问:
Regarding definition of cuban primes
While considering the relationship between $6n-1$ (OEIS A002476) and generalized cuban primes(OEIS A007645) I came across something I thought was interesting:

Seems like the description of generalized cuban primes as: "primes of the form $x^2 + xy + y^2$; or primes of the form $x^2 + 3y^2$; or primes equal to 0 or 1 mod 3."  is a bit heavy considering:

$$f(x)=\left \lfloor 3x \over 2 \right \rfloor$$Where: $ \left \lfloor x \right \rfloor = floor(x)$

When $x > 3$ and $f(x)$ is prime, it is a generalized cuban prime.

Confirmed terms $n = 1 \to 1000$ against OEIS A007645

Seems a bit more elegant?

EDIT:

Also:

$$ f(x) = {3x \over 2}-{1\over 2}  = \left \lfloor 3x \over 2 \right \rfloor$$

答:
"When $f(x)$ is prime, it is a Cuban prime."

This doesn't seem to be true.  For example, take $x=2$.  Then $f(x)= \lfloor 3 \rfloor=3$, which is prime.  On the other hand, $3$ is not a Cuban prime, as the smallest Cuban prime is $7$.

It appears, by your references, that you are instead considering generalized Cuban primes.  That is, whereas Cuban primes are simply those arising as the difference of successive cubes, generalized Cuban primes take the form
$$p=(x^3-y^3)/(x-y),$$
i.e. as a difference of powers, divided by one trivial factor (without this division, we would find no generalized Cuban primes that are not Cuban themselves).

Now turning to your problem, suppose that $f(x)$ is prime.  If $x=2k$ is even, then
$$f(2k)= \lfloor 3(2k)/2 \rfloor = \lfloor 3k \rfloor = 3k,$$
which is composite unless $k=1$ (i.e. $x=2$ and $f(x)=3$).  Otherwise, suppose that $x=2k+1$ is odd.  Then
$$f(2k+1)= \lfloor (3(2k+1)/2\rfloor=\lfloor 3k+3/2 \rfloor =3k+1.$$
Thus $f(x)$ prime implies $p \equiv 0,1 \!\!\mod 3$, so that prime values of $f$ are in fact (generalized) Cuban primes.  Of course, since $f$ surjects onto the set of integers congruent to $1$ modulo $3$, the two sets you consider are in fact equal.

Why, then, is your definition non-standard?

The floor function implicitly introduces cases, in situations that may not require casework.

The condition $p \equiv 0,1 \!\!\! \mod 3$ makes the properties of $p$ very clear.

The other two definitions for (generalized) Cuban primes are given in terms of binary quadratic forms.  These have widespread use in number theory, and it is useful (yet sometimes non-obvious) to know when a set is related to such functions.

In short, while your definition is elegant (but perhaps less so than $p \equiv 0,1 \!\!\mod 3$), it is far from transparent, which is often more useful in mathematics.

Title: Regarding definition of cuban primes


问:
Help with solving a first order inexact initial value problem.
Been stuck on this for quite a bit. The goal is to find the solution to the initial value problem:

$$(e^{x+y}+e^y)dx + (e^{x+y}+e^x)dy = 0, where...$$
$$y(0)=0$$

So the first thing I did was check for exactness - the equation is in the form $M(x,y)dx + N(x,y)dy =0$, so taking the partial derivative with respect to y of #M(x,y)# and the same with respect to x of $N(x,y)$ doesn't change either expression. They're not equal, so the equation is inexact. Okay, cool.

Next thing I did was look for an integrating factor - we were taught to check two expressions, $(Nx-My)/N$, and $(My-Nx)/M$. I did that, and both of the resulting expressions were a function of both x and y. In this case, I'm not sure what to do - our textbook only covers the case where one or the other is either a function of x or y only.

Have I made a mistake somewhere? How can I move forwards here? I feel like I need to utilize the initial condition somehow but I haven't been able to figure out how.

Thanks for any advice you can give!

ContentLicense: CC BY-SA 4.0

Body:
It is separable.

$(e^{x+y}+e^y)dx + (e^{x+y}+e^x)dy = 0$

Dividing by $e^{x+y}$,

$(1 + e^{-x})dx + (1 + e^{-y})dy = 0$

$(1 + e^{-x})dx = -(1 + e^{-y})dy$


问:
On a local existence theorem for ODEs
I have a question about the proof of Theorem 3.10 in Applied Analysis by Hunter and Nachtergaele. The theorem is stated as follows:

Theorem.
Let $f: I \times \overline{B}_R(u_0) \to \mathbb{R}^n,$ where
$$
I = \{t \in \mathbb{R}: |t- t_0| \leq T \}
$$
is an interval in $\mathbb{R}$ and
$$
\overline{B}_R(u_0) = \{u \in \mathbb{R}^n : ||u - u_0|| \leq R \}
$$
is the closed ball of radius $R > 0$ centered at $u_0 \in \mathbb{R}^n.$ Suppose $f(t,u)$ is continuous on $I \times \overline{B}_R(u_0)$ and Lipschitz continuous with respect to $u$ uniformly in $t.$ Set
$$
M = \sup \{||f(t,u)|| : t \in I \text{ and } u \in \overline{B}_R(u_0) \} < \infty.
$$
Then the initial value problem
\begin{align*}
    u'(t) &= f(u)\\
    u(t_0) &= u_0.
\end{align*}
has a unique continuously differentiable local solution $u(t),$ defined in the time interval $|t-t_0| < \delta,$ where
$$
\delta = \min(T, R/M).
$$

The proof involves showing that the map defined by
$$
Tu(t) = u_0 + \int_{t_0}^t f(s, u(s)) ds
$$
is a contraction that maps $X \to X$, where
$$
X = \{ u : \lbrack t_0 - \eta, t_0 + \eta \rbrack \to \overline{B}_R(u_0) : u \text{ is continuous}\}
$$
for some $\eta \in (0, \delta).$ I understand how to show that $T$ maps $X \to X$, but I don't understand the last parts of the proof that $T$ contracts. I understand how we obtain the estimate
$$
|| Tu - Tv ||_\infty \leq C \eta ||u-v||_\infty,
$$
where $C$ is the Lipschitz constant for $f$. Next, the authors write that we should choose $\eta = C/2,$ and this proves $T$ is a contraction. How does this show $T$ contracts? So far as I know, the only information I have about about $C$, being the Lipschitz constant for $f$, is that $C \geq 0.$ How does it follow that $C^2/2 < 1?$ (which is what I believe we need for $T$ to be a contraction).

Lastly, the authors write that "since $\eta$ depends only on the Lipschitz constant of $f$ and on the distance $R$ of the initial data from the boundary of $\overline{B}_R(u_0)$, repeated application of this result gives a unique local solution defined for $|t-t_0| < \delta.$" How am I to understand what is meant by this last part, the "repeated application of this result?"

Thanks in advance for any and all replies.

答: Appears that the book has a typo: a newer version of this same chapter has $\eta = 1/2C$, which makes much more sense. https://www.math.ucdavis.edu/~hunter/book/ch3.pdf

Title: On a local existence theorem for ODEs


问:
Does Godel 1st theorem make sense?
It seems to me that there is 2 family of statements that I call "logically undecidable" (to distinguish with computationally undecidable which is define by turing machines), i.e. of statements for which it doesn't exist any proof that the statement is true nor a proof it is false :

The first family is defined like this : If you take a statement of this family, if you add it to the axioms as an axiom, you get a new consistent theory. And if you add its negation you get an other new consistent theory. A proof of such a statement has been done for the axiom of choice regarding ZF as theory.

The second family is defined like this : If you take a statement of this family, if you add it to the axioms as an axiom, you get an inconsistent theory (you can prove false), and if you add its negation you get an inconsistent theory too. I think any paradoxal statement, like the liar paradox, falls in this family.

A first question is : Does it really make sense to consider the statements of the second family as well defined ?

It seems to me that by accepting such a statement as well defined we are adding some incompleteness in the mathematics. Let's me explain, what arises to Godël 1st theorem if indeed these statements are considered as not well defined.

Let's say Godël is proving his 1st theorem from ZF, actually doesn't matter we could say he is doing the proof from axioms A and do the same reasoning with very acceptable hypothesis on A. Let's consider any problem proved as computationally undecidable, still from ZF, not necessarily the one Godel exhibits in his proof, let's consider the Halting Problem.

Let's show by contradiction that some part of the Halting Problem is actually independent of ZF. Assume that for any TM, there exists a proof that the TM halts or there exists a proof that the TM doesn't halt. But then you could solve the Halting Problem just like this : enumerate the proofs of ZF until you find either a proof that your input machine halts or a proof that your input machine doesn't halt. This is a contradiction since the Halting Problem is proved computationally undecidable. So, there exist a TM, let's call it U, such that the statement S : 'U halts on input x' is "logically undecidable". By the same method, for any computationally undecidable problem, it exists such a logically undecidable statement.

So now I am asking, in which of my 2 categories fall such a statement (any of them actually) ? If you add the statement to ZF, you get a theory $ZFS$ in which you can both prove S trivially and prove that you can't prove S (since the proof of undecidability still apply, you can still "write" it using the axioms of ZF in $ZFS$), so $ZFS$ is inconsistent. Well and if you add the opposite, you get a theory $ZF\overline{S}$ where you can prove both 'not S' and that that you can't prove 'not S', by exactly the same argument, so $ZF\overline{S}$  is inconsistent too. By the same method, all of these logically undecidable statement that come from computationally undecidable problems fall in the second family.

So what happens if we consider all the statement of the second family as not well define (which actually is pretty natural regarding the liar paradox) ? By contraposition, there is no longer something well define on which 1st Godël theorm apply ! Thus there is no longer any undecidable problem which is fully well defined, but you can restrict any problem to the inputs for which it is logically decidable and then the resulting restricted problem is both well defined and computationally decidable (still by enumeration of the proofs).

Even better : if we consider all the statement of the second family as not well define there is no longer anything on which the 2nd theorem of Godël apply and we could have the hope of proving the self-consistency of a well defined theory from this theory.

So my second question is, why, why are we considering such strange paradox like liar paradox as well define ? What does it add to mathematics except incompleteness ? Are you sure we really need that for anything ?

Philosophically, it seems to me that we are unnecessarily authorizing some local inconsistency and, since this inconsistency can not interact with anything, the all mathematics stay consistent, but become incomplete.

答: If the original theory is consistent, your second family is empty. If the original theory is inconsistent, all potential new axioms belong to the second family.</p>

Title: Does Godel 1st theorem make sense?


问:
Entire functions satisfying a given condition
"n" is a given positive integer. Find all entire functions satisfying $f^n(z)=z$ for every z in $\mathbb C$

Not getting any ideas. Can someone help please?

答: $f$ is one-to-one, so by Picard's "great" theorem it can't have an essential singularity at $\infty$.  We conclude that $f$ is a polynomial of degree $1$, and then it's easy to show $f(z) = c z + d$ where $c$ is an $n$'th root of unity (with $d=0$ in the case $c = 1$).


问:
Curvature of a non-compact complete surface
Assume $\Sigma$ is a non compact, complete surface. Assume the integral $$\int_{\Sigma}K$$ is convergent, where K is the Gauss curvature of $\Sigma$. Is it always true that $$\frac{1}{2\pi}\int_{\Sigma}K$$ is an integer?</p>

答: Better luck on this for complete minimal surfaces. In that case, $K$ can't change sign and complex variable theory tells you the Gauss map can't miss many values.

ContentLicense: CC BY-SA 3.0


问:
Relation of eigenspace of A to eigenspace of $A^k$
Let $A\in \mathbb{C}^{n \times n}$ with $\lambda\in\sigma(A).$ I have shown that $\lambda^k\in\sigma(A^k)$ using an induction argument. That is;

We know $Ax=\lambda x$ is true. Assume $A^kx=\lambda^k x$ is true, then by multiplying both sides by $A$ and some manipulation we obtain $A^{k+1}x=\lambda^{k+1}x.$ Hence by induction we obtain the result.

I now need to show that the eigenspace of A corresponding to $\lambda,$ $E_A(\lambda),$ satisfies the relation $E_A(\lambda)\subseteq E_{A^k}(\lambda^k)$ and I am not quite sure how to.

答:
Suppose $\;x\in E_A(\lambda)\;$ , then $\;Ax=\lambda x\;$ , and from this, by what you've proved by induction:

$$A^kx=\lambda^kx\implies x\in E_{A^k}(\lambda^k)$$


问:
Using the fundamental theorem of calculus part 1
Find $\dfrac{dy}{dx}$ if $y = \Big( \displaystyle \int_{0}^{x} (t^3+1)^{10} dt \Big)^3$.

My try at the answer:

$\dfrac{dy}{dx} = 3\cdot \Big( \displaystyle \int_{0}^{x} (t^3+1)^{10} dt \Big)^2 \cdot ( x^3+1)^{10}$

I'm using chain rule here. Would that be the correct way to go about this? Also, how would you evaluate the integral $\Big( \displaystyle \int_{0}^{x} (t^3+1)^{10} dt \Big)^2$?

答:
The chain rule does indeed apply here, although perhaps what is confusing you is what we set $u$ to. The chain rule states that
$$
\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx} \, .
$$
If $y=\left(\int_{0}^{x} (t^3+1)^{10} dt\right)^3$, then set $u$ equal to the inner function $\int_{0}^{x} (t^3+1)^{10} dt$. Then we have
\begin{align}
\frac{dy}{dx}&=\frac{d}{du}(u^3) \cdot (x^3+1)^{10} \\
&=3u^2(x^3+1)^{10} \\
&=3\left(\int_{0}^{x} (t^3+1)^{10} dt\right)^2 \cdot (x^3+1)^{10} \, .
\end{align}
So your approach was indeed correct.


问:
Interchange derivative and expectation operator
I have a function $$f_\Sigma(x) = c\det(\Sigma^{-1})\exp(-0.5\Vert\Sigma^{-1}x\Vert^2_2)$$ and a function $$g_\Sigma(x - u) = f_\Sigma(x - u)\exp(-0.5\Vert AW_\Sigma a\Vert^2_2),$$
where $W_\Sigma$ is a $n\times n$ diagonal matrix whose $(j,j)$ element is given by $$w_{jj} = \frac{f_\Sigma(x - X_j)}{\sum_{i=1}^nf_\Sigma(x - X_i)}.$$
$A$ is matrix and $a$ some vector. $c$ is a constant and $X_1,X_2,\dots,X_n$ is a collection of iid random variables. I want to verify whether $$\frac{\partial}{\partial\Sigma}\mathbb E[g_\Sigma(x - U)] = \mathbb E\left[\frac{\partial}{\partial\Sigma}g_\Sigma(x - U)\right],$$
where $U$ is a random variable having the same distribution as the $X_i$.
I found it quite difficult to compute the derivative. Since I only want to verify whether derivative and expectation can be interchanged, I was wondering whether there is an approach to bind the derivative so that the conditions of the DCT are satisfied.

Edit: To add some background, $g_\Sigma$ is the multivariate local polynomial density estimator in case of linear fitting (https://projecteuclid.org/euclid.aos/1032298287). I want to study properties of the derivative of this estimator with respect to the bandwidth parameter $\Sigma$. Eventually, I guess the background is not too important as its more related to an application of DCT to a very very complicated function. I was hoping that someone with more calculus experience has may encountered similar functions and knows a "trick" to handle the difficulties arising from the $\exp$ terms.

答:
This is a matter of interchanging of derivative and integration. Let me simplify the expressions a bit, say you want to calculate
$$
\frac\partial{\partial p} \mathbb E[ g(p,\cdot)] =\frac\partial{\partial p} \int_\Omega g(p,X) d\mu(X)
$$
for some real parameter $p$.
Perturb $p$ by some $h$, then
$$
\frac1t \left( \int_\Omega g(p+th,X) -g(p) d\mu(X) \right)
=\int_\Omega \frac{g(p+th,X) -g(p)}t d\mu(X)
=\int_\Omega \int_0^1 \frac{\partial g}{\partial p}(p+sth,X)h d\mu(X).
$$
Now one needs to pass to the limit for $t\searrow0$. Dominated convergence will help. Hope this gives an idea, how to tackle this problem.


问:
Statistics Discrete random variable question
I have a question and I don't know where to start, could you please help me?</p>
```<code>At a computer store, the monthly demand for a particular
software package is denoted by the discrete random variable X which is assumed
to follow a Poisson distribution with λ = 10.
1. Find the probability that monthly demand for this particular software package is between 8 and 12.
2. What is the expected annual number of demand for this particular software
package?
</code>```

Body:
P( 8 <= X <= 12) = P(X=8) + P(X=9) + P(X=10) + P(X=11) + P(X=12)

E( 12X ) = 12*E(X) = 12 * $\lambda$ = 120


问:
difference of 2 partial computable algorithms
I have 2 algorithms

Algorithm 1:

```<code>if( Condition1(input)==true )
   print(input);
else
   loop forever;
</code>```

Algorithm 2:

```<code>if( Condition2(input)==true )
   print(input);
else
   loop forever;
</code>```

for any arbitrary fully-computable condition 1 and 2.

We know both of this algorithm is partial computable.

i want to write a new algorithm (algorithm 3) by this definition.

Algorithm 3:

```<code>if (Condition1( input ) ==true) and (Condition2( input ) ==false)
    print input;
</code>```

Can i write algorithm3??? and how???

HINT: main problem is: can i write if( Condition2( input ) ==false )???

答:
If I didn't misunderstood you, what you require is not possible. If x is not printed by algorithm 1, then it will loop forever, never getting into condition 2, much less 3.
Now, if you want to write algorith 3 by itself, I think the way you did it is correct.</p>


问:
Smallest group for finite groups.
Does it make sense to talk about a smallest group imaginable in terms of this group being unique? I know that given the prerequisites for a group (closure, identity, inverse, associativity) and if we don't necessarily require that the subgroup be a generator for the set I could construct a group for example: $\langle \mathbb{Z}_n, +\rangle$ for the smallest possible n, but would there not be other groups containing only 1 element? Also if we are dealing with non-finite groups what would be the possible smallest group imaginable?

(Sorry beforehand if this is poorly formulated as I have only recently started to learn about group theory.)

答:
Of course there's a group with a single element $\{e\}$ and it is unique up to isomorphism. The group operation is given by $(e,e)\mapsto e$, no other choice by the way. It is also known as the trivial group. And every group contains it as a subgroup.

Except for the trivial group every finite group has a minimal subgroup. Simply because finite groups have finitely many subgroups, and nontrival groups have at least $2$ subgroups.

Infinite groups on the other hand need not have minimal subgroups, e.g. $\mathbb{Z}$ has an infinite chain of descending groups, never ending, never stabilizng, i.e.

$$\mathbb{Z}\supseteq 2\mathbb{Z}\supseteq 4\mathbb{Z}\supseteq\cdots\supseteq 2^n\mathbb{Z}\supseteq\cdots$$

Title: Smallest group for finite groups.


问:
Show that $A$ is diagonalisable when $P(A)$ is diagonalisable and $P'(A)$ is invertible
Let $n \geq 1$ and $A \in M_n(\mathbb{C})$.

Assume there exists $P \in \mathbb{C}[X]$ such that:

$P(A)$ is diagonalisable

$P'(A)$ is invertible

I have to show that $A$ is diagonalisable.

My try:

I solved the problem when $\textrm{deg}(P) \leq 1$. I also know that $A$ is trigonalisable, then I deduced that: $\forall \lambda \in \textrm{Sp}(A), P'(\lambda) \neq 0$. Also, there exists a diagonal matrix $D$ and $S \in GL_n(\mathbb{C})$ such that $P(A) = SDS^{-1}$. But I can't say anything more. Any help is welcome.

答:
Let $\sigma(A) = \{\lambda_1, \ldots, \lambda_k\}$ be the eigenvalues of $A$. We know:

$p(A)$ is diagonalizable so its minimal polynomial $m_{p(A)}$ splits into linear factors. The eigenvalues of $p(A)$ are precisely $\sigma(p(A)) = \{p(\lambda_1), \ldots, p(\lambda_k)\}$ so
$$m_{p(A)}(x) = (x-p(\lambda_1))\cdots(x-p(\lambda_k)).$$

$p'(A)$ is invertible so $0 \notin \sigma(p'(A)) = \{p'(\lambda_1), \ldots, p'(\lambda_k)\}$ or $p'(\lambda_i) \ne 0$ for all $1 \le i\le k$.

Notice that the polynomial $m_{p(A)} \circ p$ annihilates $A$ so the minimal polynomial $m_A$ of $A$ divides it, i.e. $m_A \mid m_{p(A)} \circ p$.

We wish to show that $A$ is diagonalizable, i.e. that $m_A$ splits into linear factors. The zeroes of $m_A$ are among $\lambda_1,\ldots, \lambda_k$ so let's assume that $(x-\lambda_i)^2$ divides $m_A$. Then it also divides $m_{p(A)} \circ p$ so there exists a polynomial $q \in \Bbb{C}[x]$ such that
$$m_{p(A)}(p(x)) = (x-\lambda_i)^2q(x).$$
Taking the derivative gives
$$m'_{p(A)}(p(x))p'(x) = 2(x-\lambda_i)q(x)+(x-\lambda_i)^2q'(x)$$
and plugging in $x = \lambda_i$ yields
$$m'_{p(A)}(p( \lambda_i))\underbrace{p'( \lambda_i)}_{\ne0} = 0$$
so $m'_{p(A)}(p( \lambda_i)) = 0$. This means that $p(\lambda_i)$ is a zero of $m_{p(A)}$ with multiplicity at least $2$ which contradicts the fact that $m_{p(A)}$ splits into linear factors.

Therefore it is not possible that $(x-\lambda_i)^2$ divides $m_A$ so it has to be
$$m_A(x) = (x-\lambda_1)\cdots (x-\lambda_k)$$
so $A$ is diagonalizable.


问:
Is that true that all the prime numbers are of the form $6m \pm 1$?
Is that true that all the prime numbers are of the form $6m \pm 1$ ?

If so, can you please provide an example?

Thanks in advance.

答: Except for $2$ and $3$, we can observe that no other primes are of $6n+2,6n+3,6n+4$, and by Dirichlet's theorem on arithmetic progressions, we not only see that not only all primes but 2 and 3 are of $6n\pm1$ but also can conclude that each of the two arithmetic progressions $6n+1$ and $6n-1$ contains infinitely many primes.


问:
2 different solutions on 2 calculators
I used Photomath and Microsoft Math to compute an equation, but they gave me two different results (-411 and -411/38) Why did that happen and which is the correct answer?

https://i.stack.imgur.com/egt5A.jpg
https://i.stack.imgur.com/RIHb0.png

答:
Note the first step in the second solution. It starts by multiplying both sides by 38.

Anyway, this is not a good way to do arithmetic. You are trying to find a solution that makes $\frac{411}{38}$ equal to $0$, which cannot be done.

Maybe you could try to solve $x = \frac{16}{19} - \frac{5 \times 81}{38} - 1$, which should give you the solution $x = -\frac{411}{38}$.

Or, depending on the calculator, just type in $\frac{16}{19} - \frac{5 \times 81}{38} - 1$ (without the $=0$). That should work.

Title: 2 different solutions on 2 calculators


问:
About measure theoretic interior and boundary
Reference:-  Evans-Gariepy, Federer, other books and notes of geometric measure thoery.

I just want to clarify whether these definitions of measure theoretic interior
and boundary are correct. Given a Borel regular measure $ \mu $ in $\mathbb{R}^n $, given a $\mu$-measurable subset $E \subset \mathbb{R}^n $, let $$ \psi(x,E) = \lim_{r\rightarrow 0}\frac{\mu(E\cap B(x,r))}{\mu(B(x,r))} $$ Here $\psi $ is defined wherever the limit exists. Then do we have interior and exterior as $$ Int_\mu(E) = \{ x \in \mathbb{R}^n\ |\ \psi(x,E) = 1\},\\ Ext_\mu(E) = \{ x \in \mathbb{R}^n\ |\ \psi(x,E) = 0\} $$ and boundary $$\partial_\mu(E) = \mathbb{R}^n\setminus (Int_\mu(E)\cup Ext_\mu(E)) $$ I wold be extremely obliged if someone can tell me whether these are correct and whether they imply $$ \mu(E\setminus Int_\mu(E))= \mu(E^c\setminus Ext_\mu(E)) = \mu(\partial_\mu(E)) = 0 $$ from Lebesgue differentiation theorem. Also how exactly the topological and measure theoretic boundaries are related ? There is also another question, pardon me if it's too lame. Given a $\mu $ can we define a topology on $\mathbb{R}^n $ such that it's topological interior and boundary are the measure theoretic interior and boundary ?

答:
Definitions are not statements; they cannot be correct or incorrect. They may be inconsistent with literature, or perhaps nonsensical. The definition of $\partial_\mu(x,E)$ that you have now is consistent with how the measure-theoretic boundary $\partial_* E$ is defined (when $\mu$ is the Lebesgue measure).

However, it has a problem: if $x$ is outside of the support of $\mu$, you have $0/0$ inthe definition of $\psi$, no matter what $E$ is. Since $\psi(x,E)$ is undefined, do you put such points into $\partial_\mu(E)$? Then you'll have some strange notion of boundary. For example, if $\mu$ is a Dirac mass at $a$, then $\partial_\mu(E)=\mathbb R^n\setminus \{a\}$, no matter what $E$ is (even if it's empty). This is not going to be consistent with any topological concepts.

The Lebesgue differentiation theorem  applies to absolutely continuous measures, that is, measures of the form $\mu(E)=\int_E w$. If $\mu$ is of this form (and $w>0$ almost everywhere), then you should get the desired conclusion  $\mu(\partial_\mu E)=0$ by writing
$$\frac{\mu(E\cap B(x,r))}{\mu(B(x,r))} = \frac{\mu(E\cap B(x,r))}{\lambda(B(x,r))}\cdot \frac{\lambda(B(x,r))}{\mu(B(x,r))}$$
and passing to the limit in each fraction separately. The first will converge to $w\chi_E$, the second to $w^{-1}$ at "most" points $x$, making the ratio $1$ or $0$.


问:
Prove that $ES \leq \sigma$, with S being a random sample
Assume we have a random sample S of (X1,...,Xn) from a population with finite variance $\sigma^2$.  Prove that $ES \leq \sigma$

How would you prove this?  Is there some way to do this using Jensen's inequality?  If not, what other faster way is there?

答:
Let $g(y) = y^{1/2}$. Since this is a concave function, we can apply Jensen's inequality if we flip the direction of the relationship: $E(g(y)) \leq g(E(y))$.
Setting $y = s^2$ the result $E(s) \leq \sigma$ follows naturally.


问:
To find the power series of the function $f(x) = \dfrac{4}{3x+2}$ centered at the point $x = 1$.

To find the power series of the function $f(x) = \dfrac{4}{3x+2}$ centered at the point $x = 1$.

$$f(x) = \dfrac{4}{3x+2} = \dfrac{4}{3(x-1)+5}  = \dfrac{\frac45}{1+\frac35(x-1)}. $$

We know that the power series expansion of $$\frac{a}{1-r} = \sum_{n=0}^{\infty} a r^n.$$ Here we take $a = \frac45$ and $r = -\frac35(x-1)$, then the power series of the function $f(x) = \dfrac{4}{3x+2}$ centered at the point $x = 1$ is given as follows

$$ f(x) = \sum_{n=0}^{\infty} \frac45 \left(-\frac35(x-1) \right)^n.$$

Is the derivation correct?

答:
Yes. One minor suggestion is to use properties of exponents to clean up the summand. It is preferable to write

$$f(x)= 4\sum_{n=0}^\infty (-1)^n\frac{3^n}{5^{n+1}}(x-1)^n.$$


问:
Proof that the coefficients of formal power series must match
Say I have $f(z) = \sum_n^\infty a_n z^n$ and $g(z)= \sum_n^\infty b_n z^n$. Now suppose that
$$f(z)=g(z)$$ for infinitely many $z$ that are not discrete. Why is it true that we can match the coefficients of the power series? I have seen answers say because we can write

$$\sum_n^\infty (a_n - b_n) z^n = 0$$

and then match coefficients with the zero power series, however, this is already using the answer to my question, which is why can we match the coefficients of formal power series (including matching coefficients with the zero power series)?

Any help would be greatly appriciated.

答:
Polynomia $z^n$ are a basis of the vectorial space of polynomia, so if two vectors are equal, they have the same coefficients in any basis.

A different way of understand it is using derivatives. Evaluating any polynomius at $z=0$, it is obvious that if $f(z)=g(z)\Rightarrow f(0)=g(0) \Rightarrow a_0=b_0$. You also can do it taking derivatives!

$\left.\frac{d^k}{dz^k}\right|_{z=0}f(z)=\left.\frac{d^k}{dz^k}\right|_{z=0}g(z)$

$\sum_{n=0}^\infty a_n\left.\frac{d^kz^n}{dz^k}\right|_{z=0}=\sum_{n=0}^\infty b_n\left.\frac{d^kz^n}{dz^k}\right|_{z=0}$

$\sum_{n=0}^\infty a_n k!\delta_{k,n}=\sum_{n=0}^\infty b_n k!\delta_{k,n}$

$k! a_k=k! b_k\Leftrightarrow a_k=b_k$

Where we introduced the Kroneker delta $\delta_{k,n}=\left\lbrace\begin{matrix}1 &k=m\\0&k\neq m\end{matrix}\right.$


问:
Proving a theorem in GL modal logic
I'm trying to prove a number of theorems in the system GL using the basic axioms of system K and the following axiom schema from GL: $\vdash_G \Box A \rightarrow \Box\Box A$ for every A.

One of theorems that is supposed to hold for GL is giving me a really tough time. It is: $$\vdash_G \neg\Box A \leftrightarrow \diamond (\neg A \wedge \Box A)$$

Going from right to left is not the hard part for me. When proving the biconditional in that direction, I can use the weak modal distributivity to put the possibility (diamond) around each of the arguments in the conjunct, giving me $\diamond \neg A \wedge \diamond\Box A$.

I can then simplify the conjunct and using the modal negation schema, turn $\diamond \neg A$ into $\neg\Box A$. The issue is in going left to right for the conditional because I can't build the consequent.

Am I going about this the wrong way? Should I even be doing it in two parts or proceeding from some sort of tautology that applies here and then use substitution? My issue is that it seems incorrect to say that if A is not necessary then it is possible to have not A and necessary A. The conjunct does not seem to follow at all from the antecedent nor does it seem consistent if we understand the boxes and diamonds as necessities and possibilities.

答:
The axiom you wrote down: $$\square A\rightarrow \square\square A$$
is usually called "4". And the modal logic obtained by extending the basic system K by this axiom is called K4. The logic K4 is strictly weaker than the system GL, which is K extended by Löb's axiom: $$\square(\square A\rightarrow A)\rightarrow \square A.$$

The sentence you want to prove: $$(\star)\quad\lnot \square A \leftrightarrow \lozenge(\lnot A\land \square A)$$
is not provable in K4. How can we see this? Well, K4 is sound on transitive Kripke frames, so every sentence provable in K4 is valid on every transitive Kripke frame. Consider the frame with a single world $w$ which is accessible from itself ($wRw$).  Then $R$ is transitive. Interpret $A$ so that it  does not hold at $w$. Then $\lnot \square A$ is true at $w$, but $\lozenge(\lnot A\land \square A)$ is not true at $w$, since the only world accessible from $w$ is $w$, and $\square A$ is not true at $w$. Thus $(\star)$ is not valid on this frame, and hence is not provable from K4.

So if you want to prove $(\star)$ in GL, you have to use Löb's axiom. Let's try to do that, and we'll start by rewriting $(\star)$ using propositional logic.
\begin{align*}
\lozenge(\lnot A\land \square A) &\equiv \lnot \square \lnot (\lnot A \land \square A)\\
&\equiv \lnot \square (A\lor \lnot \square A)\\
&\equiv \lnot \square (\square A\rightarrow A)
\end{align*}

Ok so to prove $(\star)$, it suffices to prove $$\lnot \square A\leftrightarrow \lnot \square (\square A\rightarrow A).$$
Taking away the $\lnot$ from both sides, it suffices to prove $$\square A\leftrightarrow \square (\square A\rightarrow A).$$ After our rewriting, we can see that the $\leftarrow$ direction is exactly Löb's axiom! And the $\rightarrow$ direction is easy (this corresponds to the easy direction you proved in your quesiton). We certainly have $A\rightarrow (\square A\rightarrow A)$, so by necessitation, $\square(A\rightarrow (\square A\rightarrow A))$, and by distributivity of $\square$ over $\rightarrow$, $\square A\rightarrow \square (\square A\rightarrow A)$.

So at the end of the day, we can see that $(\star)$ was just a fancy way to rewrite Löb's axiom (modulo propositional logic and K).

Title: Proving a theorem in GL modal logic


问:
Extension lemma for measures
Lemma 2.1.4 on Knill's book states the following: if $\mathcal{I}\subset \Omega$ is a $\pi$-system and $\mu, \nu$ are measures on $(\Omega, \sigma(\mathcal{I}))$ such that $\mu(\Omega),\nu(\Omega) < \infty$ and $\mu(A) = \nu(A)$ for every $A\in \mathcal{I}$, then $\mu = \nu$. Now, the proof consists in defining $\mathcal{D}:=\{A \in \sigma(\mathcal{I}): \hspace{0.1cm} \mu(A) = \nu(A)\}$ and show this is a Dynkin system. The first property of a Dynkin system is that it must contain $\Omega$, but I don't see why does $\Omega \in \mathcal{D}$ from the hypotheses. Any help?

Note: A $\pi$-system is just a (non-empty) set of subsets of $\Omega$ which is closed under finite intersections.

答:
I think it is implied by the given statements. Since we start off by saying:

$\mu, \nu$ are measures on $(\Omega, \sigma(\mathcal{I}))$ such that $\mu(\Omega),\nu(\Omega) < \infty$

It is implied that $\Omega \in \sigma(\mathcal{I})$.

Title: Extension lemma for measures


问:
Prove that $n^2 \leq \frac{c^n+c^{-n}-2}{c+c^{-1}-2}$.
Let $c \not= 1$ be a real positive number, and let $n$ be a positive integer.

Prove that $$n^2 \leq \frac{c^n+c^{-n}-2}{c+c^{-1}-2}.$$

My initial thought was to try and induct on $n$, but the negative powers annoyed be badly.

I still think that induction is the way to go, but maybe it is something more clever than I am doing. Although perhaps induction is unhelpful, and I simply need to use inequalities to do this, like AM-GM and Cauchy-Schwarz. But even there I get very stuck. I have a feeling that there is some neat thing I should do to make this a lot simpler.

Any help would be really appreciated. Thanks in advance.

答:
Hint: the RHS is $$\frac{-(1-c^n)(1-c^{-n})}{-(1-c)(1-c^{-1})}$$

To clarify my bonus hint, we may rearrange the AM-HM inequality to state $$(a_1+a_2+\cdots+a_n)(\frac 1a_1+\frac 1a_2+\cdots+\frac 1a_n)\ge n^2$$


问:
Show that $(1+\sqrt{3})^{2n}+(\sqrt{3}-1)^{2n} = \lfloor(1+\sqrt{3})^{2n}+1\rfloor$.
Show that $(1+\sqrt{3})^{2n}+(\sqrt{3}-1)^{2n} = \lfloor(1+\sqrt{3})^{2n}+1\rfloor$.

I know that adding the binomial expansions in LHS will cancel out the terms having radical. And we will be left with $2(1+ ^{2n}C_2 3+ \cdots \cdots +^{2n}C_{2n}3^{2n})$. Rest I don't know how to equate these.

答:
Let $a_n+b_n\sqrt{3}=(1+\sqrt{3})^{2n}$ so $a_n-b_n\sqrt{3}=(1-\sqrt{3})^{2n}=(\sqrt{3}-1)^{2n}$. (Edit, to address @diracdeltafunk: since calculating $a_n,\,b_n$ in $a_n+b_n\sqrt{3}=(1+\sqrt{3})^{2n}$ for a specific $n$ only requires using $\sqrt{3}^2=3$, it also works if we globally replace $\sqrt{3}$ with $-\sqrt{3}$, so $a_n-b_n\sqrt{3}=(1-\sqrt{3})^{2n}=(\sqrt{3}-1)^{2n}$.)

So $(1+\sqrt{3})^{2n}+(\sqrt{3}-1)^{2n}=2a_n$ is an integer greater than $(1+\sqrt{3})^{2n}$, but only slightly because $|\sqrt{3}-1|<1$. Therefore, it's $\lceil(1+\sqrt{3})^{2n}\rceil=\lfloor(1+\sqrt{3})^{2n}+1\rfloor$.


答: Both calculators gave you the correct answer: false.

Title: 2 different solutions on 2 calculators

Body:
I used Photomath and Microsoft Math to compute an equation, but they gave me two different results (-411 and -411/38) Why did that happen and which is the correct answer?

https://i.stack.imgur.com/egt5A.jpg
https://i.stack.imgur.com/RIHb0.png


答: From that expression that you are describing you can discover that the LHS is indeed an integer. To show that $\lfloor x\rfloor =y$ you have to show that $y\leq x<y+1.$ notice that $\sqrt{3}-1\leq 1$ hence $(\sqrt{3}-1)^n\leq 1\leq 1+(\sqrt{3}-1)^n.$

Body:
Show that $(1+\sqrt{3})^{2n}+(\sqrt{3}-1)^{2n} = \lfloor(1+\sqrt{3})^{2n}+1\rfloor$.

I know that adding the binomial expansions in LHS will cancel out the terms having radical. And we will be left with $2(1+ ^{2n}C_2 3+ \cdots \cdots +^{2n}C_{2n}3^{2n})$. Rest I don't know how to equate these.


问:
If $f(nx)=f(x)^n$ then $f(x+y)=f(x)f(y)$
If $f:\mathbb{R}\to\mathbb{R}_{+}$ is injective monotone function and if, $f(nx)=f(x)^n$ for all $n\in\mathbb{Z}$ and for all $x\in\mathbb{R}$, then,
$$f(x+y)=f(x)f(y),\text{ for all }x,y\in\mathbb{R}$$

It is easy to check the equality if $y$ is a multiple of $x$.

I thought about proving the inequalities, $f(x+y)\le f(x)f(y)$ and $f(x+y)\ge f(x)f(y)$, but I couldn't get to them.

Assuming that $f$ is increasing and $x<y$, then $y=kx+r$, for some $k\in\mathbb{Z}$ and $0< r\le \lvert x \rvert$, so

$$f(x+y)=f(x+kx+r)=f(x(k+1)+r)\ge f(x(k+1))=f(x)^{k+1}=f(x)f(x)^k=f(x)f(kx)$$

That is, $f(x+y)\ge f(x)f(kx)$, but the problem is that $f(kx)<f(y)$ so I can't continue the demonstration.

答:
Following the ideas presented by Mohamed Cheddadi and Martin van IJcken, it is possible to ask the question again in the following way,

Let $f:\mathbb{R}\to\mathbb{R}_{+}$ be a monotone injective function. The statements are equivalent:

$f(nx)=f(x)^n$, for all $n\in\mathbb{Z}$ and for all $x\in\mathbb{R}$

$f(x)=a^x$, for all $x\in\mathbb{R}$, where $a=f(1)$

$f(x+y)=f(x)\cdot f(y)$, for all $x,y\in\mathbb{R}$

The implication of the original question is $1. \Rightarrow 3.$, but for that we must proof $1. \Rightarrow 2.$ and then $2. \Rightarrow 3.$

$\boxed{1. \Rightarrow 2.}$

Assuming that $1.$ is valid, note that if $r =\frac{m}{n}, m\in\mathbb{Z},n\in\mathbb{N}$, we have $f(rx)=f(x)^r$ for all $x\in\mathbb{R}$. Indeed, with $nr=m$,
$$f(rx)^n=f(nrx)=f(mx)=f(x)^m$$
So
$$f(rx)^n=f(x)^m\Rightarrow (f(rx)^n)^{\frac{1}{n}}=(f(x)^m)^{\frac{1}{n}}\Rightarrow f(rx)=f(x)^{\frac{m}{n}}$$

So, if $a=f(1)$, we have to

$$f(r)=f(r\cdot 1)=f(1)^r=a^r \text{ for all } r\in\mathbb{Q}$$

Assuming without loss of generality that $f$ is an increasing function, therefore

$$1=f(x)^0 = f(0\cdot x)=f(0)<f(1)=a\Longrightarrow 1<a$$

Let's suppose by absurdity that there is $x_0 \in\mathbb{R}$, such that $f(x_0)\neq a^{x_0}$. Say $f(x_0)<a^{x_0}$. Then we can choose an $r\in\mathbb{Q}$ such that
$$f(x_0)<a^r<a^{x_0}\Rightarrow f(x_0)<f(r)<a^{x_0}$$

As $f$ is increasing, $x_0<r$, but on the other hand, like $a>1$, then if $a^r < a^{x_0}$ we have $r<x_0$, absurd!.

$\boxed{2. \Rightarrow 3.}$

Assuming that $2.$ is valid,
$$f(x+y)=f(1\cdot (x+y))=f(1)^{x+y}=f(1)^x \cdot f(1)^y = f(x)\cdot f(y)$$

$\boxed{3. \Rightarrow 1.}$

Is trivial


答:
Let $a = (1 + \sqrt{3})^{2n} + 1$ and $b = (\sqrt{3} - 1)^{2n} - 1$.

First, $$(1 + \sqrt{3})^{2n} + (\sqrt{3} - 1)^{2n} = a + b,$$

and we want to show that this equals $$\lfloor (1 + \sqrt{3})^{2n} + 1 \rfloor = \lfloor a \rfloor,$$ or in other words that $a+b$ is the largest integer less than or equal to $a$. Equivalently, we must show three things:

$a+b$ is an integer

$b \leq 0$ (so that $a+b \leq a$)

$b > -1$ (so that there cannot be any integers strictly between $a+b$ and $a$)

Each of these is not too hard! (see the other excellent answers here for hints)

Title: Show that $(1+\sqrt{3})^{2n}+(\sqrt{3}-1)^{2n} = \lfloor(1+\sqrt{3})^{2n}+1\rfloor$.


问:
How can I solve this matrix?
I have a set of RGB colour values detected by a camera $C_{i_{RGB}}$ which are to be described by the following:

$C_{i_{RGB}} = X F_{i_{rgb}}$

where $F_{i_{rgb}}$ is the component incident light at the detector, where the components $rgb$ are fixed at narrow bands and $X$, an $(n\times 3)$ matrix, is a description of the colour band overlap (cross talk if you will) for specific frequencies.  In a simplified RGB scheme, $X$ would be described:

$\begin{align}
\mathsf X &= \begin{bmatrix}
  1 & R_{GR} & R_{BR} \\
  R_{RG} & 1 & R_{BG} \\
  R_{RB} & R_{G} & 1
\end{bmatrix}\end{align}$

That is, the diagonal entries of $X$ are $1$'s, and the off-diagonals are rational.

How can I solve for the values of $X$ only knowing $C_{i_{RGB}}$? Eventually, given $C_i$ I need to find the corresponding $F_i$ using the constant $X$.

Eventually added colour bands (band passes) will be added which will turn the above relation into something like this (and I fear the case where I have 5, 6, and 7 band passes to deal with):

$\begin{bmatrix}
  C_{i_R} \\
  C_{i_G} \\
  C_{i_B}
\end{bmatrix} = \begin{align}\begin{bmatrix}
  X_{aR} & X_{bR} & X_{cR} & X_{dR} & X_{eR} & X_{fR} \\
  X_{aG} & X_{bG} & X_{cG} & X_{dG} & X_{eG} & X_{fG} \\
  X_{aB} & X_{bB} & X_{cB} & X_{dB} & X_{eB} & X_{fB}
\end{bmatrix}\end{align}\begin{bmatrix}
  F_{i_a} \\
  F_{i_b} \\
  F_{i_c} \\
  F_{i_d} \\
  F_{i_e} \\
  F_{i_f}
\end{bmatrix}$

This is for a graphics programming application and I lament that my first year algebra was 20 years ago.

答: If I understand correctly, you're trying to correlate $n$ separate equations for $n$ different values of $C_i$ and $F_i$ into a solution for $X$; is that correct?  If so, then the easiest way to go is to turn your set of vector equations into a matrix equation!  Simply write $C = [ C_1 C_2 C_3 \ldots C_n ]$ (assuming that the $C_n$ are being written as column vectors), $F = [ F_1 F_2 F_3 \ldots F_n ]$, and then by linearity you wind up with a single matrix equation $C = X F$ which can be solved by inverting F (if F isn't invertible then you have no guaranteed solution) and writing $X = C F^{-1}$.</p>

ContentLicense: CC BY-SA 2.5

Title: How can I solve this matrix?


答:
Replaced:  Sure you can do that.  You have two conditions.  The easiest way is to evaluate them first, evaluate the conjunction, print the input if the conjunction is true, then go on to the two loops you show.  You alternate between steps of each algorithm.  If they both terminate, you will evaluate the conjunction.  Otherwise, you will be looping before the conjunction, but it appears that is what you want anyway.

Added: for the new main problem, sure you can.  It will never be satisfied, because if Condition2 is false you loop forever in algorithm2.  But you can write it.  It is not a syntax error.

Title: difference of 2 partial computable algorithms

Body:
I have 2 algorithms

Algorithm 1:

```<code>if( Condition1(input)==true )
   print(input);
else
   loop forever;
</code>```

Algorithm 2:

```<code>if( Condition2(input)==true )
   print(input);
else
   loop forever;
</code>```

for any arbitrary fully-computable condition 1 and 2.

We know both of this algorithm is partial computable.

i want to write a new algorithm (algorithm 3) by this definition.

Algorithm 3:

```<code>if (Condition1( input ) ==true) and (Condition2( input ) ==false)
    print input;
</code>```

Can i write algorithm3??? and how???

HINT: main problem is: can i write if( Condition2( input ) ==false )???


问:
Does every Vector space have at least one linearly independent subset?
It might sound like a strange question to ask, but I want to know because I'm trying to prove that every infinite dimensional vector space V has a basis. The proof consists of considering the set of all linearly independent subsets of V, together with set-inclusion relation. The first step would have to be to show that this set is non-empty. Is there a trivial subset that is linearly independent or is it impossible to say? How would I argue that the set of linearly independent subsets is non-empty?

Thanks!

答:
The empty set is linearly independent: no vector in it is a linear combination of the other elements in the set (because there's no element to begin with).

And indeed it's a basis of the trivial vector space. You want to show that every vector space has a basis: so also the trivial vector space must.

If you're afraid of the empty set (but you shouldn't), prove the statement for nontrivial vector spaces. Such a space has a nonzero vector $v$ and $\{v\}$ is linearly independent.


问:
Prove that $C([0,1]),||.||)$ with $||x||= \text{sup}_{t \in [0,1]}|tx(t)|$ is not complete using equivalence of norms
Prove that $C([0,1]),||.||)$  with $||x||= \text{sup}_{t \in [0,1]}|tx(t)|$ is not complete

In https://math.stackexchange.com/posts/3908897/edit I presented this exercise solved as in my lecture notes: by taking a Cauchy sequence and proving that it does not converges to something that is not in $C([0,1]),||.||)$. I 've made another solution inspired in other proofs I've seen and I'd like to know if it is correct and ask you some doubts about which where the same I had in the inspiration of the proof

Try: If I can proof this norm is not equivalent to the ||.||∞ norm, that means that the space C([0,1]),||.||) is not complete.

In $0 \leq t \leq 1 $   $|tf(t)| \leq |f(t)|$ so $||f(t)|| \leq ||f(t)||_{\infty}$

Now the question is if there exists an $c>0$ such that  $||f(t)||_{\infty} \leq c||f(t)||$

We take a sequence defined as follows:

$$ f_n(t)= \begin{cases} \frac{1}{ t} && ,\frac{1}{n} \leq t \leq 1 \\ n &&, 0 \leq t \leq \frac{1}{n}  \end{cases}   $$

$||f_n||_{\infty}= \text{sup}_{t \in [0,1]} |f_n(t)|=n$
$||f_n||=\text{sup}_{t \in [0,1]} |t f_n(t)|=\text{max} \{\text{sup}_{0 \leq t \leq \frac{1}{n}} |t f_n(t)|,\text{sup}_{\frac{1}{n} \leq t \leq 1} |t f_n(t)| \}=\text{max} \{\text{sup}_{0 \leq t \leq \frac{1}{n}} nt,\text{sup}_{\frac{1}{n} \leq t \leq 1} \frac{1}{t}t \}= \text{max} \{1,1\}=1 $

Then $\frac{||f_n||_{\infty}}{||f_n||} \to \infty$, when $n \to \infty$. So the norms are not equivalent

My questions are, in addition to knowing if this is correct :

In this approach I don't need the sequence to be Cauchy, right?.  in order to prove tht no such c exists , why do we consider a sequence in the first place? what's the idea?

Why do they take the limit of $\frac{||f_n||_{\infty}}{||f_n||} \to \infty$ at the end, how  does this proves that no constant c exists that makes the norms equivalent ?

答:
You can prove that $(C([0,1]),||.||)$ is not complete by comparing the norm $||f||=\sup_{0\leq t\leq 1}|tf(t)|$ with the usual norm $||f||_{\infty}$ but first let me answer to your questions:

No, you dont need to show that $(f_n)$ is Cauchy in neither of two norms in order to show that there is no constant $c>0$ such that
$$\tag{1}||f||_{\infty}\leq c||f||$$

We consider the sequence $(f_n)$ because it has the property that $||f_n||=1$ whereas $||f_n||_{\infty}=n$. In other words, we are trying to find some elements $f_n$ that belong to both spaces such that the norm $||f_n||_{\infty}$ is big whereas the norm $||f_n||$ is small so that we make $(1)$ fail.

Now, as you said, we have $\lim_{n\to \infty}\frac{||f_n||_{\infty}}{||f_n||}\to \infty$ but we dont have necessarily have to take the limit in order to prove that $(1)$ is false. To see this, if there was some $c>0$ with
$$||f||_{\infty}\leq c||f||$$
for every $f$, then in particular, for the $f_n's$ we would have that $n\leq c$ for every $n$, which leads us to a contradiction.

Now why the non-equivalance of the two above norms implies that $(C([0,1]),||.||)$ is not complete its because of the open mapping theorem.

To see this, take the identity operator $I:(C([0,1]),||.||_{\infty})\to (C([0,1]),||.||)$ with $I(f)(t)=f(t)$ for $f\in C([0,1])$ and $0\leq t\leq 1.$

Since $||f||\leq ||f||_{\infty}$ it follows that $I$ is continuous, one to one and onto. Now, if $C([0,1])$ was complete with the norm $||f||=\sup_{0\leq t\leq 1}|tf(t)|$, then, by the open mapping theorem
$$I^{-1}:(C([0,1]),||.||)\to (C([0,1],||.||_{\infty})$$
would also be continuous.  But this is equivalent to the existence of some $c>0$ that satisfies $(1)$ which you have shown there cant be such $c>0$.


问:
Finding the constant coefficients of a third-order DE knowing one solution?
First off - this isn't a homework question, just something I got wrong on a midterm and can't quite figure out.

The question:

Given that $e^t cos(2t)$ is a solution of the constant coefficients differential equation $y''' + ay''-3y'+by=0$, find a and b.

Here's what I've done so far:

Since $e^t cos(2t)$ is a solution, I know $e^t sin(2t)$ must also be a solution. This means that $(1+2i)$ and $(1-2i)$ must be roots of the characteristic equation. We clearly aren't done though, as if we use $ \lambda $ for $y'$, our characteristic equation would be:

$$ ( \lambda  - ((1+2i))( \lambda  - ((1-2i))$$ or, expanded,

$$ \lambda^2 - 2\lambda -5 $$ or

$$y''-2y'-5y.$$

I know I need another simple root (I know now that it's $\lambda - 4$, giving a,b=2,20) but I can't for the life of me figure out how to find that root. I've tried setting $y''' + ay''-3y'+by= ( \lambda  - ((1+2i))( \lambda  - ((1-2i))(\lambda - x)$ and solving for x algebraically, but I keep getting stuck with too many variables.

What am I missing?

Thanks in advance!

答:
Instead of worrying about the characteristic equation, you can substitute the two solutions you know into the ODE and solve the resulting system of two equations with the two unknowns $a$ and $b$.

For instance,

$$y=e^t\cos(2t)\implies\begin{cases}y'=e^t(\cos(2t)-2\sin(2t))\\
y''=e^t(-3\cos(2t)-4\sin(2t))\\y'''=e^t(-11\cos(2t)+2\sin(2t))\end{cases}$$

$$y'''+ay''-3y'+by=0\implies (-3a+b-14)\cos(2t)-4(a-2)\sin(2t)=0$$

Similarly, $y=e^t\sin(2t)$ gives

$$(-3a+b-14)\sin(2t)+4(a-2)\cos(2t)=0$$

Now just solve for the coefficients.

If you want to continue with your method, you know the characteristic polynomial has roots $1\pm2i$, so $(\lambda-1-2i)(\lambda-1+2i)=\lambda^2-2\lambda+5$ divides it exactly and leaves no remainder. We have

$$\frac{\lambda^3+a\lambda^2-3\lambda+b}{\lambda^2-2\lambda+5}=\lambda+a+2+\frac{(2a-4)\lambda-5a+b-10}{\lambda^2-2\lambda+5}$$

For the remainder to be $0$, you need

$$\begin{cases}2a-4=0\\-5a+b-10=0\end{cases}$$


答:
Neither attempt are trying to tell you what $\frac {16}{19} -\frac {5\times 81}{38} -1$ is.  There are both trying to tell you why it is not equal to $0$.  And the both giving you correct reasons why it isn't $0$.

Consider I asked Amy and Bill  "Is $\frac 73 = 2$" and Amy said "No, if $\frac 73 = 2$ that would mean $7 = 2\times 3$ which would mean $7 = 6$ and that isn't true".  But Bill said "No, if $\frac 73 = 2$ then $\frac 7{2\times 3} = 1$ and so $\frac 76 = 1$ annd that isn't true."

Are those two different answers?  Or are they the same answer giving different but both correct arguments?  Note, neither of them actually evaluated that $\frac 73 = 2\frac 13$ and $2\frac 13 \ne 2$.

If you want to evaluate $\frac {16}{19} -\frac {5\times 81}{38} -1=\frac {16\times 2}{19\times 2} -\frac {5\times 81}{38} -\frac {38}{38}$

$\frac {16\times 2 - 5*81 -38}{38} = \frac {32-405-38}{38} = -\frac {411}{38}$.

But you don't have to evaluate it just to find out that it isn't $0$.

====

If it were me I'd argue

$\frac {16}{19} -\frac {5\times 81}{38} -1= 0 \iff$

$\frac {16}{19} -\frac {5\times 81}{38} = 1\iff$

$32 - 5*81 = 38\iff$

$32-38 = 5*81\iff$

$-6 = 5*81=405$

And as $405\ne -6$ then $\frac {16}{19} -\frac {5\times 81}{38} -1\ne 0$.

Title: 2 different solutions on 2 calculators


问:
Given the product and sum of three variables , how can we find the maximum possible value that any variable can attain?
We've three numbers $a, b, c$ such that $a+ b+c =13$
and $abc=36$. What can the largest value of a variable amongst the three be in this case?

I tried to use calculus but that didn't work neither does eliminating a variable help. Id be glad if someone pointed out or hinted out anything.
Thanks

答:
Let's suppose $a,b,c \in \Bbb N$ and $1 \le a \le b \le c$.

By the first equation we have $c \le 11$. Let's do it manually: $11 \not \mid 36$ so $c \neq 11$, moving on to the next value, $10 \not \mid 36$ so $c \neq 10$, but $9 \mid 36$ so $c = 9$ could work: we have $ab=4$ and $a+b=4$ so $a=b=2$. Thus the largest value of a variable amongst the three is $9$.


问:
Examples of inner products on a polynomial vector space
So far, I have only seen examples like the following:

$$
\langle p, q \rangle = \int_0^1 p(x)q(x)\ dx,
$$

where $p$ and $q$ are elements (polynomials) of a finite-dimensional polynomial vector space.

I'm wondering if there other kinds of inner products involving polynomials, without involving integrals or more interesting ones of the same kind.

答:
The product $\langle p|q\rangle=\sum_n\alpha_n\beta_n$ can be viewed in a fancy way setting the map

$|p\rangle=\sum_n\alpha_nx^n\rightarrow\langle p|=\sum_n\frac{\alpha_n}{n!}\left.\frac{d^n}{dx^n}\right|_{x=0}$

Then,

$\langle p|q\rangle=\sum_{n,m}\frac{\alpha_n\beta_m}{n!}\underbrace{\left.\frac{d^nx^m}{dx^n}\right|_{x=0}}_{n!\delta_{n,m}}=\sum_n \alpha_n\beta_n$.

This is clearly a map between the vector space and its dual,

$x^n\rightarrow \frac{1}{n!}\left.\frac{d^n}{dx^n}\right|_{x=0}$ and thus the basis $\left\lbrace{1,x,x^2,...}\right\rbrace$ is orthonormal.


问:
how to get the vector solution with a system of 3 variables and 4 equations
I'm doing an exercise of solid mechanics in which I got these 4 equations

Own vector associated with eI = 7,391
(i) System of equations:

$1.39116499156263*nx + 0.5*ny + 3.0*nz  = 0$

$0.5*nx - 7.39116499156263*ny + 1.5*nz  = 0$

$3.0*nx + 1.5*ny - 7.39116499156263*nz  = 0$

$nx^2 + ny^2 + nz^2 - 1  = 0$

The solutions is $ \vec{nI} = ( 0.907 ) e_1 + ( 0.142 ) e_2 + ( 0.397 ) e_3.$

I put the system in a Linear Equation Systems Calculator and I got system impossible (and I agree with the solution)

However I can't understand why the vector solution is $\vec{nI} = ( 0.907 ) e_1 + ( 0.142 ) e_2 + ( 0.397 ) e_3$ ... Could someone explain it to me? Shouldn't  the solution be the vector $(0,0,0)$?

答:
The first three homogeneous equations have a non trivial solution (i.e. not all zeros) if the determinant of its coefficients is zero.

If the first equation has the first coefficient $-1.39116499156263$ then the matrix of coefficients has determinant almost zero
$$\left|
\begin{array}{rrr}
 -1.39116 & 0.5 & 3 \\
 0.5 & -7.39116 & 1.5 \\
 3 & 1.5 & -7.39116 \\
\end{array}
\right|\approx 0
$$
Matrix rank is $2$ so we can solve the system formed by the first two equations
$$
\begin{cases}
0.5ny + 3.0nz  &= 1.39116499156263nx \\
- 7.39116499156263ny + 1.5nz & = -0.5nx\\
\end{cases}
$$
we get the non trivial solutions:
$$ny= 0.156466nx,nz= 0.437644nx$$
Plugging in the third equation we get
$$nx^2+(0.156466nx)^2+(0.437644nx)^2=1$$
therefore the results are (I suppose that $nx>0$)
$$nx= 0.907;\;ny=0.142;\; nz= 0.397$$


答:
First, assume that $0$ is an accumulation point. Since power series are continuous, this gives us $f(0)=g(0)$, so $a_0=b_0$. But then we can consider $f_1(z):=f(z)-f(0)$ and $g_1(z):=g(z)-g(0)$. Their power series expansions are $\sum_{k=1}^\infty a_k z^k$ and the same with $b_k$, respectively. We can now write $f_1(z)=z\sum_{k=1}^\infty a_k z^{k-1}$ and $g_1(z)=z\sum_{k=1}^\infty b_k z^{k-1}$. The two power series in this equation must be equal at the exact same points (except possibly at $0$) as the original ones of $f$ and $g$, so we can apply the same argument to obtain $a_1=b_1$. Repeat inductively to obtain $a_k=b_k$ for all $k\in\mathbb N$.

Now if $0$ isn't actually an accumulation point, we have to do some more work. I suggest you try drawing the discs I mention on paper. Let $U_R(0)$ be the disc of convergence (and $R$ the radius of convergence). Somewhere in this disc (not on its boundary, which you should probably state more clearly instead of just saying that the set is not discrete, because otherwise this doesn't work), there is an accumulation point $z_0$ of the set where $f=g$. We take this point and expand $f$ and $g$ around this point $z_0$. We will get two different power series for $f$ and $g$, with new coefficients. But to these two new power series we can apply the argument from above, so their coefficients are equal. We also know that their disc of convergence is at least $U_{r_0}(z_0)$, where $r_0$ the largest possible radius such that $U_{r_0}(z_0)\subseteq U_R(0)$. On this disc, the power series expansions of $f$ and $g$ are equal, so $f$ and $g$ are equal. So every boundary point of this disc $U_{r_0}(z_0)$ is an accumulation point of the set where $f=g$. And apart from the single shared boundary point with $U_R(0)$, they are also contained in $U_R(0)$. Now choose the boundary point of $U_{r_0}(z_0)$ which is closest to $0$. Call it $z_1$. Around this point we can also expand $f$ and $g$ as power series, again with a disc of convergence that is the largest possible one such that it is still contained in $U_R(0)$ (or larger). Its radius, let's call it $r_1$, will be twice that of $U_{r_0}(z_0)$. And on this disc, the functions are again equal, so its boundary points are again accumulation points of the set where $f=g$. Repeat the argument until $0$ is contained in the disc (which happens after a finite number of steps). See Maths Wizzard's excelent visualization below. At this point $0$ must be an accumulation point of the set where $f=g$, so we can finally apply our original argument to the power series.

Title: Proof that the coefficients of formal power series must match

Body:
Say I have $f(z) = \sum_n^\infty a_n z^n$ and $g(z)= \sum_n^\infty b_n z^n$. Now suppose that
$$f(z)=g(z)$$ for infinitely many $z$ that are not discrete. Why is it true that we can match the coefficients of the power series? I have seen answers say because we can write

$$\sum_n^\infty (a_n - b_n) z^n = 0$$

and then match coefficients with the zero power series, however, this is already using the answer to my question, which is why can we match the coefficients of formal power series (including matching coefficients with the zero power series)?

Any help would be greatly appriciated.


问:
Proving a result on divisibility
Let $n \ge 3$ be an integer , I need to show

(a) $6| n^{n^{n}}- n^{n}$ , and

(b) $9|n^{n^{n^{n}}}- n^{n^{n}}$

I am trying to prove both these by induction  but unable to proceed as I am confused by the endless series of algebraic manipulations that follow to prove the induction step.

Can someone tell me a better way to solve these? it would be helpful to know the technique for solving (a), so I can use it for (b) as well.

Thank you.

答:
Observe that $$1^1\equiv 1\pmod{6}$$
$$2^2\equiv 4\pmod{6}$$
$$3^3\equiv 3\pmod{6}$$
$$4^4\equiv 4\pmod{6}$$
$$5^5\equiv 5\pmod{6}$$

I can be observed that for all $n\leq 6,\space n\neq 2$, we have $n^n\equiv n\pmod{6}$

This means that $n^{n^n}\equiv n^n\pmod{6}\ \forall n\not\equiv 2\pmod{6}$.

Now when $n\equiv 2\pmod{6}$, $n^n\equiv 2^2\equiv 4\pmod{6}$ and $n^{n^n}\equiv n^4\equiv 2^4\equiv 4\pmod{6}$, thus $n^{n^n}\equiv n^n\pmod{6}\ \forall n\equiv 2\pmod{6}$

Thus $6|n^{n^n}-n^n\ \forall n\in \mathbb Z,\space n\geq 3$

I think you can solve part $2$ now

Title: Proving a result on divisibility


问:
What exactly is $\mathbb{Z_n}$?
My impression has been that $\mathbb{Z_n}$ is the set $\{0,1,...,n-1\}$ under binary operation addition modulo $n$. However I'm also coming across this notion that $\mathbb{Z_n}$ is actually a set of equivalence classes of equivalence relation $x\sim y \iff x \equiv y$ mod $n$ and the addition here is actually addition of equivalence classes rather than simply addition of integers modulo $n$. Is this correct? So would it be correct to say $\mathbb{Z_n} = \{n,n+1,...,2n-1\}$? if we are considering these elements as being equivalence classes?

答:
To put it shortly: yes.

What isn't important is what $\mathbb{Z}_n$ "is" as a set. Rather, what's important is that all these different formulations of $\mathbb{Z}_n$ (equivalence classes of integers, select integers with a certain operation, a different set of integers with a certain operation) all give rise to an algebraic structure (groups, rings, fields, depending on what you're working with) which are all isomorphic.

Though, if you're saying that $\mathbb{Z}_n$ is equivalence classes, then the accepted notation would probably look more like $\mathbb{Z}_n = \{[n]_n, [n + 1]_n, \dots\}$, or perhaps even $\mathbb{Z}_n = \{n + n\mathbb{Z}, (n + 1) + n\mathbb{Z}, \dots\}$, depending on your background and your personal aesthetic concerns. $n$, after all, is an integer, not an equivalence class of integers.

As Qiaochu Yian says in his comment, the equivalence class definition is usually preferred, as this makes certain proofs much easier.

Title: What exactly is $\mathbb{Z_n}$?


问:
Find the change of coordinates matrix
Another homework problem is asking me the following:

Given basis B = {[2 -3 0], [0 1 1], [4 0 -1]} and standard basis E = {[1 0 0], [0 1 0], [0 0 1]}, find the change of coordinates matrix M^B B sub E. Then show how you would use this to help calculate the basis B coordinates of vector [1 3 -2].

So far, I have used the formula in Theorem 15 for this basis to try and set [x]_E equal to P_E<-B, but I am missing [x].

An example I found online helped me to realize I need to use the basis B in this, but then it said something about multiplication with another matrix that wasn't an identity matrix (E in this problem). Don't I need to have an original matrix for this? If not, am I supposed to multiply basis B by the identity?

Any pointers would be greatly appreciated.

*Side note, I'm not sure if it matters, but I was unable to get the vectors to look like a column vector as they do in the problem on the handout I received.

答:
Getting the hang of changing basis can be really tricky so no worries about asking this question at all! Might I recommend watching the Youtuber "3blue1brown" and looking at his playlist essence of algebra. This video gives a really nice explanation for what is going on when we change basis: https://www.youtube.com/watch?v=P2LTAUO1TdA

I will just give you answers for your problem with a little explanation:

Firstly the matrix B that has columns equal to your basis vectors is :
$$
   B  = \begin{pmatrix}
    2 & 0 & 4  \\
    -3 & 1 & 0 \\
    0 & 1 & -1 \\
    \end{pmatrix}
$$
Let us think about what B, as a map/function/transformation does. It takes some vector $v_B$ ( a vector writen in our B basis ) and gives us the vector $v_E$ (the vector written in the standard basis B)
Try it, pick any vector, something simple like $(1,0,0)$ or $(1,1,0)$ and notice that $B(v_B) := Bv_B = v_E$ So "B takes us from our weird basis B and gives us a vector in the standard basis"

However we are after something a little different indeed. We wish to travel from the standard basis into the the world of basis B. We want to go the other way around than what the matrix $\textbf{B} $ does. So how can we do this? Why invert the matrix of course! A little algebra gives:

$$
   B^{-1}  = \frac{1}{14}\begin{pmatrix}
    1 & -4 & 4 \\
    3 & 2 & 12 \\
    3 & 2 & -2 \\
    \end{pmatrix}
$$

And hence $B^{-1}(v_E) := B^{-1}v_E = v_B $.

$B^{-1}$ takes a vector in the standard basis and gives us the same vector but in basis B So to find the basis B coordinates of vector $(1, 3, -2)^T$ We do $B^{-1} (1, 3, -2)^T $ =

$$
    \frac{1}{14}\begin{pmatrix}
    1 & -4 & 4 \\
    3 & 2 & 12 \\
    3 & 2 & -2 \\
    \end{pmatrix}
     \begin{pmatrix}
    1  \\
    3  \\
    -2  \\
    \end{pmatrix} = \frac{1}{14} \begin{pmatrix}
    -19  \\
    -15  \\
    13  \\
    \end{pmatrix} = v_B
$$

A simple check of our answer for $v_B $ shows it yields the correct result!

Hope this helped. Please if you have an issues with my soloution post them below! Good luck with studies

Oskar

Title: Find the change of coordinates matrix


问:
How to prove that $|u_{n+1} - \alpha | \le \frac{1}{2} | u_n - \alpha |$
I'm preparing for a high school math exam and I came across this question in an old exam.

Let $f(x) = \dfrac{1}{2(1+x^3)}$.

$\alpha \in (0, \frac{1}{2})$ is the only real number such that $f(\alpha) = \alpha$.

$(u_n)$ is the series such that
$$
\begin{cases}
u_0 = 0 \\
u_{n+1} = f(u_n) \quad \forall n \in \mathbb{N}
\end{cases}
$$
Prove that $|u_{n+1} - \alpha| \le \frac{1}{2} |u_n - \alpha|$

The end goal is to prove that $(u_n)$ converges to $\alpha$. But we have to do it this way instead of finding that both $(u_{2n})$ and $(u_{2n+1})$ converge to $\alpha$.

It's supposed to be a very easy question. But I have been trying for the last two hours and I couldn't find it.

It's easy to prove that $u_n \in [0, \frac{1}{2}]$ by induction.

Here's what I tried:

I tried separating the question into two cases $u_n \le \alpha$ and $u_n \ge \alpha$.
I got that I just need to prove that $u_{n+1} + \frac{1}{2} u_n \ge \frac{3}{2} \alpha$. But I didn't know how to go from here. Substituting $u_{n+1}$ with $\dfrac{1}{2(1+u_n^3)}$ seems to only make the problem more complicated.

I tried squaring both sides. It only made the expression more complicated.

The problem vaguely reminds me of the epsilon-delta definition of $\lim_{x \to \alpha} f(x) = \alpha$. But that seems like it won't lead anywhere.

答:
Hint: Since $f(\alpha)=\alpha$, you are trying to prove
$$
|u_{n+1}-\alpha|\leq \frac 12|u_n-\alpha|,
$$
which is equivalent to showing
$$
|f(u_n)-f(\alpha)|\leq \frac 12|u_n-\alpha|
$$
or
$$
\frac{|f(u_n)-f(\alpha)|}{|u_n-\alpha|}\leq \frac 12
$$
By the mean value theorem, what does the left hand side equal and how can you show it is bounded above by $1/2$?


答:
Hint:

Prove that for $x,y \in \left(0,\frac12\right)$ we have
$$|f(x)-f(y)| \le \frac12 |x-y|.$$
and then apply this to $x = u_n$ and $y = \alpha$.

For example, you can use the mean value theorem since
$$f'(x) = -\frac{3x^2}{2(1+x^3)^2}$$
is strictly decreasing on $\left(0,\frac12\right)$ and $f'\left(\frac12\right) =-\frac8{27} =-0.296296.$

Title: How to prove that $|u_{n+1} - \alpha | \le \frac{1}{2} | u_n - \alpha |$

Body:
I'm preparing for a high school math exam and I came across this question in an old exam.

Let $f(x) = \dfrac{1}{2(1+x^3)}$.

$\alpha \in (0, \frac{1}{2})$ is the only real number such that $f(\alpha) = \alpha$.

$(u_n)$ is the series such that
$$
\begin{cases}
u_0 = 0 \\
u_{n+1} = f(u_n) \quad \forall n \in \mathbb{N}
\end{cases}
$$
Prove that $|u_{n+1} - \alpha| \le \frac{1}{2} |u_n - \alpha|$

The end goal is to prove that $(u_n)$ converges to $\alpha$. But we have to do it this way instead of finding that both $(u_{2n})$ and $(u_{2n+1})$ converge to $\alpha$.

It's supposed to be a very easy question. But I have been trying for the last two hours and I couldn't find it.

It's easy to prove that $u_n \in [0, \frac{1}{2}]$ by induction.

Here's what I tried:

I tried separating the question into two cases $u_n \le \alpha$ and $u_n \ge \alpha$.
I got that I just need to prove that $u_{n+1} + \frac{1}{2} u_n \ge \frac{3}{2} \alpha$. But I didn't know how to go from here. Substituting $u_{n+1}$ with $\dfrac{1}{2(1+u_n^3)}$ seems to only make the problem more complicated.

I tried squaring both sides. It only made the expression more complicated.

The problem vaguely reminds me of the epsilon-delta definition of $\lim_{x \to \alpha} f(x) = \alpha$. But that seems like it won't lead anywhere.


问:
Interesting binomial coefficients sum: $\sum\limits_{i = 0}^b(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$
Any idea on whether or not $$\sum_{i = 0}^b(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$$
has a closed formula on $a$ and $b$ (and on what it is, in case it does)?

It is supposed that $b \le a$.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\sum_{i = 0}^{b}\pars{-1}^{i}
{b \choose i}{a + b - i - 1 \choose a - i}}
\\[5mm] = &\
\sum_{i = 0}^{b}\pars{-1}^{i}
{b \choose i}\bracks{z^{a - i}}
\pars{1 + z}^{a + b - i - 1}
\\[5mm] = &\
\bracks{z^{a}}\pars{1 + z}^{a + b - 1}\sum_{i = 0}^{b}
{b \choose i}\pars{-\,{z \over 1 + z}}^{i}
\\[5mm] = &\
\bracks{z^{a}}\pars{1 + z}^{a + b - 1}\,\,
\pars{1 - {z \over 1 + z}}^{b}
\\[5mm] = &\
\bracks{z^{a}}\pars{1 + z}^{a - 1} =
\bbx{\large 0} \\ &
\end{align}


答:
We can write
$$
|u_{n+1} - \alpha| \le \frac{1}{2} |u_n - \alpha| \iff |f(u_{n}) - f(\alpha)| \le \frac{1}{2} |u_n - \alpha|
$$
which reminds us of a famous theorem in analysis (mean value theorem). Since $f \in \mathcal{C^\infty}$, we only have to show that $\vert f' \vert \le \frac 1 2$ to apply it. But
$$
f'(x)= \frac{-3x^2}{2(1+x^3)^2}
$$
so
$$
\vert f'(x) \vert = \frac{3}{2} \left(\frac{x^2}{1+x^3}\right)^2 \le \frac{1}{2} \iff P(x)\ge0
$$
where $P(x)=\frac{x^3+1}{\sqrt{3}}-x^2$. After computing $P'$, you see that $P(x) \ge 0$. So by induction, we have
$$
|u_{n+1} - \alpha| \le \frac{1}{2} |u_n - \alpha| \le \left(\frac{1}{2}\right)^2 |u_{n-1} - \alpha| \le \cdots \le \left(\frac{1}{2}\right)^{n+1} |u_{0} - \alpha|.
$$


答:
$$A=n^{n^n}-n^n=n^n[(n^n)^{n-1}-1]$$

case 1: $n-1=2k$ the we have:

$$A=n^n[(n^{kn})^2-1]=n^n(n^{kn}-1)(n^{kn}+1)$$

If $kn=2m+1$ then:

$n^n-1|(n^{kn}-1)$

$n^n+1|(n^{kn}+1)$

Therefore:

$(n^n-1)(n^n)(n^n+1)|A$

This product of three consecutive numbers which is divisible bt $6$.

Similar thing happens when $n-1=2k+1$.

Title: Proving a result on divisibility

Body:
Let $n \ge 3$ be an integer , I need to show

(a) $6| n^{n^{n}}- n^{n}$ , and

(b) $9|n^{n^{n^{n}}}- n^{n^{n}}$

I am trying to prove both these by induction  but unable to proceed as I am confused by the endless series of algebraic manipulations that follow to prove the induction step.

Can someone tell me a better way to solve these? it would be helpful to know the technique for solving (a), so I can use it for (b) as well.

Thank you.


问:
Use identities to find the exact values of the remaining five trigonometric functions at alpha. $\sin(\alpha/2) = 3/5$ and $3\pi/4 < \alpha/2 < \pi$
I tried to solve this trig identity problem from my Trigonometry textbook and I can't seem to get the correct answer. I've solved similar trig identity problems and have successfully found the correct answers for other problems; however, I'm not familiar with being given "$\sin(\alpha/2)$". (I have the answers for this math question, but my textbook doesn't give an explanation for how to solve this particular problem).

Using the half-angle or double-angle formula was recommended to me on another math forum, but I'm unsure how to properly apply those trigonometric formulas to this specific math problem. I'm used to applying the Pythagorean Identity "$\sin^2 x + \cos^2x = 1$" to tackle this kind of problem, but since I've never been given "$\sin(\alpha/2)$" for my homework assignments or quizzes I don't have proper notes to reference for this specific math problem. :/

Correct Answers from the textbook:
$$\sin a = -24/25$$
$$\cos a = 7/25$$
$$\tan a = -24/7$$
$$\csc a = -25/24$$
$$\sec a = 25/7$$
$$\cot a = -7/24$$

Any help is appreciated ~ Thank you :)

答:
Let's begin by finding the value of $\cos(\frac{\alpha}{2})$.

$\sin(\frac{\alpha}{2}) = \frac{3}{5}$
so $\sin^2(\frac{\alpha}{2}) = \frac{9}{25}$.

$\cos^2(\frac{\alpha}{2}) + \sin^2(\frac{\alpha}{2}) = 1 \implies \cos^2(\frac{\alpha}{2}) = 1 - \frac{9}{25} = \frac{16}{25}$

$\frac{\alpha}{2} \in (\frac{3\pi}{4}, \pi)$ so $\cos(\frac{\alpha}{2}) < 0$.

As such, $\cos(\frac{\alpha}{2}) = - \sqrt{\frac{16}{25}} = -\frac{4}{5}$.

Now that we have both the values of $\sin$ and $\cos$. We can just use the half angle identities:

$\sin(\alpha) = 2 \sin(\frac{\alpha}{2}) \cdot \cos(\frac{\alpha}{2})$

And $\cos(\alpha) = \cos^2(\frac{\alpha}{2}) - \sin^2(\frac{\alpha}{2}) = 1 - 2 \sin^2(\frac{\alpha}{2})$. And deduce the rest from that.


问:
derivate by definition $f(z)=\frac{8z^2-i}{z^2+1}$
derivate by definition of taking the limit $z\to z_0$ to this funciton $f(z)=\frac{8z^2-i}{z^2+1}$

I have tried to derivate by definition but always got stucked. my last attempt got me this far: $\lim_{z\to z_0}=\frac{5z^2-11z_0^2}{(z^2+1)(z_0^2+1)(z-z_0)}$

Body:
$$\frac{\frac{8 z^2-i}{z^2+1}-\frac{8 z_0^2-i}{z_0^2+1}}{z-z_0}=[\ldots]=\frac{\frac{(8+i) (z-z_0) (z+z_0)}{\left(z^2+1\right) \left(z_0^2+1\right)}}{z-z_0}\\=\frac{(8+i) (z+z_0)}{\left(z^2+1\right) \left(z_0^2+1\right)}$$
as $z\to z_0$ we get
$$f'(z_0)=\frac{2z_0(8+i)}{(z_0^2+1)^2}$$


答:
(a)

It is not difficult to show that $n^{n^n}-n^n$ and $n^n-n$ is divisible by $2$. Write $n^n-n=2m$

Now
$$n^{n^n}-n^n=n^n(n^{n^n-n}-1)=n^n(n^{2m}-1)$$
If $n$ is divisible by $3$, then $n^{n^n}-n^n$ is also divisible by $3$

If not then we have $(n,3)=1, $ then Fermat's theorem implies $n^2 \equiv 1 \mathop{mod} 3$. Thus

$$n^2 \equiv 1 \mathop{mod} 3 \\
 n^{2m} \equiv 1 \mathop{mod} 3 \\
n^{2m}-1 \equiv 0 \mathop{mod} 3
$$

Thus we conclude that $n^{2m}-1=n^{n^n-n}-1$ is divisible by $3$. This proves that $n^{n^n}-n $ is divisible by $6$

(b)

$$n^{n^{n^n}}-n^{n^n}=n^{n^n}(n^{n^{n^n}-n^n}-1)$$
Using the previous problem, we know that $n^{n^n}-n^n$ is divisible by $6$. Write $n^{n^n}-n^n=6k$

$$n^{n^{n^n}}-n^{n^n}=n^{n^n}(n^{6k}-1) $$
Then when $n$ is divisible by $3$, then it is trivial to show that $n^{n^{n^n}}-n^{n^n}$ is divisible by $9$.

if not,then $(n,9)=1$ and Euler's theorem implies
$$n^6 \equiv 1 \mathop{mod} 9 \\
 n^{6k} \equiv 1 \mathop{mod} 9 \\
n^{6k}-1 \equiv 0 \mathop{mod} 9
$$

This proves that $n^{n^{n^n}}-n^{n^n}$ is divisible by $9$.

Title: Proving a result on divisibility


问:
Intersection of collection of sets not in the collection of sets?
Hello I was doing a practice problem and I came across something that kind of tripped me up. The question says:

Suppose the Collection $F$ is given by $F=\{ [1, 1+ 1/n] : n\in \mathbb{N}\}$

Find the union of this collection of sets and find the intersection of these sets (I'm struggling to type this).

I looked at the solution and I don't understand why the intersection is $\{1\}$. This element is not actually contained in any of the sets since all the elements of the set are intervals. Can anyone please explain why this is? Intuitively on some level it makes sense since $1$ is the only common "thing" for lack a of a better, word but isn't the intersection of a collection of sets be the elements that are common to all the sets?

答: The interval $[1,1+1/n]$ is a closed interval that contains both endpoints.  Normally an open interval is denoted as $(a,b)$, although some people use $]a,b[$.</p>


问:
Does the following hold: $\sup_{x\in X} \frac{d(x,Y)}{\Vert x \Vert} \leq 1$?
I was thinking about the following inequality and could not come up with either a counterexample or a valid proof... Some help would be highly appreciated! I would like to use this in a proof.

Thanks a lot!

Consider $(X,\Vert \cdot\Vert)$ a normed vector space and $Y\subset X$ a subspace.

Does the following hold in general ?

$\sup_{x\in X} \frac{d(x,Y)}{\Vert x \Vert} \leq 1$ where $d(x,Y)=
   \inf_{y\in Y}\Vert(x-y) \Vert $ and $x\neq0$?

答: Well, yes, because, since $0\in Y$ (I presume subspace means linear subspace?), $d(x,Y)\le \lVert x\rVert$ for all $x\in X$. Of course, there should be the caveat that sup should be taken for $x\in X\setminus\{0\}$.


问:
Units of expected value and moments of random variables
Might be a straightforward question, but I've been getting some mixed answers and I'd like to know for sure if my understanding is correct. Suppose I've some random variable $X$ and the expected value of this random variable is to be used in an equation. For example, let $X$ be the amount of apples per person, and I'm interested in calculating the amount of apples eaten by a group of people over time. A differential equation for the amount of apples in the population, $A$, may look something like this:

$$\dot{A} = - \alpha A \sum_x xp(x),$$ where each individual in the population has probability $p(x)$ of possessing $x$ apples, and $\alpha$ is the apple-eating rate. The sum is just the expected value of the number of apples in the population, and we let this be denoted by $\mu$, so our differential equation becomes $$\dot{A} = -\alpha \mu A$$ but now the units don't quite work out, as $[\alpha] = [1/time]$ and $[A] = [apples]$. Is $\mu$ unitless or am I formulating the equation incorrectly? I've even seen instances where the mean is dependent on a state variable, e.g. $\mu = N/A$ for some other state variable $N$, which would make the above differential equation $$\dot{A} = -\alpha A \left(\frac{N}{A}\right) = -\alpha N,$$ which can be a bit confusing dimension-wise, especially when $N$ has nothing to do with apples! This question applies to higher moments as well, e.g. problems of the form $$\dot{A} = -\alpha A \sum_x x^2 p(x) = -\alpha A E(X^2).$$

答:
The mean has the same units as the random variable, period. Then, for a dimensionally consistent equation, you should have $[\alpha]=([X][{\rm time}])^{-1}$

The same goes for the units of the moments, i.e. $[ E(X^k)] = [X]^k$


问:
Remove curl from vector field
I have a discretized real-valued vector field on a 3D Cartesian grid as an input, which possesses some non-zero curl. I would like to find the closest possible approximation to this vector field which can be expressed as the gradient of a potential. Or more precisely, I am looking for the potential whose gradient is the closest possible approximation to the input vector field (all on the same discrete grid). I understand that only an approximate solution can exist, because the gradient of a potential is curl-free. Can you recommend a numerical method to achieve this?</p>

答:
I think I found a method which comes quite close to what I require; it is called Helmholtz decomposition:

https://en.wikipedia.org/wiki/Helmholtz_decomposition

It expresses an arbitrary smooth vector field as a sum of a curl-free part and a divergence-free part. I could not find a proof that the curl-free part is indeed the closest possible approximation to the original vector field... Still, it looks pretty promising for my task.

Title: Remove curl from vector field


问:
On finding $2$ numbers in a set such that one divides the other.
Here is the original problem

Let $p > 5$ be a prime number and $$X = \{p - n^2| n \in \mathbb{N}, n^2 < p\}.$$ Prove that $X$ contains two distinct elements $x, y$ such that $x \neq 1$ and $x$ divides $y$.

I found solutions to this but they involved the use of floor function$\lfloor\cdot\rfloor$.

I am very much a beginner to number theory, and I wish solutions that avoid the use of these and modular arithmetic. What I essentially know is theorems in divisibility, and some that relate to primes.

For a special note, one tends to conjecture on the first few observations that $p - (n')^2\ |\ p - 1$, but this fails for $23$, for example, where $n'$ is the greatest value of $n$.

So now, okay, the hope to prove that conjecture is gone, but still I hoped that $p - (n')^2$, would have to do some thing with the proof.

But again, one can see the example for $17$, where $p - (n')^2 = 1$, which is not as what the question asks. ($x \neq 1$)

Now how should I relate the elements in the original set so that one divides the other, I don't know.

I provided what my thoughts were; though I have no idea how to exploit them, or how to create another stream of algebra that says something more about the problem. I provided, from my view, what not works for the problem.

Helping is welcome, and thanks in advance!

答:
For simplicity of notation, let $P$ be the largest integer smaller than or equal to $ \sqrt{p-2}$. This is the only use of the floor function, which you can otherwise ignore by just memorizing the definition of $P$.

We want to find a $n, m$ such that $ p -n^2 \mid p - m^2 $, subject to $ 1 \leq m < n \leq \sqrt{ p - 2 }$, or that $ 1 \leq m < n \leq P$.

Notice that $ p - n^2 \mid p - m^2 \Rightarrow p-n^2 \mid (p-m^2)  - (p-n^2) = n^2-m^2 = (n-m)(n+m)$.

Wishful thinking: It would be very nice if we had $ p-n^2  = n+m$. If so, the divisibility result is immediate.

What can we do to help ensure this could happen? What issues might arise (and can we fix them)?

Firstly, since $n + m < 2P$ is "small", let's aim to make $ p-n^2$ small. What's the smallest it can be?

 With $n = P,$ we get $p - P^2$.

Secondly, if $ p -n^2 < n$, then $m$ will be negative.

 Well, in this case, we have $ p -n^2 = n - m$, so we just need to modify our wishful thinking slightly to $ p - n^2 = n-m$ or $n+m$.

Finally, can we always have $ p - n^2 = n-m$ or $n + m$?

 Since $ 0 < m < P$, the only requirement is that $ 0 < |p - P^2 -P | < P $.

(Fill in the gaps, use the fact that $p>5$ is prime.)

Verify that this holds true unless $ p  = (P+1)^2 + 1 $ for some odd $P \geq 3$.

For this case, setting $ n = P, m = 1$ gives us $ p - P^2 = 2P + 2, p - 1 = P^2 + 2P + 1$, so $ \frac{p-m^2}{p-n^2} = \frac{P+1}{2}$ which is an integer as $P$ is odd, so we are done.

Corollary: $n = P$ always works.

Note

You could have guessed at the conclusion $n=P$ by checking small (but large enough) cases and listing out all of the $(n, m)$ that worked. You came close to this, but didn't properly deal with the $p = k^2+1$ scenario.

With this guess, it's direct to prove that it works (which likely is similar to the above).


问:
What are the eigenvalues of the matrix $A+2I$ when $A$ is given?
Question: What are the eigenvalues of the matrix $A+2I$ if
$A =\begin{bmatrix}3 & 2\\5 & 0\end{bmatrix}$?

My attempted solution: I'm actually not quite sure how I should approach this problem. I know that that the augumenting matrix $A+2I$ is $\begin{bmatrix}5 & 2\\5 & 2\end{bmatrix}$, but other than that I'm not quite sure how to go about solving this problem. I'd be much appreciated if someone could explain this to me.

Thanks in advance!

答:
Let $c_1,...,c_n$  be eigenvalues of $A$ then

$t=c_i$ $\Rightarrow$ $\det(A-tI)=0$

$\det((A+2I)-tI)=\begin{vmatrix} 5-t & 2 \\ 5 & 2-t \end{vmatrix}=(t-5)(t-2)-10=t(t-7)$

roots of determinants are $7$ and $0$ so our eigenvalues are $0$ and $7$.


问:
Give the proof of the inequality $0.5(|A+B|+|A-B|)= \max (|A|,|B|)$
$$0.5(|A+B|+|A-B|)= \max (|A|,|B|)$$
where $A$ and $B$ are any values (real number)

I have tried to solve this using triangle inequality but got $0.5(|A+B|+|A-B|)> |A|$
How to prove the given inequality??

答:
WLOG, $A\ge B\ge0$ because swapping $A,B$ and/or changing one or two signs has no effect.

Then the identity reduces to

$$\frac{A+B+A-B}2=A,$$ which is true.


问:
Show that $\exp: \mathfrak{sl}(n,\mathbb R)\to \operatorname{SL}(n,\mathbb R)$ is not surjective
It is well known that for $n=2$, this holds. The polar decomposition provides the topology of $\operatorname{SL}(n,\mathbb R)$ as the product of symmetric matrices and orthogonal matrices, which can be written as the product of exponentials of skew symmetric and symmetric traceless matrices. However I could not find  out the proof that  $\exp: \mathfrak{sl}(n,\mathbb R)\to\operatorname{SL}(n,\mathbb R)$ is not surjective for $n\geq 3$.</p>

答: Over $\mathbb{R}$, for a general $n$, a real matrix has a real logarithm if and only if it is nonsingular and in its (complex) Jordan normal form, every Jordan block corresponding to a negative eigenvalue occurs an even number of times. So, you may verify that $\pmatrix{-1&1\\ 0&-1}$ (as given by rschwieb's answer) and $\operatorname{diag}(-2,-\frac12,1,\ldots,1)$ (as given by Gokler's answer) are not matrix exponentials of a real traceless matrix.</p>


问:
How to show that closed balls is not bounded in the space of all bounded functions?
Show that closed balls is not bounded in the space of all bounded functions.

Let X be infinte. And let B(X) be the set of all bounded functions f:X$\rightarrow\mathbb{R}$ on X.

and equip it with the metric

d(f,g)=sup $\left \{ |f(x)-g(x)| : x\in X \right \}$

Show that for r>0 and $f\in$ B(X) the closed ball $B_{r}(f)$ is not compact.

I think I have to find a sequence of functions such that its limit is not bounded .
If I consider $f_n:(0,1)\rightarrow\mathbb{R}$ such that $f_n$(x)= $\frac{n}{nx+1}$ then

|$f_n$|$\leq$n and the point wise limit f(x)=1/x is not bounded in (0,1) but can't put it into a closed ball. Please help me to find this closed ball or if there is any other example you can also provide that. Thank you.

答:
Let $\{x_n\}_{n\in\Bbb{N}}$ be a countably infinite subset of $X$ and for $n \in \Bbb{N}$ let $f_n : X \to \Bbb{R}$ be a function such that
$$f(x_n) = r, \quad f(x) = 0, \forall x \in X \setminus \{x_n\}.$$
Then $(f_n)_n$ is a sequence in the closed ball $B_r(X)$ which has no convergent subsequence since $d(f_m,f_n) = r$ for all $m \ne n$.


问:
Bound on $e^{-ix}$?
Is the following bound true for all $x\in\mathbb{R}$?
$$|e^{-ix}| = |\cos(x)+i\sin(x)| \leq 2.$$
Any confirmations or counter-examples are appreciated!

答: That's fine. But it gets better:\begin{align}\left|e^{-ix}\right|&=|\cos(x)-i\sin(x)|\\&=\sqrt{\cos^2x+\sin^2(x)}\\&=1.\end{align}

Title: Bound on $e^{-ix}$?


答:
YEs it is true, but, in fact, the "bound" is better:

$$|\cos x\pm i\sin x|=\sqrt{\cos^2 x+\sin^2x}=\sqrt{1}=1.$$

Title: Bound on $e^{-ix}$?

Body:
Is the following bound true for all $x\in\mathbb{R}$?
$$|e^{-ix}| = |\cos(x)+i\sin(x)| \leq 2.$$
Any confirmations or counter-examples are appreciated!


问:
Solve $y'+a(x)y=b(x)$ where $b(x)$ is not continuous

Find all the solutionsof the equation:
$$y'+ay=b(x),\ 0<x<\infty,\ $$
where $a$ is a constant and $b(x)=1$ for $0\le x\le \alpha$, and $b(x)=0$ for $x\gt \alpha$ and $\alpha$ here is a positive constant.

Well, I really got stuck on this problems where the right hand side of  the differential equation $b(x)$ is not continuous. Could anyone give me some hints on this, please?

答: Solve the two different intervals  , $0 \le x \le \alpha$ and $x>\alpha$, separately.</p>


问:
I literally can't understand the Independence Probability
They say the coin flipping are independent events each other. (Gambler's fallacy).

And they also say that a set of probabilities is independent each other, if they satisfy P(A∩B) = P(B│A)×P(A) = P(A)×P(B).

But let's go back to the coin flipping for a second. There exists two elements in the S set, Head and Tail, and the probability of flipping Head, P(H), has only 1 element, H, so that P(H) = 1/2, which means H out of H and T.

Here, flipping a coin once again should be independent to the first one. But if they really are independent each other, it should satisfy P(H∩H) = P(H│H)×P(H) = P(H)×P(H) = 1/4, which is wrong, because P(H∩H) = P(H) = 1/2. What am I missing?

答: When you say $P(H \cap H)$ you really mean $P(H_1 \cap H_2)$ with $H_1$ indicating that the first toss is heads, and $H_2$ indicating that the second toss is heads. In this case, indeed, $P(H_1 \cap H_2) = 1/2*1/2 = 1/4$


问:
How to show symmetric product of complex space is homeomorphic to the complex space
Suppose $\def\Sym{\operatorname{Sym}} \def\C{\mathbb{C}} \Sym^n\C$ is the $n$-fold symmetric product space defined by the quotient map $q: \C^n\to \Sym^n\C$. To show $\C^n$ is homeomorphic to $\Sym^n\C$.

My effort:

I want to show: $g: \Sym^n\C \to \C^n$ defined as
$$
g \bigl( [r_0, \dots, r_{n-1}] \bigr) = \prod_{i=0\,}^{n-1}(z-r_i)
$$
is a homeomorphism, where $z$ is a formal variable. (We write element in $\C^n$ in the form of a $n$-th order polynomial in $z$.)

I have shown that $g$ is a continuous bijective by universal property. I know that one can show the inverse of $g$ is continuous. But the proof is rather long. I would like to know if there is a short proof that $g$ is open or closed.

答: The question Constructing explicit connected path for coefficients of monic polynomials with roots lying in the open left half plane? has at first glance nothing to do with your present question, but in my answer you can find a proof that $g$ is a closed map.


问:
Are piecewise functions continuous?

Determine which of three $2\pi$-periodic functions are continuous:

$f_1(x)=  \begin{cases}
                                   0 & \text{for $x\in[0,\pi[$} \\
                                   4\pi-x & \text{for $x\in[3\pi,4\pi[$} \\
  \end{cases}$

$f_2(x)=  \begin{cases}
                                   \cos(\frac{x}{2}) & \text{for $x\in[0,\pi[$} \\
                                   \frac{1}{\pi^2}(x-3\pi)^2 & \text{for $x\in[3\pi,4\pi[$} \\
  \end{cases}$

$f_3(x)=  \begin{cases}
                                   \sin(\frac{x}{2}) & \text{for $x\in[0,\pi[$} \\
                                   -\cos(x) & \text{for $x\in[3\pi,4\pi[$} \\
  \end{cases}$

Attempt

By definition, a function is continuous if and only if $\lim_{x\rightarrow x_0}f(x)=f(x_0) $ for all points in the functions range.

I tried plotting the three piecewise function to see for which the above definition holds, and it seems like it's only $f_2(x)$ that follows the definition. However, the function still has "gaps" and I'm therefore in doubt if it actually is a continuous function. Can someone clarify this for me?

Here are the plots: -

答:
You're failing to account for the fact that $f_1$, $f_2$, $f_3$ all have periods of $2\pi$. As such, each of the functions is defined on all of $\mathbb R$. Nevertheless, your question opens up a point I think is important to address.

By your definition of continuity, none of your plotted functions are continuous. This is because in order for a limit $\displaystyle{\lim_{x\to x_0}f(x)}$ to exist, the function must be defined in some open interval containing $x_0$. This won't happen in any of your functions at $x_0=\pi$.

However, there are other definitions of continuity, whereby a function is also continuous at $x_0$ if either one-sided limit exists, and the function is undefined on the other side. Under these definitions, all three of your plotted functions would be continuous.

So, the answer to "which of the functions corresponding to these three plots is continuous?" can be either "none" or "all" depending on the fine points of your definition of continuity. As to the original question, I'll leave that up to you.

Title: Are piecewise functions continuous?


答:
I will assume here that you meant $0.5(|A+B|+|A-B|) \le \max \{|A|,|B|\}$ rather than $0.5(|A+B|+|A-B|) = \max \{|A|,|B|\}.$

As often happens with absolute values, it is useful to consider two cases:

$A,B$ have the same sign (i.e. both are positive or both negative).

$A,B$ have different signs.

No generality is lost by assuming $|A|\ge|B|,$ i.e. whichever one of $A,B$ has the larger absolute value will be called $A$ and the other $B.$  Then $\max\{|A|,|B|\} = |A|.$

If $A,B$ are both positive you have
\begin{align}
& 0.5(|A+B|+|A-B|) \\[6pt]
= {} & 0.5(A+B) + 0.5|A-B| \\[6pt]
\le {} & 0.5(A+B) \le 0.5(A+A) = A = \max\{|A|,|B|\}.
\end{align}
If both are negative then a similar argument works.

Now suppose one is positive and one negative. Then $|A-B| = |A|+|B|,$ and so
\begin{align}
& 0.5(|A+B|+|A-B|) \\[6pt]
= & 0.5(|A|+|B|) + 0.5|A+B| \\[6pt]
\le {} & 0.5(|A|+|A|) = |A| = \max\{|A|,|B|\}.
\end{align}

Title: Give the proof of the inequality $0.5(|A+B|+|A-B|)= \max (|A|,|B|)$

Body:
$$0.5(|A+B|+|A-B|)= \max (|A|,|B|)$$
where $A$ and $B$ are any values (real number)

I have tried to solve this using triangle inequality but got $0.5(|A+B|+|A-B|)> |A|$
How to prove the given inequality??


问:
How to solve the limit $\lim_{x \to \infty} \frac{e^{\log(x)^{c_1}}}{2^{c_2 x}}$
$$
\lim_{x \to \infty} \frac{e^{\log(x)^{c_1}}}{2^{c_2 x}}
$$

I tried to use L'hopital rule by differentiating top and bottom, but it ended up getting more and more complicated. Here, $c_1$ and $c_2$ are constants.

答:
Suppose $a,b>0$
$$\log \left(\underset{x\to \infty }{\text{lim}}\frac{e^{\log ^a(x)}}{2^{b x}}\right)=\underset{x\to \infty }{\text{lim}}\log\frac{e^{\log ^a(x)}}{2^{b x}}=\underset{x\to \infty }{\text{lim}}\left(\log e^{\log ^a(x)}-\log2^{bx}\right)=$$
$$=\underset{x\to \infty }{\text{lim}}\left(\log^a x-bx \log 2\right)=\underset{x\to \infty }{\text{lim}}x\left(\frac{\log^a x}{x}-b\log 2\right)=-\infty$$
As the $\log$ of the limit is $-\infty$ the limit is $0$, provided that $a,b>0$.


问:
A question regarding a recursively defined sequence: $s_{n+1} = \frac{1}{2}(s_{n} + \frac{a}{s_{n}})$
I'm reading Sherbert and Bartle's Introduction to Real analysis. In the context of the study of sequences, they give as an example the following sequence, defined recursively by:

$$s_{n+1} = \frac{1}{2}(s_{n} + \frac{a}{s_{n}}),$$

where $a$ is a positive real number. $s_1$ is defined to be any arbitrary positive number. The purpose is to give an illustration of a sequence that converges to $\sqrt{a}$.

One of the things they show to prove the convergence of this sequence is that it is decreasing for $n \geq2$. To show this, they rewrite the equation above as follows:

$$s_{n}^2-2s_{n}s_{n+1}+a=0.$$

Next, they note the following: "Since $s_n$ satisfies the quadratic equation $s_{n}^2-2s_{n}s_{n+1}+a=0,$ this equation has a real root."

My question is: why does it follows that the equation mentioned has a real root? Because the sequence is defined for every $n\in \mathbb{N}$ (actually, one can show that $s_{n}$ is strictely positive for all $n\in \mathbb{N}$)?

答: $s_n$ is the real root mentioned, which is a real number.


问:
What is the splitting field of the polynomial $X^2-X+3 \in \mathbb{Q}[X]$ equal to?
What is the splitting field of the polynomial $X^2-X+3 \in \mathbb{Q}[X]$ equal to?

What I have tried:

I factorised the polynomial to get $X= \pm \frac{i\sqrt{11}}{2}$. Does this mean the answer is $\mathbb{Q}[i\sqrt{11}]$?

答: Yes, it does. $\mathbb{Q}[i\sqrt{11}]$ is the splitting field of your polynomial $p(x)$ over $\mathbb{Q}$ since it is the smallest field containing $\mathbb{Q}$ and the roots of $p(x)$. As $\mathbb{Q}[i\sqrt{11}]$ contains $\mathbb{Q}$, it contains $\frac{1}{2}$ and, because it is a field it contains $\frac{1\pm i \sqrt{11}}{2}$.


问:
How to solve this partial derivative?
The function
$u(x,t) = \frac{2}{\sqrt{\pi}}$$\int_{0}^\frac{x}{\sqrt{t}} e^{-s^2}ds$
satisfies the partial differential equation

$$\frac{\partial u}{\partial t} = K\frac{\partial^2u}{\partial x^2}$$

where $K$ is a positive constant number. Find the value of $K$.

The value of $K$ that I calculated isn't a constant. Can anyone please help with this question?
Thank you so much!

答:
Just compute the derivatives: For fixed $t>0$ consider the function $v(x)=u(x,t)$. Then, by the fundamental theorem of calculus and the chain rule, $v'(x)=\frac{2}{\sqrt{\pi}\sqrt{t}}e^{-\frac{x^2}{t}}$. Differentiating once again yields $v''(x)=\frac{2}{\sqrt{\pi}\sqrt{t}}\frac{-2x}{t}e^{-\frac{x^2}{t}}=-\frac{4x}{\sqrt{\pi} t^{\frac{3}{2}}}e^{-\frac{x^2}{t}}$.
In the same way, for fixed $x$ consider $h(t)=u(x,t)$. Once again, by the fundamental theorem of calculus and the chain rule, $h'(t)=\frac{2}{\sqrt{\pi}}\frac{-x}{2t^{\frac{3}{2}}}e^{-\frac{x^2}{t}}=-\frac{x}{\sqrt{\pi}t^{\frac{3}{2}}} e^{-\frac{x^2}{t}}$. Now, note that $v''=\partial_{xx}u$ and $h'=\partial_t u$ so that comparing the expression yields $K=\frac{1}{4}$.

Title: How to solve this partial derivative?


问:
Does the set of open sets in $\mathbb{R}$ is countable?
I want to prove that the cardinality of Borel $\sigma $ algebra is $\aleph$, using the next proposition:

If $ E \subset P(X) $ is infinite, and the cardinality of E is $\aleph $ , the $\sigma$ algebra generated by E is of cardinality $\aleph$.

Im pretty stuck with the step of proving that the generators of that $\sigma$ algebra, which are the open sets of Borel, is of the requested cardinality.
Here, I'm taking the standard topology on R, where $\aleph$ is the cardinality of continoum.

答: An open subset of $\mathbb{R}$ is a countable union of open intervals, so the open intervals generate the Borel sets.  There are $\aleph$ open intervals, so the theorem you quote completes the proof.


答:
Use the Leibniz rule:

$$u(x,t)=\frac2{\sqrt\pi}\int_0^{\frac x{\sqrt t}}e^{-s^2}\,\mathrm ds$$

$$\implies\frac{\partial u}{\partial x}=\frac2{\sqrt{\pi t}}e^{-\frac{x^2}t}\implies\frac{\partial^2u}{\partial x^2}=-\frac{4x}{t\sqrt{\pi t}}e^{-\frac{x^2}t}$$

$$\implies\frac{\partial u}{\partial t}=-\frac x{t\sqrt{\pi t}}e^{-\frac{x^2}t}$$

Then

$$\frac{\partial u}{\partial t}=K\frac{\partial^2u}{\partial x^2}\iff-1=-4K\implies K=\frac14$$

Title: How to solve this partial derivative?

Body:
The function
$u(x,t) = \frac{2}{\sqrt{\pi}}$$\int_{0}^\frac{x}{\sqrt{t}} e^{-s^2}ds$
satisfies the partial differential equation

$$\frac{\partial u}{\partial t} = K\frac{\partial^2u}{\partial x^2}$$

where $K$ is a positive constant number. Find the value of $K$.

The value of $K$ that I calculated isn't a constant. Can anyone please help with this question?
Thank you so much!


问:
Given the product and sum of three variables , how can we find the maximum possible value that any variable can attain?
We've three numbers $a, b, c$ such that $a+ b+c =13$
and $abc=36$. What can the largest value of a variable amongst the three be in this case?

I tried to use calculus but that didn't work neither does eliminating a variable help. Id be glad if someone pointed out or hinted out anything.
Thanks

答:
If you want to do it using Lagrange multiplier method, here is how you may -

We want to maximize $f(a) = a$

Given the constraint $ab (13 - a - b) = 36 \,$ ....(i)

Using Lagrange multiplier method,

$a = \lambda (13ab-a^2b-ab^2-36)$

Differentiating wrt to $a, b$,

$1 = \lambda (13b - 2ab - b^2)$

$0 = \lambda (13a - a^2 - 2ab)$

From second equation, we have $13a - a^2 - 2ab = 0$ ($\lambda \ne 0$)

i.e $a(13-a-2b) = 0$ leads to $a = 0, a + 2b = 13$.

As $a \ne 0$ given our constraint, $a + 2b = 13 \implies b = c, b = \frac{13-a}{2}$.

Substituting back in our constraint (i),

$a \times \frac{13-a}{2} \times \frac{13-a}{2} = 36$

or $\, a^3 -26a^2 + 169a - 144 = 0$

or $\, (a-1)(a-9)(a-16) = 0$.

Testing these values, we  get max and min of $9, 1$ for $a, b, c \gt 0$. $a = 16, b = c = -\frac{3}{2}$ is a solution though in real.


答:
$f(x)$ is continuos at $x=c$ given that:
$$
\lim_{x\to c^-} f(x) = \lim_{x\to c^+} f(x) = \lim_{x\to c} f(x) = f(c)
$$
Hence, $f(c)$ must be defined. $f_2(x)$ is not defined for every real number $x$ so it is not a continuous function.

Title: Are piecewise functions continuous?

Body:
Determine which of three $2\pi$-periodic functions are continuous:

$f_1(x)=  \begin{cases}
                                   0 & \text{for $x\in[0,\pi[$} \\
                                   4\pi-x & \text{for $x\in[3\pi,4\pi[$} \\
  \end{cases}$

$f_2(x)=  \begin{cases}
                                   \cos(\frac{x}{2}) & \text{for $x\in[0,\pi[$} \\
                                   \frac{1}{\pi^2}(x-3\pi)^2 & \text{for $x\in[3\pi,4\pi[$} \\
  \end{cases}$

$f_3(x)=  \begin{cases}
                                   \sin(\frac{x}{2}) & \text{for $x\in[0,\pi[$} \\
                                   -\cos(x) & \text{for $x\in[3\pi,4\pi[$} \\
  \end{cases}$

Attempt

By definition, a function is continuous if and only if $\lim_{x\rightarrow x_0}f(x)=f(x_0) $ for all points in the functions range.

I tried plotting the three piecewise function to see for which the above definition holds, and it seems like it's only $f_2(x)$ that follows the definition. However, the function still has "gaps" and I'm therefore in doubt if it actually is a continuous function. Can someone clarify this for me?

Here are the plots: -


问:
Solution for $y(x^2+y^2)dx + x(3x^2-5y^2)dy = 0$.
I am teaching assistant of an elementary course of ODE and a student asked me to solve the following equation:

$$y(x^2+y^2)dx + x(3x^2-5y^2)dy = 0$$

The standard procedure to solve it consists in multiplying the equation by some function $\mu = \mu(x,y)$. Once we do this we must solve
$$-8\mu(x,y)\cdot (x^2-y^2) +\partial_y\mu(x,y)\cdot \left(yx^2 + y^3\right) - \partial_x\mu(x,y)\left(3x^3-5xy^2\right) = 0.$$
In general one assumes that $\mu$ is constant in one of each variable, but this can not be the case here.

I tried lots of manipulations, but could not handle it. I do appreciate any hints.

答:
It is a First Order homogenous ODE (observe that all monomials have the same degree -exactly 3-). So the substitution $u=\dfrac{y}{x}$ will work.

Indeed, your ODE is $$\frac{dy}{dx}=\frac{y(x^2+y^2)}{x(5y^2-3x^2)} $$ by dividing each part by $x^3$ (remember the degree of all monomials), you get
$$\frac{dy}{dx}=\frac{y/x-(y/x)^3}{5(y/x)^2-3}$$
Now, the substitution $u=y/x$ or equivalently $y(x)=xu(x)$, whence $\frac{dy}{dx}=u+x\frac{du}{dx}$. So
$$u+x\frac{du}{dx}=\frac{u-u^3}{5u^2-3}
$$
and you can separate variables to solve it.