 1 \Rightarrow \bigg (\frac{8}{5} \bigg) ^ {n} \rightarrow \infty \text{ as } n \rightarrow \infty$$

And because of this I concluded that the sum of the series is $\infty$, thus the series is divergent. However, I am not very satisfied with this last part. It doesn't feel right to jump from

$$\bigg (\frac{8}{5} \bigg) ^ {n} \rightarrow \infty \text{ as } n \rightarrow \infty$$

to

$$\sum_{n = 0}^{\infty} \frac{2^{3n}}{5^{n-1}} = \infty$$

What extra steps am I missing?

答:
is divergent and can be proven with the quotient test, assuming
$ a_n = \frac{2^{3n}}{5^{n-1}} $ and showing that $ \frac{a_{n + 1}}{a_n } = \frac{8}{5}> 1$ and therefore the series does not converge


问:
Diophantine equations and recurrence sequences
I was working on the diophantine equation $7x^2+y^2=2^n$ for each natural $n\geq 3$. I proved by induction that $(x_n,y_n)$ is an integer solution for  $n\geq 3,$ where $x_3=y_3=1$ and
$$x_{n+1}=\frac{x_n+y_n}{2} $$
$$y_{n+1}=\frac{7x_n-y_n}{2} $$
So I obtained a family of infinite solutions for each $n$ given by a system of recurrence sequences.

I wonder if there is a general theory about this, which relate diophantine equations with recurrence sequences.

Thank you in advance.

答:
In the ring $\mathbb Q[\sqrt{-7}]$ the algebraic integers are of the form $\frac a2+\frac b2\sqrt{-7}$ where $a,b$ are integers of the same parity (both odd or both even.)

It’s a bit long to describe in an answer, but your $(x_n,y_n)$ can be seen as $$\frac{x_n}2+\frac{y_n}2\sqrt{-7}=\left(\frac12+\frac12\sqrt{-7}\right)^{n-2}$$

More generally, if:

$$\frac x2+\frac y2\sqrt{-7}=\left(\frac a2+\frac b2\sqrt{-7}\right)^k$$ then
$\frac{x^2+7y^2}4=\left(\frac{a^2+7b^2}4\right)^k.$

So you can solve $x^2+7y^2=11^k$ using $a=4,b=2.$

This kind of argument works for any $\sqrt{D},$ where $D$ is square-free. If $D\equiv 1\pmod 4$ you have to use the half-integers, like for $-7,$ but for other $D$ you only are allowed to take $a+b\sqrt D$ where $a,b$ are integers.

Study Quadratic Fields.


问:
prove $a^6 + b^6 + c^6 + 3a^2b^2c^2 > 2(a^3b^3+b^3c^3+c^3a^3)$ using Popoviciu's inequality.
I am trying to prove $a^6 + b^6 + c^6 + 3a^2b^2c^2 > 2(a^3b^3+b^3c^3+c^3a^3)$ for $a, b, c$ positive numbers not all equal using Popoviciu's inequality.

The equality states the following: Let $f : I \rightarrow \mathbb{R}$ be a continuous function on the real inverval $I$. $f$ is convex if and only if
$\frac{f(x)+f(y)+f(z)}{3} + f\left(\frac{x+y+z}{3}\right) \geq \frac{2}{3}\left[f\left(\frac{x+y}{2}\right) + f\left(\frac{y+z}{2}\right)+ f\left(\frac{z+x}{2}\right)\right]$
for all $x, y, z \in I.$

Can anyone get me a hint how to find a convex function with the desired property? Thanks a lot.

答:
Hint: The form of Popoviciu's inequality is very similar to your inequality to be proven in this form: $$f(x) + f(y) +f(z) + 3f\left(\frac {x+y+z}3\right) \ge 2 \left[f\left(\frac {x+y}2\right)+f\left(\frac {y+z}2\right)+f\left(\frac {z+x}2\right)\right]$$

Comparing this to

$$a^6+b^6+c^6+3a^2b^2c^2>2(a^3b^3+b^3c^3+c^3a^3)$$

It seems that this $f$ takes the power $6$ and divides it across the variables, and addition becomes multiplication:

e.g. $f(x)=a^6, f(y) = b^6, f\left(\frac x2+\frac y2\right) = a^3b^3$

Do you know a kind of functions that does this? The constraint $a,b,c>0$ is another hint.

Hint 2: Consider the function $f(x)=e^{6x}$. We have

$$e^{6x}+e^{6y}+e^{6z}+3e^{2x}e^{2y}e^{2z}\ge2(e^{3x}e^{3y}+e^{3y}e^{3z}+e^{3z}e^{3x})$$

What happens if you take $x = \ln \dots$?


问:
Deriving AM-GM in a way that reflects my intuition.
One of my core intuitions with inequalities (especially the olympiad kind in which this is very explicit, but this has more profound influences on how I understand the relationship between inequalities) is that in a homogenous inequality, on average, terms that are more mixed ($x^2yz$) are smaller than their less-mixed counterparts ($x^4$).

With this bit of intuition, many inequalities seem trivially true, like am-gm:

$$x^3+y^3+z^3\geq xyz + xyz +xyz$$

Or this one:

$$x^2y + y^2z + z^2x\geq xyz+xyz+xyz$$

And a generalisation of part of this idea: Murihead's Inequality.

I want to find a proof (by which I mean not particularly formal derivation) of AM-GM, that adequately captures 'mixed terms being smaller on average' as being the reason for the inequality. I think this would satisfy some innate and inexplicable urge.

There may not exist a satisfying answer to this question (though I would certainly no one if I saw it), but hopefully, I have communicated the kind of thing I'm looking for.

答:
Here is one approach to thinking about these inequalities that I found.

It's not exactly a proof, but built off a more explicable intuition.

Consider, for example, the inequality:

$$x^3+y^3+z^3\geq xyz + xyz +xyz$$

On each side, we have the sum of the volumes of three rectangular prisms. Since the inequality is homogenous, we can interpret this as meaning that both sides have the same total surface area.

But, of any rectangular prism, a cube has the highest ratio of volume : surface area, so the left-hand-side is maximal.

This generalises easily to inequalities like:

$$x^2y+y^2z+z^2x\geq xyz + xyz +xyz$$

The less mixed terms on the left-hand-side have a higher volume : surface area ratio. It also genralises to higher dimensions.


问:
Does infinite equal infinite?
I have a question.

Let $x$ be infinite.

$$2x=\infty\times2, \quad 2x=\infty$$

So actually, does $2x=x$?

答:
This depends on what definition for "infinite" you compare.

For instance, if we generalize the Laplace transform to divergent integrals, we will see that

$$\int_0^\infty 1 dx=\int_0^\infty\frac1{x^2}\ne\int_0^\infty x \,dx$$

All three integrals are divergent and infinite and have the regularized value zero, but two of them are equal but not equal to the third one.

In other cases of divergent integrals or series, the regularized value and/or growth rate (germ at infinity) or behavior at a singularity can differ as well or the differences can compensate for each other as in the example above.

So no, at least in terms of divergent integrals and series, the infinities may differ.

Title: Does infinite equal infinite?


问:
$n$th derivative of $e^{ax^m}$

I am seeking an explicit expression for the $n$th derivative of $e^{ax^m}$.

To achieve this I consider the $n$th coefficient of the Taylor series with $t\rightarrow 0$.
\begin{align}
    e^{a(x+t)^m} &= \sum^\infty_{n=0}\lim_{t\rightarrow 0}\frac{d^n}{dt^n}e^{a(x+t)^m}\frac{t^n}{n!}\\
    &=\sum^\infty_{n=0}\lim_{t\rightarrow 0}\frac{d^n}{dt^n}e^{a\left[\sum\limits_{k=0}^{m}\binom{m}{k}x^kt^{m-k}\right]}\frac{t^n}{n!}\\
    &=e^{ax^m}\sum^\infty_{n=0}\lim_{t\rightarrow 0}\frac{d^n}{dt^n}e^{a\sum\limits_{k=0}^{m-1}\binom{m-1}{k}x^kt^{m-k-1}}\frac{t^n}{n!}
\end{align}
But we can also consider the left hand side as
\begin{align}
e^{a(x+t)^m} &= \sum_{k=0}^\infty\frac{(a(x+t)^m)^k}{k!}\\
&= \sum_{k=0}^\infty\frac{a^k}{k!}\sum_{j=0}^{mk}\binom{mk}{j}x^{j}t^{mk-j}\\
&= \sum_{k=0}^\infty\frac{a^k}{k!}x^{mk}\sum_{j=0}^{mk-1}\binom{mk-1}{j}x^{j}t^{mk-1-j}\\
&= \sum_{k=0}^\infty\frac{(ax^m)^k}{k!}\sum_{j=0}^{mk-1}\frac{(mk-1)!}{j!(mk-1-j)!}x^{j}t^{mk-1-j}
\end{align}
Although I can't really take it further. My next step would be to extract the $n$th coefficient and, upon comparison to the first formula, obtain an expression for the $n$th derivative. When $m=2$ we obtain the Hermite polynomials. Related to

$n$th derivative of $e^{1/x}$

Find an expression for the $n$-th derivative of $f(x)=e^{x^2}$

and links therein. With thanks to BillyJoe from the comments, Wolfram gives
\begin{equation}
\frac{d^n}{dx^n}e^{ax^m}=e^{ax^m}x^{-n}\sum^n_{k=0}\sum^k_{j=0}\frac{(-1)^j(ax^m)^k(1-jm+km-n)_n}{j!(k-j)!}
\end{equation}

I have also seen that this is equivalent to n exponent rial Riordan array, $B_m=[1,(1+x)^m+1]$.
\begin{equation}
\frac{d^n}{dx^n}e^{ax^m}=\left( \sum_{k\in \mathbb Z}b_{n,k;m}(ax)^{km-n}\right)e^{ax^m}
\end{equation}

答:
Hint:

Experimentally, the expression is of the form

$$(e^{x^m})^{(n)}=P_{mn}(x^m)x^{(-n)\bmod m}e^{x^m}$$ where $P_{mn}$ is a polynomial of degree $n-\left\lfloor\dfrac nm\right\rfloor$. You can arrange the coefficients in a triangular table, with a recurrence relation between the rows.

$$P_{m.n+1}x^{(-n-1)\bmod m}=\\mx^{m-1}(P'_{mn}(x^m)+P_{mn}(x^m))x^{(-n)\bmod m}+((-n)\bmod m)P_{mn}(x^m)x^{(-n)\bmod m-1}$$

which simplifies differently when $m|n$ or not.

Title: $n$th derivative of $e^{ax^m}$


问:
Probability and the first uncountable ordinal number
Let's suppose we can put a probability measure on the set of countable ordinals. (which is the same as the first uncountable ordinal). Let's now play a game. I pick a countable ordinal, say $\alpha$. Now you pick one. (Clearly our choices are independent). But $\alpha$ has countably many ordinals less than it and for you the number of choices greater than $\alpha$ are uncountable. So with a large probability, possibly one, your choice is bigger than mine. But the choices were independent!  How to explain? Of course, you could take this as a proof that no such measure exists but intuitively it seems to make sense that such a game could exist.

答:
This is indeed a nonexistence proof in disguise: it shows that there is no "nice" measure on $\omega_1$ according to which the set $\{\langle \alpha,\beta\rangle:\alpha<\beta\}$ is also "nice" as a subset of $(\omega_1)^2$ with respect to the product measure $\mu^2$.

Specifically, once we strip away the game aspect of the question, what you're really doing is pushing back against Fubini's theorem. We partition $(\omega_1)^2$ into three pieces: the diagonal $\{\langle\alpha,\beta\rangle: \alpha=\beta\}$, and the two "triangles" $T_1=\{\langle \alpha,\beta\rangle: \alpha<\beta\}$ and $T_2=\{\langle \alpha,\beta\rangle: \alpha>\beta\}$. As long as $\mu$ is "reasonable," the product  measure will satisfy $\mu^2(D)=0$ and $\mu^2(T_1)=\mu^2(T_2)$. And of course we'll have $\mu^2((\omega_1)^2)=1$.

You now want to argue that this gives a contradiction: again as long as $\mu$ is "reasonable" we'll have $\mu(\{x: x<\eta\})=0$ for all $\eta<\omega_1$, but by Fubini's theorem this should give $\mu^2(T_1)=\mu^2(T_2)=0$ (a special case of Fubini is: "Any measurable subset of the square of the space, each of whose 'fibers' has measure zero, itself has measure zero."). But in order to apply Fubini here, we need that $T_1$ and $T_2$ are in fact $\mu^2$-measurable.

So really what this line of thought shows is that there is no measure on $\omega_1$ satisfying a couple basic properties and making the ordering relation on $\omega_1$ well-behaved.

For example, suppose $\mathsf{CH}$ holds. Then we can "port over" the usual Lebesgue measure $\lambda$ on $[0,1]$ to a measure $\mu$ on $\omega_1$ via some bijection $f:[0,1]\rightarrow\omega_1$. This $\mu$ satisfies all the relevant reasonability conditions, but $\{\langle a,b\rangle: f(a)<f(b)\}$ is a non-Lebesgue-measurable subset of $[0,1]^2$.

If you want to keep the "sequential" flavor of the question, we can think in terms of order of integration (and to be fair this is how Fubini is usually presented). Let $\chi:(\omega_1)^2\rightarrow\{0,1\}$ be the characteristic function of the set $\{\langle \alpha,\beta\rangle: \alpha<\beta\}$. Then as long as we have a "reasonable" notion of measure,  the maps $$\alpha\mapsto \int_{\omega_1} \chi(\alpha,x)dx\quad\mbox{and}\quad\beta\mapsto\int_{\omega_1} \chi(x,\beta)dx$$ are each the constant-zero function. However, this does not mean that $$\iint_{\omega_1\times\omega_1}\chi(x,y)d(x,y)=0:$$ for that equality to hold we'd need that the whole function $\chi$ is integrable with respect to the product measure, and just because each of its "sections" is doesn't mean it itself is. In fact, the non-integrability of $\chi$ with respect to any "reasonable" measure is exactly what your argument shows.


问:
Showing that the r-th moment of $u_t$ is finite
I am a reading a paper, in which $u_t = \pi(L)\epsilon_t$, where $L$ is the lag operator and $\pi(z)=\sum_{k=0}^{\infty}\pi_kz^k$.
Also,

It is said that the Marcinkiewicz-Zygmund inequality for martingales implies that the r-th moment of $u_t$ exists and is uniformly bounded in t. However, $u_t$ is not a martingale, thus I wonder how I could derive this result. Any comment would be appreciated even if it would not be about the MZ inequality.

答:
You have $$
u_t=\sum_{k=0}^{\infty} \pi_k \epsilon_{t+k}
$$
The partial sums of the right hand side are the martingale that you apply MZ to, separately for each $t$.

EDIT: More details:
Define $u_{t, s}:= \sum_{k=0}^{s}\pi_k \epsilon_{t+k}$. Then $(u_{t, s})_{s}$ is a martingale because
$$
E[u_{t, s+1} | \mathcal{F}_s] = u_{t,s} + E[\pi_{s+1}\epsilon_{t+s+1} | \mathcal{F}_{s}]  = u_{t,s}
$$


问:
Solution to Riccati differential equation using transformations
How to solve this Riccati differential equation?

$$\frac{d y}{d x}=\frac{-a}{2(bx-x^3)}-\frac{y^
2(bx-x^3)}{(a^2)}$$

Where $a, b $ are real constants.

答:
Your equation has the form $y'+py^2=q$. The transformation is in this case $y=\frac{u'}{pu}$. Alternatively $y=\frac{qv}{v'}$ can be used. Continuing with the first variant
$$
\frac{u''}{pu}-\frac{p'u'}{p^2u}-\frac{u'^2}{pu^2}+p\frac{u'^2}{p^2u^2}=q,
$$
so that then without fractions
$$
u''-\frac{p'}{p}u'=pqu.
$$
Now insert $p=\frac{bx-x^3}{a^2}$ and $q=-\frac{a}{bx-x^3}$ to get
$$
u''-\frac{b-3x^2}{bx-x^3}u'+\frac1au=0.
$$


问:
What is the geometric meaning of this vector equality? $\vec{BC}\cdot\vec{AD}+\vec{CA}\cdot\vec{BD}+\vec{AB}\cdot\vec{CD}=0$
I was doing some exercises for linear algebra. One of them was to prove that for any four points $A, B, C, D \in \mathbb{R}^3$ the following equality holds:
$$\overrightarrow{BC} \cdot \overrightarrow{AD}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ = 0$$
The proof is easy; you just make three vectors starting in $A$ and then see that all the terms cancel out.

My question is: what is the geometric interpretation of this equality? How can I visualize it or understand its deeper meaning? Does this equality have a name or where can I read more about it?

I'm asking this because it turns out that it is not just a random equality and is rather useful. For example, if we want to prove the existence of orthocenter, we can do it surprisingly easily and quickly using this equality.

答:
Let $O$ be the orthocenter $O$ of $\triangle ABC$. Then
\begin{align}
&\overrightarrow{AB}\cdot\overrightarrow{CD}
\ +\ \overrightarrow{BC}\cdot\overrightarrow{AD}
\ +\ \overrightarrow{CA}\cdot\overrightarrow{BD}\\
=\ &\left(\overrightarrow{AB}\cdot\overrightarrow{CO}
\ +\ \overrightarrow{BC}\cdot\overrightarrow{AO}
\ +\ \overrightarrow{CA}\cdot\overrightarrow{BO}\right)
+ \left(\overrightarrow{AB}\cdot\overrightarrow{OD}
\ +\ \overrightarrow{BC}\cdot\overrightarrow{OD}
\ +\ \overrightarrow{CA}\cdot\overrightarrow{OD}\right)\\
=\ &\left(\overrightarrow{AB}\cdot\overrightarrow{CO}
\ +\ \overrightarrow{BC}\cdot\overrightarrow{AO}
\ +\ \overrightarrow{CA}\cdot\overrightarrow{BO}\right)
+ \left(\overrightarrow{AB}\ +\ \overrightarrow{BC}\ +\ \overrightarrow{CA}\right)\cdot\overrightarrow{OD}\tag{$\dagger$}\\
=\ &0+0=0.\\
\end{align}
The first bracket on line $(\dagger)$ is zero because every side of $\triangle ABC$ is perpendicular to the altitude dropped from the opposite vertex. The second bracket is zero because it is the sum of directed edges of a closed circuit.

In short, the identity is basically a cyclic sum of expressions of the form "side dot altitude" on $\mathbb R^2$, but another cyclic sum of the form "side dot $\overrightarrow{OD}$" has been added to conceal the significance of the orthocenter and make the identity present in $\mathbb R^3$.


问:
If $ A, B \subset K $ closed non-empty disjoint with $K$ then exist $ K_A, K_B \subset K $ mutually separated such that $ K = K_A \cup K_B $
Let $ (X, \tau) $ be a Hausdorff topological space and $ K \subset X $ compact. If $ A, B \subset K $ closed non-empty disjoint such that no component of $ K $ intersects $ A $ and $ B, $ show that there exist $ K_A, K_B \subset K $ mutually separated such that $ K = K_A \cup K_B $, $ A \subset K_A $ and $ B \subset K_B $

Proof: Let $ A, B \subset K $ closed non-empty disjoint of a Hausdorff space then they are compact and disjoint of a Hausdorff space, therefore there exist $ K_A $ and $ K_B $ open and disjoint such that $ A \subset K_A $ y $ B \subset K_B $ and let $ x \in K $ such that $ C_{K} (x) $ the component connected with $ C_{K} (x) \cap A = \emptyset = C_{K} (x) \cap B $ then $ C_{K} (x) \cap K_A = \emptyset = C_{K} (x) \cap K_B $, so $K_A, \ K_B$ they are mutually separate ...

How could the demonstration continue?

Thanks!

答:
SKETCH: Work in the compact Hausdorff space $K$. For each $x\in K$ let $C(x)$ be the component of $K$ containing $x$. $C(x)$ is also the quasicomponent of $K$ containing $x$, so it is the intersection of the clopen nbhds of $x$ in $K$.

Use that to show that each $x\in A$ has a clopen nbhd $H(x)$ disjoint from $B$..

$A$ is compact, and $A\subseteq\bigcup_{x\in A}H(x)$, so there is a finite $F\subseteq A$ such that $A\subseteq\bigcup_{x\in F}H(x)$; let $K_A=\bigcup_{x\in F}H(x)$ and $K_B=K\setminus K_A$, and verify that these sets have the desired properties.


问:
Inequality with integers
I am trying to find an analytical way to prove that for integers $n\ge 6$,

$$\sqrt[n]{n+1}\le \frac{1}{n}+\frac{1}{\sqrt[n]{n+1}}+\frac12$$

答:
Here is an analytic proof using derivatives.

Let

$$f(x)=(x+1)^{1/x}-(x+1)^{-1/x}-{1\over x}=2\sinh\left({\ln(x+1)\over x}\right)-{1\over x}$$

It's enough to show that $f(6)\lt1/2$ and $f'(x)\lt0$ for $x\ge6$. The inequality $f(6)\lt1/2$ is not difficult to verify with a calculator. (It amounts to checking that $\sqrt[6]7\approx1.383$ is less than $(1+\sqrt{10})/3\approx1.387$.) As for the derivative, we have

$$\begin{align}
f'(x)&=2\left({1\over x(x+1)}-{\ln(x+1)\over x^2}\right)\cosh\left({\ln(x+1)\over x}\right)+{1\over x^2}\\
&={1\over x^2}\left(1-2\left(\ln(x+1)+{1\over x+1}-1\right)\cosh\left({\ln(x+1)\over x}\right) \right)\\
&\lt{1\over x^2}\left(1-2\left(\ln(x+1)+{1\over x+1}-1 \right)\right)\quad\text{if }x\ge2\\
&=-{1\over x^2}\left(2\ln(x+1)+{1\over x+1}-3 \right)
\end{align}$$

and it's easy to see that $g(x)=2\ln(x+1)+{1\over x+1}-3$ is an increasing function with $g(3)=2\ln4-{11\over4}\approx0.0226\gt0$, so $f'(x)\lt0$ for $x\ge3$. (It would have been enough to show $g(6)\gt0$, but it doesn't hurt to get a better bound on where the function starts decreasing; the actual maximum of $f$ occurs at around $2.005$.)

Remark: Computationally, the trickiest step here is showing $f(6)\lt1/2$; the inequality $g(3)\gt0$ amounts to showing $\ln2\gt11/16=0.6875$, and there are various ways to show that $\ln2\approx0.69$. To show $f(6)\lt1/2$ without simply letting a computer do all the work, let $u=\sqrt[6]7$, and note that

$$u+{1\over u}-{1\over6}\lt{1\over2}\iff3u^2-2u-3\lt0\iff{1-\sqrt{10}\over3}\lt u\lt{1+\sqrt{10}\over3}$$

so one need "merely" show that $7\cdot3^6\lt(1+\sqrt{10})^6$. Expanding the binomial and doing some arithmetic, this becomes $1226\lt403\sqrt{10}$, and this inequality is easily verified from $3.1^2=9.61\lt10$, so that

$$1226\lt1240=400\cdot3.1\lt403\sqrt{10}$$

Title: Inequality with integers


问:
Minimize expected value from deck of cards containing numbers from 1 to 6.
Part 1: You have infinite cards and on each of them there is written one of the following numbers: {1, 2, 3, 4, 5, 6}. You can build a deck in the way you want and after you build it, you need to tell your opponent how many cards of each type you put in the deck. You shuffle the deck. Your opponent will guess a number from 1 to 6 and pick a card at random. If he guesses 1 and picks 1 he wins 1 dollar from you, if he guesses 2 and picks 2 he wins 2 dollars from you, etc. If he does not guess correctly he wins nothing. How would you build a deck so that your opponent wins the least amount of money from you? You know your opponent plays optimally, i.e. he always tries to maximize his profit.

Part 2: How does the answer to the same question change if you need to pay 0.01 dollar for each card you put in the deck?

答:
Suppose that you took $a_i$ cards with number $i$ written on them, for a total of $a$ cards. Denote $p_i = a_i / a$ — the probability of card with number $i$ ending up on top of the deck after shuffling.

If your opponent guesses $i$ then his expected winnings are $i \cdot p_i$. The only thing he is able to control is $i$, so he maximizes over it and gets $\max\{p_1, 2p_2, \dots, 6p_6\}$ as the result.

As in every other minimax game, your goal now is to minimize this expression over $p_1, p_2, \dots, p_6$. Because $p_i$ have a fixed sum of $1$ it follows that you should choose $$p_1 = 2 p_2 = 3 p_3 = 4 p_4 = 5 p_5 = 6 p_6,$$ and therefore $$a_1 = 2 a_2 = 3 a_3 = 4 a_4 = 5 a_5 = 6 a_6$$ with the minimal integer solution being $a_1 = 60, a_2 = 30, a_3 = 20, a_4 = 15, a_5 = 12, a_6 = 10$, $a = 147$.

The expected winnings of your opponent will then be $\tfrac{60}{147} \approx 0.41$, i.e. 41 cent.

In the second part, however, it may be beneficial to select smth like $a_i' = \left\lfloor \frac{a_i}{K} \right\rfloor$ instead, with some fixed $K$. In this way you maintain the close-to-optimal ratio $a_1 : a_2 : a_3 : a_4 : a_5 : a_6$ but save a lot because the deck shrinks significantly.

I don't think that there is an analytical solution to such a problem, I would do some programming (and I suppose that the exact implementation does not belong to math.se): given a value of $a$ you know the optimal way to distribute it into $a_i$, and also optimal $a \le 147$, which leaves not that many choices


答:
If you have a curve $y=f(x)$, the length of the curve between points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is given by
$$L(x_1,x_2) = \int_{x_1}^{x_2} \sqrt{(dx)^2 + (df(x))^2} = \int_{x_1}^{x_2}\sqrt{1+\left(\dfrac{df}{dx} \right)^2} dx$$
In your case, $f(x) = x^2$. Hence, $\dfrac{df}{dx} = 2x$. Hence, the length of curve between the points $(x_1,x_1^2)$ and $(x_2,x_2^2)$ is
$$L(x_1,x_2) = \int_{x_1}^{x_2} \sqrt{1+(2x)^2} dx = \int_{x_1}^{x_2} \sqrt{1+4x^2} dx$$
Can you evaluate this integral?

$$\int_{x_1}^{x_2} \sqrt{1+4x^2} dx = \left.\dfrac{2x \sqrt{4x^2+1} + \log(2x+\sqrt{1+4x^2})}4 \right \vert_{x_1}^{x_2}$$

EDIT

Added picture to highlight the different lengths.

Title: How do you find the distance between two points on a parabola

Body:
So I've been wanting to figure out a formula for an odd pattern I found... but to write a proof, I need to know one thing...

How do I find the distance between two points on a parabola?

Like, if I have y=x^2, how do I find out how much actual line is between say, (4,16) and (2,4)? If you use the normal line distance thing, it tells you how much space is between each point, but not how much of the actual parabola is between the points.

So can you help me out?

Thanks.


问:
Finding volume of a cone with integration
I am very confused on where to start with the question, do I start by trying to create a triangle on the top somehow?

答:
Hint

Let $ R $ be the ray of the base and $ Z $ be the height of the cone.

By Pythagoras, we have

$$(2R)^2=5^2+5^2\implies R=\sqrt{12.5}$$
and
$$\tan(45°)=\frac RZ=1$$
The volume of revolution  will be
$$V=\int_0^Zs(z)dz=$$
$$\int_0^Z(\pi r^2)dz$$

with

$$\frac rz=\frac RZ=1$$


问:
Calculating the flux of a vector field.
Let $S$ be the unit sphere $x^2+y^2+z^2=1$ with the outward pointing normal vector n. Calculate the flux for the vector field $\mathbf{f}(\mathbf{r})=4\mathbf{r}$ through $S$.

What I have done so far:

I have rearranged the equation to get $z=\sqrt{1-x^2-y^2}$ and thus $N(x,y)=(\frac{x}{\sqrt{1-x^2-y^2}},\frac{y}{\sqrt{1-x^2-y^2}},1)$

Now to calculate the flux the equation is $\iint\limits_{S}\mathbf{f}\cdot\mathbf{n}dS$.

Hence I now have $$4\iint\limits_{S}(x,y, \sqrt{1-x^2-y^2} )\cdot (\frac{x}{\sqrt{1-x^2-y^2}},\frac{y}{\sqrt{1-x^2-y^2}},1)dS$$

$$=4\iint\limits_{S}\frac{1}{\sqrt{-x^2-y^2+1}}dxdy$$

However I don't know how to get the limits for the integral.

答:
In spherical coordinates,

$x = r \cos \theta \sin \phi, y = r \sin \theta \sin \phi, z = r \cos \phi$

Surface area element $dS = r^2 \sin \phi \ d \theta d \phi = \sin \phi \ d \theta d \phi \, $ (as $r = 1$)

Please note the outward normal vector should be a unit vector pointing directly away from the origin for this surface. So using, $\vec{r} = x \hat{i} + y\hat{j} + z\hat{k} = \cos \theta \sin \phi \, \hat{i} + \sin \theta \sin \phi \, \hat{j} + \cos \phi \,\hat{k}$

$\hat{n} = \frac{\vec{r}}{||r||} = \cos \theta \sin \phi \, \hat{i} + \sin \theta \sin \phi \, \hat{j} + \cos \phi \,\hat{k}$

$\vec{F} = 4r = 4(\cos \theta \sin \phi \, \hat{i} + \sin \theta \sin \phi \, \hat{j} + \cos \phi \,\hat{k})$

$Flux = \displaystyle \iint_S \vec{F} \cdot \hat{n} dS$

$ \displaystyle = 4 \int_0^{\pi} \int_0^{2\pi} (\cos \theta \sin \phi \, \hat{i} + \sin \theta \sin \phi \, \hat{j} + \cos \phi \,\hat{k}) \cdot (\cos \theta \sin \phi \, \hat{i} + \sin \theta \sin \phi \, \hat{j} + \cos \phi \,\hat{k}) \sin \phi \, d\theta d\phi$

$ \displaystyle = 4 \int_0^{\pi} \int_0^{2\pi} \sin \phi d\theta d\phi = 16 \pi$

Title: Calculating the flux of a vector field.


问:
Evaluate $\lim_{(x,y)\to(2,0)}\frac{\cos(xy)-1}{x^2y^2}$
normally, I would just use substitution, since I'm pretty sure the limit does exist. However, I'm not sure if I can use substitution, since x does not equal y.

Can I use substitution, or am I thinking about this problem in the wrong way?

Thank you!

答:
Hint: let $t=xy$ and work on
$$ \lim_{t\to0}\frac{\cos t-1}{t^2}. $$


问:
$O\left((n^{\frac{3}{4}}+2)\log(n^{\frac{3}{4}}+2)\right)=O(n^{\frac{3}{4}}\log(n))$ for $n\rightarrow\infty$
In a book, the following is used without explanation:

$O\left((n^{\frac{3}{4}}+2)\log(n^{\frac{3}{4}}+2)\right)=O(n^{\frac{3}{4}}\log(n))$ for $n\rightarrow\infty$

proof:

\begin{align*}
O\left((4n^{\frac{3}{4}}+2)\log(4n^{\frac{3}{4}}+2)\right)&=O\left(4n^{\frac{3}{4}}\log(4n^{\frac{3}{4}}+2)+2\log(4n^{\frac{3}{4}}+2)\right)\\
&=O\left(4n^{\frac{3}{4}}\log(4n^{\frac{3}{4}}+2)\right)\\
&=O\left(n^{\frac{3}{4}}\log(4n^{\frac{3}{4}}+2)\right)
\end{align*}

What is the exact argument to get rid of the $+2$ inside of the logarithm? ... because than it is clear that for $n>1$:

\begin{align*}
O\left(n^{\frac{3}{4}}\log(4n^{\frac{3}{4}})\right)&=O\left(n^{\frac{3}{4}}\log(4)+n^{\frac{3}{4}}\log(n^{\frac{3}{4}})\right)\\
&=O\left(n^{\frac{3}{4}}\log(n^{\frac{3}{4}})\right)\\
&=O\left(\frac{3}{4}n^{\frac{3}{4}}\log(n)\right)\\
&=O\left(n^{\frac{3}{4}}\log(n)\right)
\end{align*}

... is it, that for $n$ large enough $\log(4n^{\frac{3}{4}}+2)\approx\log(4n^{\frac{3}{4}})$ ?
Or am I overlooking the real arumgent?

答:
"the exact argument"?  I don't know about that.  Here's an exact argument.

\begin{align*}
O &{}\left( \left( n^{3/4} + 2 \right)
\log \left( n^{3/4} + 2 \right) \right)  \\
    &= O \left( n^{3/4} \left( 1 + 2n^{-3/4} \right) \log \left( n^{3/4}(1 + 2n^{-3/4}) \right) \right)  \\
    &= O \left( n^{3/4} \left( 1 + 2n^{-3/4} \right) \left( \log \left( n^{3/4} \right) + \log \left(1 + 2n^{-3/4} \right) \right) \right)  \\
    &= O \left( n^{3/4} \left( 1 + 2n^{-3/4} \right) \left( \frac{3}{4} \log(n) + \log \left(1 + 2n^{-3/4} \right) \right) \right)  \\
    &= O \left( \frac{3}{4} n^{3/4} \log(n) \left( 1 + 2n^{-3/4} \right) + \right.  \\
    & \qquad \left.{}+ n^{3/4} \left( 1 + 2n^{-3/4} \right)\log \left(1 + 2n^{-3/4} \right) \right)  \\
\end{align*}

Then, for $n \geq 1$, $1 \leq 1+2n^{-3/4} \leq 3$, so for $n \geq \mathrm{e}$,
\begin{align*}
\frac{3}{4} &n^{3/4} \log(n)  \\
    &\leq \frac{3}{4} n^{3/4} \log(n) \left( 1 + 2n^{-3/4} \right) + n^{3/4} \left( 1 + 2n^{-3/4} \right)\log \left(1 + 2n^{-3/4} \right)  \\
    &\leq \frac{9}{4} n^{3/4} \log n + 3 \log(3) n^{3/4}  \\
    &\leq \frac{9}{4} n^{3/4} \log n + 3 \log(3) n^{3/4} \log n  \\
    &= \left( \frac{9}{4} + 3 \log 3 \right) n^{3/4} \log n, \quad n \geq \mathrm{e}
\end{align*}

Having shown bounded above and below by $[\text{constant}] n^{3/4} \log n$, it is $O(n^{3/4} \log n)$.

Here's a (perhaps) more direct way, although it is the same "factor out the big part" method.
\begin{align*}
\lim_{n \rightarrow \infty} &  \frac{\left( n^{3/4} + 2 \right)
\log \left( n^{3/4} + 2 \right)}{n^{3/4} \log(n)}  \\
    &= \lim_{n \rightarrow \infty} \left( 1 + 2n^{-3/4} \right) \frac{
\log \left( n^{3/4} + 2 \right)}{\log(n)}  \\
    &= \lim_{n \rightarrow \infty} \left( 1 + 2n^{-3/4} \right) \frac{
(3/4)\log(n) + \log \left( 1 + 2n^{-3/4} \right)}{\log(n)}  \\
    &= \frac{3}{4} \lim_{n \rightarrow \infty} \left( 1 + 2n^{-3/4} \right) \left( 1+ \frac{\log \left( 1 + 2n^{-3/4} \right)}{\log(n)}\right)   \\
    &= \frac{3}{4}(1+0)(1+0)  \\
    &= \frac{3}{4}  \text{.}
\end{align*}
Since this is a constant,
$$  O \left( \left( n^{3/4} + 2 \right)
\log \left( n^{3/4} + 2 \right) \right) = O\left( n^{3/4} \log n \right)  \text{.}  $$


问:
Find the sum of the series $\displaystyle \sum_{n \ge 0} \frac{2^{3n}}{5^{n-1}}$ and specify if it is convergent or divergent.
I have to find the sum of the following series and specify if it is convergent or divergent:

$$\sum_{n \ge 0} \frac{2^{3n}}{5^{n-1}}$$

This is what I have so far:

$$\sum_{n \ge 0} \frac{2^{3n}}{5^{n-1}}
= \sum_{n \ge 0} \frac{5 \cdot 2^{3n}}{5^{n}}
= \sum_{n \ge 0} \frac{5 \cdot (2^{3})^{n}}{5^{n}}
= 5 \cdot \sum_{n \ge 0} \bigg (\frac{8}{5} \bigg )^n$$

From here I said that:

$$\frac{8}{5} > 1 \Rightarrow \bigg (\frac{8}{5} \bigg) ^ {n} \rightarrow \infty \text{ as } n \rightarrow \infty$$

And because of this I concluded that the sum of the series is $\infty$, thus the series is divergent. However, I am not very satisfied with this last part. It doesn't feel right to jump from

$$\bigg (\frac{8}{5} \bigg) ^ {n} \rightarrow \infty \text{ as } n \rightarrow \infty$$

to

$$\sum_{n = 0}^{\infty} \frac{2^{3n}}{5^{n-1}} = \infty$$

What extra steps am I missing?

答:
Your way is fine, as an alternative

$$\frac{2^{3n}}{5^{n-1}} =\frac{8^n}{5^{n-1}}\ge \frac{8^n}{8^{n-1}}=8$$

and therefore the series diverges.


问:
Is it possible to calculate log10 x without using log?
Is it possible to calculate $\log_{10} x$ without using $\log_{10}$? I'm interested because I'm working with a framework that has some simple functions, but log is not one of them.

The specific platform is capable of doing addition, subtraction, multiplication and division. I can write a formula of a finite, predefined length; e.g. while loops or for loops that continue indefinitely are not available. It's a black-box project with an interface capable of creating formulas, so basically, I need to write the expression to calculate $\log_{10} x$ in one line, and I can't write my own method or something to do more dynamic calculations.

An approximation of $\log_{10} x$ is perfectly acceptable.

答:
Let x = 1545 and proceed as follows:

Divide x by base 10 until x becomes smaller than base 10.
1545/10 = 154.5
154.5 / 10 = 15.45
15.45 / 10 = 1.545
As we divide 3 times the value of x by 10, the whole digit of our logarithm will be 3.

Raise x to the tenth power:
1.545 ^ 10 = 77.4969905 ...

Repeat the procedures from step 1.

As we divide only 1 time, the next digit of our logarithm will be 1.

Repeat step 2.

7.74969905 ... ^ 10 = 781.354.964,875286221 ...

Repeat step 1.

Since we divide x by 10 by 8 times, the next digit in our log will be 8.

And so on, as many digits as you want.


问:
Prove that there are infinitely many natural numbers $n$, such that $n(n+1)$ can be expressed as sum of two positive squares in two distinct ways.

Prove that there are infinitely many natural numbers $n$, such that $n(n+1)$ can be expressed as sum of two positive squares in two distinct ways.($a^2+b^2$, is same $b^2+a^2$), $n \in \mathbb{N}.$

I have proved the above question which appeared in one of the Math-Olympiad. And I do know the solution. Sharing the question only because the question has a cute solution.

答:
If $n=t^2$ is a square, all odd prime factors of $n+1$ are equal to $1 \mod 4$, and by using the Chinese remainder theorem for equations $t^2+1=0 \mod p_i$ it can be arranged that $n+1$ is divisible by many different such primes.

The conclusion is that for any $k$, there is an arithmetic progression of values of $t$ such that $n(n+1)$ has more than $k$ different representations as sum of two squares, when $n=t^2$.  This is a denser set of $n$ than the other solutions, and it would be interesting to see if there is an explicit construction that has higher density, possibly as high as linear (ignoring logarithmic factors).


Body:
In a book, the following is used without explanation:

$O\left((n^{\frac{3}{4}}+2)\log(n^{\frac{3}{4}}+2)\right)=O(n^{\frac{3}{4}}\log(n))$ for $n\rightarrow\infty$

proof:

\begin{align*}
O\left((4n^{\frac{3}{4}}+2)\log(4n^{\frac{3}{4}}+2)\right)&=O\left(4n^{\frac{3}{4}}\log(4n^{\frac{3}{4}}+2)+2\log(4n^{\frac{3}{4}}+2)\right)\\
&=O\left(4n^{\frac{3}{4}}\log(4n^{\frac{3}{4}}+2)\right)\\
&=O\left(n^{\frac{3}{4}}\log(4n^{\frac{3}{4}}+2)\right)
\end{align*}

What is the exact argument to get rid of the $+2$ inside of the logarithm? ... because than it is clear that for $n>1$:

\begin{align*}
O\left(n^{\frac{3}{4}}\log(4n^{\frac{3}{4}})\right)&=O\left(n^{\frac{3}{4}}\log(4)+n^{\frac{3}{4}}\log(n^{\frac{3}{4}})\right)\\
&=O\left(n^{\frac{3}{4}}\log(n^{\frac{3}{4}})\right)\\
&=O\left(\frac{3}{4}n^{\frac{3}{4}}\log(n)\right)\\
&=O\left(n^{\frac{3}{4}}\log(n)\right)
\end{align*}

... is it, that for $n$ large enough $\log(4n^{\frac{3}{4}}+2)\approx\log(4n^{\frac{3}{4}})$ ?
Or am I overlooking the real arumgent?

Body:
Expanding upon my argument in the comments, since $n^{3/4} <n^{3/4} + 2 < 3n^{3/4} $ for $n>1$, we have

$$n^{3/4} \log (n^{3/4}) < (n^{3/4} + 2) \log (n^{3/4}+2)< (3n^{3/4}) \log (3n^{3/4})$$

$$\begin{align}\frac34n^{3/4} \log n < (n^{3/4} + 2) \log (n^{3/4}+2)&< \frac94 n^{3/4} \log n + 3n^{3/4} \log 3\\&<\frac94 n^{3/4} \log n + 3n^{3/4} \log (n^2)\\&=\frac{33}4n^{3/4} \log n\end{align}$$

$(n^{3/4} + 2) \log (n^{3/4}+2)$ is bounded by $Cn^{3/4} \log n$ from both sides.

This shows that $O((n^{3/4} + 2) \log (n^{3/4}+2)) = O(n^{3/4} \log n)$.


问:
Why is the function $\mu(E)=\sum_{k=1}^{\infty}(1/2^k)\delta_{q_k}(E)$ a probability measure over $\mathbb{Q}$?
In this question, the OP asks how to construct a probability measure $\mu(k):[\mathbb{Q},\mathcal{F},P]\to\mathbb{R}$ over the set $\mathbb{Q}$ of Rational numbers, where $[\mathbb{Q},\mathcal{F},P]$ is a probability space. One of the answers suggests to let $\{q_n\}_{n=1}^\infty$ be an enumeration over $\mathbb{Q}$ and then defines the measure
\begin{gather}
    \mu(E)=\sum_{k=1}^{\infty}\frac{1}{2^k}\delta_{q_k}(E),
\end{gather}
for each subset $E\subset\mathbb{Q}$, where $\delta_{q_k}(E)$ is the point mass (i.e., Dirac delta). The answer does not provide much context, and it does not explain why such a measure satisfies countable additivity nor why it returns results in the unit interval $[0, 1]\cap\mathbb{R}$, returning $0$ for the empty set and $1$ for the entire space. Could anybody please explain why such a measure is a probability measure over $\mathbb{Q}$?

BONUS QUESTION: Since the Cartesian product of any two countable spaces is also countable, could such a measure be extended to the $n$-dimensional set $\mathbb{Q}^N$?

Thank you all for your help.

答:
I think the easiest way to think about this question is to start by constructing a nice measure on the set of positive integers. Assign to the singleton $n$ the probability $1/2^n$. Then define the probability $P$ of any subset to be the sum of the probabilities of the elements it contains.

Then
$$
P(\mathbb{N}) =
 \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = 1.
$$

The countable additivity of $P$ is straightforward - you're just summing subsets of the terms in that absolutely convergent geometric series.

To define a probability measure on a countably infinite set (like the rationals) just use a bijection to the positive integers to transfer that one. That settles the bonus question.

Edit in response to comment.

To "transfer" this measure from $\mathbb{N}$ to $\mathbb{Q}$, start with your favorite bijection
$$
b: \mathbb{Q} \to \mathbb{N}.
$$
Then for each $q \in \mathbb{Q}$ let
$$
P(\{q\}) = \frac{1}{2^{b(q)}}.
$$


问:
Is principle of mathematical induction an example of inductive or deductive reasoning?
Is principle of mathematical induction an example of inductive or deductive reasoning? According to wikipedia it says it is deductive as it is just a mathematical proof but according to the definition of inductive reasoning it should be inductive instead of deductive. Can someone help?</p>

答:
Proof by induction is indeed inductive. What is the difference between inductive and deductive reasoning? Inductive reasoning starts with a specific case and generalizes it. Deductive reasoning starts with a general concept and makes it more specific.

So what happens in proof by induction? Let's step back before the proof. I'm adding up numbers 1, 2, and 3. I get 6. Okay. 1, 2, 3, 4 and I get 10. Okay. 6 = 3(4)/2. Cool! 10 = 4(5)/2. Oh cool!!

So what if it's just 1? Well, that's just 1(2)/2 = 1. Hey that works. So let's say it works for SOME k. Would that imply that it works for k+1? Yes. Proof omitted for brevity. But what this shows is we can generalize our pattern to all integers greater than or equal to one.

So what did we do? We found a few specific examples. We then showed it worked for a VERY specific example (the base case) and then assumed that it worked for a less specific example, and showed that if it does, then it works for the next example. That then shows us that it works in general. And so we went from specific to general (inductive reasoning).

On the other hand, now that we know that $\sum_{i=1}^{n}{i} = \frac{n(n+1)}{2}$, we can use it to find the specific sum, $\sum_{i=1}^{100}{i}$. Or any other specific case. And that would be deductive reasoning.


问:
Sum binomial coefficients
Show that
$$\sum_{k=0}^{n}\binom{n}{k}\left(\binom{m+k+1}{m-n}+(-1)^{k+1}\binom{m+2(n-k)+1}{m}\right)=0$$
for all integers $m$ and $n$ with $m\ge n\ge 0$.
I tried induction on $n$, but there's not a very nice way to change the LHS from the $n$ case to $n+1$.

Title: Sum binomial coefficients

Body:
We get for the first of the two pieces

$$\sum_{k=0}^n {n\choose k} {m+k+1\choose m-n}
= [z^{m-n}] (1+z)^{m+1} \sum_{k=0}^n {n\choose k} (1+z)^k
\\ = [z^{m-n}] (1+z)^{m+1} (2+z)^n.$$

The second piece is

$$\sum_{k=0}^n {n\choose k} (-1)^{k+1} {m+2(n-k)+1\choose m}
\\ = - [z^m] (1+z)^{m+2n+1}
\sum_{k=0}^n {n\choose k} (-1)^k (1+z)^{-2k}
\\ = - [z^m] (1+z)^{m+2n+1}
\left(1-\frac{1}{(1+z)^2}\right)^n
\\ = - [z^m] (1+z)^{m+1} (2z+z^2)^n
= -[z^m] z^n (1+z)^{m+1} (2+z)^n
\\ = -[z^{m-n}] (1+z)^{m+1} (2+z)^n.$$

This is the claim.


问:
Upper triangular matrix iff transformation applied to vector is in span
Let $B = v_1, . . . , v_n$ be an ordered basis for $V$ and $T : V → V$ be a linear transformation. Prove that $[T]_B$ is upper triangular if and only if $T(v_j ) \in span(v_1, . . . , v_j)$
for all $j = 1, . . . , n$.

My attempt:

Suppose that $[T]_B$ is upper-triangular. Then,

$$[T(v_j)]_B = \begin{pmatrix}
A_{1j}\\
.\\
.\\
.\\
A_{nj}
\end{pmatrix}$$

$$\implies T(v_j) = \sum_{i = 1}^{n} A_{ij}v_i$$

Since $A_{ij} = 0$ for $i>j$, it follows that $T(v_j ) \in span(v_1, . . . , v_j)$.

So, $[T]_B$ is upper-triangular $\implies T(v_j ) \in span(v_1, . . . , v_j)$.

Now, suppose $T(v_j ) \in span(v_1, . . . , v_j)$, then,

$$T(v_j) = c_1v_1 + ... + c_jv_j$$ for scalars $c_1,...,c_n \in F$

[FROM HERE, I HAVE NO IDEA WHAT TO DO]

So, can anyone explain or show what is meant to be done. I can't figure out how the matrix is formed. Also, I'm not sure if the former part of the proof is correct. Any help is much appreciated.

答: If $T(v_j)\in\operatorname{span}\{v_1,v_2,\ldots,v_j\}$, then $T(v_j)$ can be written as $a_{1j}v_1+a_{2j}v_2+\cdots+a_{jj}v_j$. But then$$[T]_B=\begin{bmatrix}a_{11}&a_{12}&a_{13}&\ldots&a_{1n}\\0&a_{22}&a_{23}&\ldots&a_{2n}\\0&0&a_{33}&\ldots&a_{3n}\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\ldots&a_{nn}\end{bmatrix},$$which is an upper triangular matrix.


问:
An interesting version of the problem "balls into bins"
Consider n people, each has k identical balls. Each people choose k different bins from m bins, constrained by the condition that there are no two people choose exactly the same k bins. For instance, there are 6 people A,B,C,D,E,F, each hold 2 balls and there are 4 different bins①，②，③，④.  If A choose ①②, B choose ①③， C choose ①④, D choose ②③，E choose ②④，and F choose③④, then we call it a proper configuration since no two people choose exactly the same 2 bins

Now each people flip a unbiased coin, if HEAD appears then he put all his ball into the k bins he has choosen, each bin with one ball. He will do nothing if TAIL appears. Here comes the problem, given a proper configration, everyone flip a coin and behave the way we described above.  Can we infer the coin result of each people based on the number of balls in each bin?

答:
Not in general.  For example, suppose there are four people $A,B,C,D$ and four bins 1,2,3,4, and suppose that $A$ chooses 1 and 2, $B$ chooses 3 and 4, $C$ chooses 1 and 4, and $D$ chooses 2 and 3. In this case, there's no way to distinguish between $A$ and $B$ flipping heads and $C$ and $D$ flipping tails, versus $A$ and $B$ flipping tails and $C$ and $D$ flipping heads.

If you want the choices of bins to be sufficient to determine the choice of people, then you need the bin choices to be linearly independent as vectors in $\mathbb{Z}_2^m$.  That is, the choices should be the basis codewords for a binary linear code of type $[m,k,2]$.

Edit: As Hagen von Eitzen points out, using a binary linear code is sufficient to determine the coin flips, but it isn't necessary.


问:
Extreme values using Taylor Expansion.

The n:th derivative of $f$ is continuous in a neighbourhood around a point $b$

$f^{'}(b)=f^{''}(b)=...=f^{(n-1)}(b)=0$

$f^{(n)}(b)\neq 0 $

From the conditions above I am supposed to come to the following conclusions:

$n\ even\ \land f^{n}(b)<0 \implies f$ has strict maxima in $b$

$n\ even\ \land f^{n}(b)>0 \implies f$ has strict minima in $b$

$n\ odd \implies f$ has no extreme in $b$

My attempt is based on Taylor's formula:

$f(x) = f(b) + \sum_{k=1}^{n-1} \frac {f^{(k)}(b)}{k!} (x-b)^{k} + R_{n}(x) \iff f(x)-f(b)= R_{n}(x)$

$R_{n}(x)$ is the Lagrange remainder $\iff R_{n}(x) = \frac {f^{(n)}(b+c \ (x-b)))}{n!} \ (x-b)^{n}$ , $0<c<1$

So $f(x)-f(b) = \frac {f^{(n)}(b+c \ (x-b)))}{n!} \ (x-b)^{n}$

I stuck here and would appreciate some hints/tips on how to carry forward.

答:
hint

By Taylor-Young formula, we can write

$$f(b+h)-f(b)=$$
$$\sum_{k=1}^n\frac{h^k}{k!}f^{(k)}(b)+\frac{h^n}{n!}f^{(n)}(b)\epsilon(h)$$

$$=\frac{h^n}{n!}f^{(n)}(b)\Bigl(1+\epsilon(h)\Bigr)$$

with $$\lim_{h\to 0}\epsilon(h)=0$$
and for small $ h $ we have $1+\epsilon(h)>0$.

Title: Extreme values using Taylor Expansion.


问:
Show $z$ can be expressed as a function of $x$ and $y$ in the equation $x+y+z=\sin(xyz)$ near $x=y=z=0$
Show $z$ can be expressed as a function of $x$ and $y$ in the equation $x+y+z=\sin(xyz)$ near $x=y=z=0$, and find the partial derivatives of the solution with respect to $x$ and to $y$.

What I tried:

Let $f(x,y,z)=x+y+z$, $g(x,y,z)=\sin(xyz)$, $F(x,y,z)=(f, g)$, I think then I should use implicit function theorem here, but

(1) The determinant of matrix of partial derivative $[DF/x, DF/y]=0$

(2) Even the determinant of (1) isn't $0$, what I got from implicit function is that functions of $x$ and $y$ in terms of $z$, not $z$ in terms of $x$ and $y$.

答:
Your $F$ is not the function you want to work with. Instead, define

$$F(x,y,z) = f(x,y,z) - g(x,y,z)$$

Note that $ \frac{\partial F}{\partial z} (0,0,0) \neq 0$ and so you can apply the implicit function theorem to conclude that there exists a function $h$ defined on a neighbourhood of $(0,0)$ in $\mathbb{R}^2$ satisfying $F(x,y, h(x,y)) =0$ and $0=h(0,0)$


问:
A gradient system
A system of the type
$$\dot{x} = f(x), \quad f(x) = -\text{grad} V(x),$$
is called a gradient system. Investigate the stability of a fixed point.
I try to compute the Lie derivative of $V$.

答: For stable, I try to use if $x_0$ is a fixed point of $f$. If there is a Liapunov function L, then $x_0$ is stable.

Title: A gradient system


问:
Laplace urn scheme

(Laplace example). We have $r + 1$ urn. In each urn, there is a $r$ of
balls, with Urn number i contains both white and $r-i$ and black
balls. The drawing takes place in the following way.

We choose one of the urns with the same probability $\frac{1}{r+1} $

We draw n + 1 times with the return from the previously selected urn. It is known that in n draws we chose only white balls. (a)
Calculate the probability that in the next (n + 1)-drawing we will
also draw white?

My solution:
Let $X_1^{i} \ldots X_n^{i} $ be outcome from ith urn.
We try compute probability
$$
P(X_1^{i} = white  \land \ldots \land X_n^{i} = white | X_{n+1}^{i}=white)= \frac{1}{r+1}(\frac{i}{r})^{n+1}
$$
It's correct thinking?
I suppose that we have independent trials.

答:
If you are new to this kind of stuff, the best way to solve a problem like this is to write down each event clearly.

For example, you have $A = \{ \text{I drew $n$ white balls consecutively from an urn I picked} \}$. This alone leads you nowhere, but you can decompose it into disjoint events like
$$
A_i = \{ \text{I drew $n$ white balls from urn #$i$ I picked}. \}
$$
Then you will find you need $D_i=\{\text{I picked urn #$i$ in the first step}\}$, as well as the conditional event $A_i | D_i$. Now you can use the total probability formula to calculate $\Pr(A)$.

Next you need $C=\{ \text{I drew $n+1$ white balls consecutively from an urn I picked}\}.$ Please start from here.

With more experiences you no longer need to write down the specifics of all the events. Instead you will be able to plugin the numbers into whatever formula in one shot.

Title: Laplace urn scheme


问:
Where is wrong in this proof
Suppose $a=b$.

Multiplying by $a$ on both sides gives $a^2 = ab$. Then we subtract $b^2$ on both sides, and get

$a^2-b^2 = ab-b^2$.

Obviously, $(a-b)(a+b) = b(a-b)$, so dividing by $a - b$, we find

$a+b = b$.

Now, suppose $a=b=1$. Then $1=2$ :)

答:
The mistake is that from the fact that $$xy = zy$$ you cannot conclude that $x=z$. You can only conclude that $x=z$, if $y \neq 0$, i.e.,
$$xy = zy \implies x=z \color{red}{\text{ or } y=0}$$

In your case, you had $$(a-b)(a+b) = b(a-b)$$
Since $a = b$, we have $a-b = 0$ and hence you cannot cancel $a-b$ from both sides, i.e.,
from the fact that
$$(a-b)(a+b) = b(a-b)$$
we can only conclude that
$$a+b = b  \color{red}{\text{ or } a-b=0}$$
The $\color{red}{\text{later}}$ is what is valid in your case, since you started with the assumption $a=b$, i.e., $a-b=0$.

Title: Where is wrong in this proof


问:
Proving all exponential properties from $b^{x+k}=b^x\cdot b^k$
Let's say that there is a function $f(x)$ which satisfies the following property:
$$f(x+k)=f(x)\space{}\cdot\space{}f(k)$$
In addition, $f(1)=b\gt1$. I am trying to prove the following property without explicitly relying on the fact that $f(x)=b^x$ (circular reasoning) or requiring that $n\in\mathbb{Z}^+$:
$$[f(x)]^n=f(nx)$$
The only way I've been able to prove it is thinking about $nx$ as being $x+x+x+x+\space{}...\space{}$ n times, and then using the assumed property of $f(x)$ to break this up as a product $n$ times which can then be written as an exponent by definition. However, this argument only really makes sense if $n$ is a natural number, but this property should hold for all $n\in\mathbb{R}$. Is there a way of extending or altering this argument so that it is still sensible for any real number value of $n$?

答:
Define $f(x+y)=f(x)\cdot f(y)$
by induction we prove that $[f(x)]^n=f(nx)$

Base case $[f(x)]^1=f(1\cdot x)=1\cdot f(x)=f(x)$

now suppose the property is true for $n$ that is $[f(x)]^n=f(nx)$

and let's prove it for $(n+1)$

$[f(x)]^{n+1}=[f(x)]^n f(x)=f(nx)f(x)=f(nx+x)=f(n+1)x)$

Proved.


问:
What is the geometric meaning of this vector equality? $\vec{BC}\cdot\vec{AD}+\vec{CA}\cdot\vec{BD}+\vec{AB}\cdot\vec{CD}=0$
I was doing some exercises for linear algebra. One of them was to prove that for any four points $A, B, C, D \in \mathbb{R}^3$ the following equality holds:
$$\overrightarrow{BC} \cdot \overrightarrow{AD}\ +\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ = 0$$
The proof is easy; you just make three vectors starting in $A$ and then see that all the terms cancel out.

My question is: what is the geometric interpretation of this equality? How can I visualize it or understand its deeper meaning? Does this equality have a name or where can I read more about it?

I'm asking this because it turns out that it is not just a random equality and is rather useful. For example, if we want to prove the existence of orthocenter, we can do it surprisingly easily and quickly using this equality.

答:
We take (assume) vector $AD$  given is directed as shown, that is, essentially it is given as the resulting sum of vectors $ AB,BC,CD  $, i.e.,

$$ \overrightarrow{AB} +\overrightarrow{BC}+ \overrightarrow{CD} =  \overrightarrow{AD}  \tag 1 $$

We verify this proposition with direct dot products of position vectors taken with $(x,y) $ components. Position vectors are shown without overhead arrows.

$$ (ab,bc,cd,ad)= [(p,q),(r,s),(u,v),(p+r+u,q+s+v)]\;\tag1$$

$$ ac= (p+r,q+s),bd=(r+u),(s+v) \tag 2 $$

then dot vector product sums of opposite sides

$$ab.cd + bc.ad =\tag 3$$

$$ p r + r^2 + q s + s^2 + p u + r u + q v + s v \tag 4$$

and dot vector product sums of diagonals

$$(ac.bd)=(ab + bc).(bc + cd)=$$

$$ p r + r^2 + q s + s^2 + p u + r u + q v + s v \tag 5  $$

Since (4) and (5) are same, the assumed sign convention in  (1) is validated and then we have the vector sign convention:

$$\overrightarrow{BC} \cdot \overrightarrow{AD}\ -\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}=0 \tag 6 $$

The central diagonal products has different signs for $ca$  and $ac$ so these relations are identical and the middle term is left as it is for consistency. Moreover the diagonal product term is given negative in Ptolemy Inequality the Wiki reference, retaining sign as it is more appropriate.

In search of a geometrical significance

The given relation is a beautiful  3-space vectorisation /generalisation of the Ptolemy theorem which deals with scalars arising from dot products.

This post has prompted me to define  New Oval shapes here in the plane allied to the Circle... that circumscribe non-cyclic quadrilaterals having the new non-zero constant as property $e$.

This approach resulted in Ptolemy theorem generalization  in the plane (where the New Ovals circumscribe plane quadrilaterals whose sides product sum and diagonals product sum bears a constant ratio $e.$

It has been verified by the OP and me that the given scalar result from dot products are valid in $\mathbb R^3 $ also. Hece it must be concluded the given vector dot products relation  is valid for a skew quadrilateral sitting inside a sphere (vertices lie on the sphere). This is a clear possible geometrical interpretation.

$$\overrightarrow{BC} \cdot \overrightarrow{AD}\ -\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ =0 \tag 7 $$

The above property has been verified in Mathematica vector computation for a zero sum taking four arbitrary points $(A,B,C,D)$ on a unit sphere computed and sketched thus:

This and what follows is strictly not part of answer, but mentioned for the sake of continuity of the subject.

It is very exciting to imagine that we could even validate in $\mathbb R^3 $ * non-spherical Ovaloid surfaces  circumscribing skew quadrilaterals* obeying the modified ( by me) Ptolemy inequality ...using the relation that includes $e$:

$$\overrightarrow{BC} \cdot \overrightarrow{AD}\ -\ \overrightarrow{CA} \cdot \overrightarrow{BD}\ +\ \overrightarrow{AB} \cdot \overrightarrow{CD}\ = e \tag 8 $$

Further on it is not difficult to demonstrate that the Ovaloid surface circumscribes a skew quadrilateral and the given scalar product including non-zero right-hand side $e$.

Finally imho the relation is having a physical significance in Mechanics rather than the asked geometrical interpretation.

Force equilibrium is easily established by zero vector sum. When right hand side  vanishes the moment equilibrium exists making for * full static equilibrium of forces and moments*.

$$ \sum F_i=0; \sum M_i=0 ;\tag9$$

When it does n't, i.e., with RHS =$e$ there is a constant  unbalanced moment in dynamic equilibrium.

$$ \sum F_i=0;\sum M_i= e ;\tag{10}$$

To establish it with finality within time available before bounty is tough for me, so this may be considered fwiw in conceptual relation to Newtonian Mechanics.

Thanks for the indulgence.. Regards


问:
Differentiating a function of x and y with respect to t if x and y are both functions of t.
Suppose I have a function such as: $$f(w) = e^{2w(x,y)}$$

where x and y are bot a function of t.

How would I calculate then the following derivative:

$$\frac{df(w)}{dt}$$

答:
Write $g(t)=e^{2t},\ \gamma(t)=(x(t),y(t))$ and $\omega:\mathbb{R^2}\to \mathbb{R}$. Then,
$$f(t)=g(\omega(\gamma(t)))=e^{2\omega(x(t),y(t))}$$

Applying the chain rule once we get
$$\tag{1}\frac{\partial}{\partial t}f(t)=J_g|_{\omega(\gamma(t))}\cdot J_{\omega\circ\gamma}|_{t}$$

where $J_g|_{\omega(\gamma(t))}$ is the Jacobian of $g$ evaluated at $\omega(\gamma(t))$ and $J_{\omega\circ\gamma}|_{t}$ is the Jacobian of $\omega\circ \gamma$ evaluated at $t$. The Jacobian of $g$ at $\omega(\gamma(t))$ is just $e^{2\omega(\gamma(t))}$. Then applying chain rule again to evaluate $J_{\omega\circ\gamma}|_{t}$ we have
$$J_{\omega\circ\gamma}|_{t}=J_{\omega}|_{\gamma(t)}\cdot J_\gamma|_t$$

Now, since $\omega:\mathbb{R^2}\to \mathbb{R},\ J_{\omega}|_{\gamma(t)}$ is just the gradient $\nabla \omega|_{\gamma(t)}=(\frac{\partial \omega}{\partial x},\frac{\partial \omega}{\partial y})|_{\gamma(t)}$. For curves $\gamma:\mathbb{R}\to \mathbb{R^2}$ the Jacobian $J_\gamma|_t$ is just the tangent vector $(x'(t),y'(t))$.

Putting all these observations in (1) we end up with
\begin{align}
\frac{\partial}{\partial t}f(t)&=e^{2\omega(\gamma(t))}\cdot \bigl(\frac{\partial \omega}{\partial x}|_{\gamma(t)},\frac{\partial \omega}{\partial y}|_{\gamma(t)}\bigr)\cdot(x'(t),y'(t))\\
&=e^{2\omega(x,y)}\cdot \bigl(x'(t)\cdot \frac{\partial \omega}{\partial x}+y'(t)\cdot \frac{\partial \omega}{\partial y}\bigr)
\end{align}