问:
Sum of two divergence series is always divergence series?
In my opinion, I still believe that sum of two divergence series can be convergence series.

But here is my proof:

First, we have an obvious result: sum of two convergence series is convergence series.

Take opposite of this statement is: if exist one series  is (non-convergence) so cannot be (non-convergence)

$==>$ if exist one series is divergence, so sum always be divergence.!!! (because non-convergence series is divergence series and vice verse)

If my proof is wrong, please tell me where and give me an example that  sum of two divergence series can be convergence series please.

Thanks :)

答:
The proof is wrong and here is why. The theorem states that if we add two convergent series then we get a new convergent series. So that statement is equivalent to the following. If a divergent series  was produced by sum of two series then these two series were not convergent series (so at least one is divergent).This is different than saying adding two divergent series produces a divergent series.
It is like having  A implies B
this is indeed equivalent to  notB implies notA
but is certainly different from notA implies notB.</p>


问:
Number of labelled trees on $n$ vertices containing $2$ fixed non-adjacent edges
Let $n\in \mathbb{N}$. Choose $e,f$ two disjoint pairs of integers in $1,...,n$. How many labelled trees on $n$ vertices ($1, 2, ..., n$) are there such that $e,f$ are edges of the tree?

I could not proceed in any way which made any significant progress. Any hints/ideas ?

答:
Here's a low-tech answer using Prüfer codes. Let's start by answering a couple of related questions.

How many trees are there containing $e$? Without loss of generality, we can assume $e = \{n, n - 1\}$. Then trees containing $e$ correspond exactly to Prüfer codes that end in either $n - 1$ or $n$. So there are $2n^{n - 3}$ of them.

How many trees are there containing adjacent edges $e$ and $f$? Without loss of generality, we can assume $e = \{n, n - 1\}$ and $f = \{n, n - 2\}$. Then these trees correspond exactly to Prüfer codes that end in $n$ and whose next-to-last entry is one of $\{n - 2, n - 1, n\}$. So there are $3n^{n - 4}$ of them.

Now for some counting. Suppose that if $e$ and $f$ are disjoint then there are $k$ trees containing both. If we list out every tree containing $e$, then this consists of $2n^{n - 2} \cdot (n - 2)$ non-$e$ edges. All the $2( n -2)$ edges adjacent  to $e$ show up on this list $3n^{n - 4}$ times, and all the $\binom{n}{2} - 2( n - 2) - 1$ edges not adjacent to $e$ show up $k$ times. This is true by the symmetry of $K_n$. So $$2n^{n - 3} \cdot (n - 2) = 2( n -2) \cdot 3n^{n - 4} + \left(\binom{n}{2} - 2( n - 2) - 1 \right) \cdot k$$

which means

$$
\begin{align}
k &= \frac{2 n^{n - 3} \cdot (n - 2) - 2( n -2) \cdot 3n^{n - 4} }{\binom{n}{2} - 2( n - 2) - 1} \\
&= 4 n^{n - 4}
\end{align}
$$
as required.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
Show that $p \ \equiv \ 1\ (mod \ \ 4)$ and $(r/p)^2 + (t/p)^2 \equiv 0 \ (mod \ \ n/p^2)$

Suppose $r^2 + \ t^2 \equiv \ 0\  (mod \ \ n)$, where $n$ is a positive integer, and suppose $p$ is an odd prime dividing $n$. Show that:

a) if $p$ divides neither $r$ nor $t$, then $p \ \equiv \ 1\  (mod \ \  4)$;

b) if $p$ divides one of $r$ or $t$, then it divides the other, and moreover, $p^2$ divides $n$, and $(r/p)^2 + (t/p)^2 \equiv 0 \ (mod \ \  n/p^2)$

For question a), my answer was "$p$ is an odd prime and $n$ is divisible by $r^2 + t^2$ then $p \ \equiv \ 1 \ (mod \ \ 4)$", but I think I was wrong.

For question b), I have no clue how to answer this.

I would be so thankful for your support to solve this problem !

Body:
$b)$ $(p|n$ so $p|r^{2} +t^{2}$ and $p|r(\implies p|r^{2})) \implies p|t^{2} \implies p|t$ therefore $p^{2} |r^{2},t^{2},n$.

From the problem statement, $r^{2}+t^{2} =nk$ and since $p^{2} |r^{2},t^{2},n$:

$\frac {r^{2}}{p^{2}}, \frac {t^{2}}{p^{2}}, \frac {n}{p^{2}} = (\frac{r}{p})^{2}, (\frac {t}{p})^{2}, \frac {n}{p^{2}}$ respectively, are all integers.

Hence, $r^{2}+t^{2} =nk \implies \frac {r^{2}}{p^{2}}+ \frac {t^{2}}{p^{2}}=\frac {n}{p^{2}}k$ so that $ (\frac{r}{p})^{2}+ (\frac {t}{p})^{2}\equiv 0 \pmod {\frac{n}{p^{2}}}$.


问:
Limit approximates $0.\infty$ to 0
A question defines $$f(x)=\begin{cases}g(x)\cos\frac{1}{x}&\text{if } x\neq0\\0&\text{if } x=0\end{cases}.$$ Here, $g$ is an even function and  differentiable at $0$, and $g(0)=0$. You're required to find $f'(0)$.

On differentiating $f(x)$ to find it, you get $$g'(x)\cos\frac{1}{x}+\frac{g(x)\sin\frac{1}{x}}{x^2}$$. The first term is obviously $0$ at $x=0$, but the second term is in a $\frac{0}{0}$ indeterminate form, as both $g$ and $x^2$ approach 0 at x=0. But all the solutions I've seen simplify that term to zero, too.

Isn't that incorrect? I tried using L'Hôpital's Rule on $\frac{g(x)}{x^2}$, but since we don't know what g'(x) is, it doesn't help much. So how does that term end up as 0?

答:
Since $g$ is even and differentiable at $0$ we must have $g'(0)=0$.

All you have to do is apply the definition of derivative. $f'(0)=\lim_{x \to 0} \frac {f(x)-f(0)} x=\lim_{x \to 0} \frac {g(x)} x \cos (\frac 1x)=0$ because $ \frac {g(x)} x \to 0$ and $\cos (\frac 1 x)$ is bounded.

Title: Limit approximates $0.\infty$ to 0


问:
Prime number generator, how to make
Can anybody point me an algorithm to generate prime numbers, I know of a few ones (Mersenne, Euclides, etc.) but they fail to generate much primes...

The objective is:
    given a first prime, generate the 'n' next primes. But thanks for the link ;-)

```<code>for example : primes( 17, 50 ) -> Generate 50 consecutive primes starting at 17
</code>```

and do not fail any prime in this 50... no holes!

Title: Prime number generator, how to make

Body:
It is possible to do this using Alpertron.  If you input

x=17;x=n(x);i-50;x.

into its Batch factorization tool, it will output

17 is prime
19 is prime

[...snipped the list...]

257 is prime
263 is prime

To make it more interesting, we could try

x=n(10^100);x=n(x);i-50;x.

which finds the first 50 primes after $10^{100}$.  It takes a few seconds on my computer, but gives:

10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000267 is prime
10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000949 is prime

[...snipped the list...]

10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000016431 is prime
10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000016443 is prime

I believe these are Rabin-Miller pseudoprimes (which would be extremely unlikely to be composite), but if this is not suitable the output can be formally checked using e.g. Primo.


问:
Are the minimal polynomials of $A$ and $A^T$ related?
I'm wondering if the minimal polynomials of $A, A^T\in\mathbb{M}_{n\times n}(\mathbb{F})$ are somehow related? Are they the same?

If $m_A(t)$ is the minimal polynomial of $A$, then $m_A(A) = 0$. So, $(m_A(A))^T = 0$. If two matrices $P,Q$ commute then $(PQ)^T = P^TQ^T$. Using this, we can see that $m_A(A^T) =0$.

Is this enough to conclude that $m_A(t)$ is also the minimal polynomial for $A^T$?

P.S.
Does this depend on whether or not the minimal polynomial splits over $\mathbb{F}$?

答:
You don't need to consider any factorisation. Let $p(x)=\sum_{i=1}^ka_ix^i$ be any monic polynomial. If $p(A)=0$, then
$$
p(A^T)=\sum_{i=1}^ka_i(A^T)^i=\left(\sum_{i=1}^ka_iA^i\right)^T=p(A)^T=0.
$$
Since $A=(A^T)^T$, it is also true that $p(A^T)=0\Rightarrow p(A)=0$. Therefore $A$ and $A^T$ share the same set of annihilating polynomials. The monic polynomial of the lowest degree in this set is the common minimal polynomial of $A$ and $A^T$.


问:
Approximation of $\Big[\Gamma(1+x)\Big]^{-1}$ for $0 \leq x \leq 1$ (for the art for art's sake).
Recently was asked on the site the question : how to compute
$$f(t)=\int_0^t\frac{\mu^x}{\Gamma(x+1)}\ dx$$ for which I assumed $0 \leq t \leq 1$. There is no antiderivative available from any CAS I tried.

My first idea was to approximate $\Gamma(x+1)$ by a more or less constrained polynomial in $x$, then to use partial fraction decomposition. This leads to a linear combination of elliptic integrals.

Later, I found that  $\Big[\Gamma(1+x)\Big]^{-1}$ looks to be very similar to a Gibbs excess energy diagram for a binary system. Some models of this physical properties are very simple (Van Laar, Margules), but they only have two parameters which is not enough for an "accurate" representation of the function. Much more accurate are  Scatchard-Hildebrand, Wilson, NRTL or Uniquac models but  their complexity would not allow the required integration.

So, with a limited choice, I decided to try with a model looking like the one proposed by Redlich-Kister which contains a pure polynomial component. So my idea was to write
$$\frac{1}{\Gamma(x+1)}\sim 1+x(x-1) \sum_{k=0}^p d_k\, x^k$$
$$f(t)=\frac{\mu ^t-1}{L}+$$ $$\sum_{k=0}^p (-1)^k\,d_k  L^{-(k+3)} (L\, \Gamma (k+2,-L t)+\Gamma (k+3,-L t)-(k+L+2) \Gamma (k+2))$$ where $L=\log(\mu)$ (provided that $\Re(\log (\mu ))<0\land \Re(k)>-2$).

Willing to stay non-empirical, in my answer, I used $p=3$ and the $d_k$'s were computed in order to  match the function and first derivative values at $x=0$, $x=\frac 12$ and $x=1$. The result were no too bad.

For the art for art's sake, I decided to use $p=6$, the $d_k$'s were computed in order to  match the function, first and second derivative values at $x=0$, $x=\frac 12$ and $x=1$. This leads to
$$d_0=-\gamma \qquad d_1=-\gamma -\frac{\gamma ^2}{2}+\frac{\pi ^2}{12}$$

$$4 \sqrt \pi\,d_2=6 (178+3 \gamma  (8+\gamma )) \sqrt{\pi }+32 \pi ^2-3 \pi ^{5/2}-64 \left(P^2+4 P+36\right)$$
$$3 \sqrt \pi\,d_3=-3 (1356+\gamma  (111+16 \gamma )) \sqrt{\pi }-144 \pi ^2+8 \pi ^{5/2}+96
   \left(3 P^2+8 P+92\right)$$
$$3 \sqrt \pi\,d_4=2 \left(3 (1376+\gamma  (61+14 \gamma )) \sqrt{\pi }+156 \pi ^2-7 \pi ^{5/2}-24
   \left(13 P^2+20 P+372\right)\right)$$
$$\frac{\sqrt{\pi }}{4}\,d_5=-\left(628+9 \gamma +6 \gamma ^2\right) \sqrt{\pi }-24 \pi ^2+\pi ^{5/2}+16
   \left(3 P^2+2 P+84\right)$$
$$3 \sqrt \pi\,d_6=-4 \left(-6 (106+(\gamma -1) \gamma ) \sqrt{\pi }-24 \pi ^2+\pi ^{5/2}+48
   \left(P^2+28\right)\right)$$ where
$P=\psi \left(\frac{3}{2}\right)=2-\gamma -2\log (2)$

The maximum absolute error is $1.5 \times 10^{-8}$ which seems to be decent.

Making the numbers rational, this would give
$$\frac{1}{\Gamma(x+1)}\sim 1+x(1-x) P_6(x)$$
$$P_6(x)=\frac{2807}{4863}-\frac{247}{3140}x-\frac{461 }{3820}x^2+\frac{66
   }{1435}x^3+\frac{11 }{3303}x^4-\frac{15 }{2726}x^5+\frac{3
  }{2750}x^6$$

A few values of the definite integral (just for comparison)
$$\left(
\begin{array}{cccc}
\mu & t & \text{approximation} & \text{exact}     \\
 0.25 & 0.2 & 0.182857067200 & 0.182857068268 \\
 0.25 & 0.4 & 0.329965795201 & 0.329965797571 \\
 0.25 & 0.6 & 0.443007596935 & 0.443007599310 \\
 0.25 & 0.8 & 0.526656210875 & 0.526656212492 \\
 0.25 & 1.0 & 0.586607844050 & 0.586607845209 \\
      &     &                &                \\
 0.50 & 0.2 & 0.195704099746 & 0.195704100926 \\
 0.50 & 0.4 & 0.376453536397 & 0.376453539149 \\
 0.50 & 0.6 & 0.535922898037 & 0.535922900791 \\
 0.50 & 0.8 & 0.671420159082 & 0.671420160584 \\
 0.50 & 1.0 & 0.782934567076 & 0.782934567750 \\
      &     &                &                \\
 0.75 & 0.2 & 0.203784473259 & 0.203784474510 \\
 0.75 & 0.4 & 0.407825165336 & 0.407825168343 \\
 0.75 & 0.6 & 0.602995831714 & 0.602995834721 \\
 0.75 & 0.8 & 0.782793915893 & 0.782793917219 \\
 0.75 & 1.0 & 0.943235581611 & 0.943235581767
\end{array}
\right)$$

I am more than happy with these results but,again, just for the art for art's sake, I would like to be even better. For sure, I could use $p=9$ and obtain the parameters in order to  match the function, first, second and third derivative values at $x=0$, $x=\frac 12$ and $x=1$; this would lead to monstreous expressions.

So, my question is : without any curve fit and avoiding so many terms, is there a way to obtain a better (mathematically based) approximation of $\Big[\Gamma(1+x)\Big]^{-1}$ for $0 \leq x \leq 1$

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]{1 \over \Gamma\pars{x + 1}} = {1 \over x!}
=
{x + 1 \over \pars{x + 1}!} =
{\pars{x + 2}\pars{x + 1} \over \pars{x + 2}!}
\\[5mm] = &\ \cdots =
{\prod_{k = 1}^{n}\pars{x + k} \over \pars{x + n}!} \\[5mm] \approx &\
{\prod_{k = 1}^{n}\pars{x + k} \over
\root{2\pi}\pars{x + n}^{x + n + 1/2}\,\,\,\expo{-n - x}} =
{\tt mySol}\pars{x,n}
\end{align}


问:
Why is showing that $ker(T_A) = \vec 0$ the same as showing that a linear transformation is one-to-one?
Determine if the standard matrix $A$ representing the linear transformation is one-to-one.

$A = \begin{bmatrix}1&-1\\2&0\\3&-4\end{bmatrix}$

Answer: Yes

My question: When determining if multiplication by A is a one-to-one linear transformation, is it equivalent to determine if $ker(A)=\vec 0$ ? In other words, is the following method always a valid way to determine if $T_A$ is a one-to-one linear transformation?

$T_A$ is one-to-one if $ker$($T_A$)$=\vec 0$ , i.e, $A\vec x=\vec 0$ only has the trivial solution $\vec x=\vec 0$

$\begin{bmatrix}1&-1&0\\2&0&0\\3&-4&0\end{bmatrix}$ $\rightarrow$ $x=0,y=0$ $\rightarrow$ $\vec x=\vec 0$

So $T_A$ is one-to-one.

I am confused about this method because if $ker$($T_A$)$=\vec 0$ then that means the only vector that is mapped to $\vec 0$ after the linear transformation is $\vec 0$. So this only shows that the there is one input vector for the output vector  $\vec 0$. However, a one-to-one linear transformation means that each unique vector in the domain maps to only one vector  in the range. So why don't I have to show all vectors have this property and just show that zero vector has this property?

So my question is why is showing that the kernel of the transformation is $\vec 0$ enough to show that the transformation is one-to-one?

答:
Like you have mentioned in your comment, the span of the column vectors of A (column space) represent the range of the linear transformation. This is because when you multiply the given standard matrix by column vector of the right number of rows,

$$A\vec x = \vec b$$

$$\begin{bmatrix}1&-1\\2&0\\3&-4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x-y\\2x\\3x-4y\end{bmatrix}$$

your resultant vector $\vec b$ is a linear system representing all possible linear combinations, i.e., all possible output vectors (range) reached by multiplying any input vector $\vec x$ (domain) by the standard matrix $A$ (transformation).

So since the column space corresponds to the range, if you end up with the trivial solution for your homogeneous augmented matrix that means your column vectors are linearly independent, in other words all output vectors in the range are also linearly independent. This then implies that your linear transformation is one-to-one because your output vectors cannot be expressed in terms of other output vectors, meaning that the preimage of each output vector was unique.


答:
You have that $f(0)=0, f(t)=f(-t)$, which implies graphically that $x=0$ is a turning point of $f$, where its derivative is zero.

You can check this is the case, as the derivative calculated from both sides is equal. In the positive case:

$$f'(0)=\lim_{t\to0}\frac{f(t)-f(0)}{t}=\lim_{t\to 0}\frac{f(t)}{t}$$
and in the negative case:
$$f'(0)=\lim_{t\to 0} \frac{f(-t)-f(0)}{-t}=\lim_{t\to 0}-\frac{f(t)}{t}\overbrace{=}^{\text{ AOL}}-\lim_{t\to 0}\frac{f(t)}{t}$$

Since $f'(0)=-f'(0)$, we have that $f'(0)=0$.

Title: Limit approximates $0.\infty$ to 0

Body:
A question defines $$f(x)=\begin{cases}g(x)\cos\frac{1}{x}&\text{if } x\neq0\\0&\text{if } x=0\end{cases}.$$ Here, $g$ is an even function and  differentiable at $0$, and $g(0)=0$. You're required to find $f'(0)$.

On differentiating $f(x)$ to find it, you get $$g'(x)\cos\frac{1}{x}+\frac{g(x)\sin\frac{1}{x}}{x^2}$$. The first term is obviously $0$ at $x=0$, but the second term is in a $\frac{0}{0}$ indeterminate form, as both $g$ and $x^2$ approach 0 at x=0. But all the solutions I've seen simplify that term to zero, too.

Isn't that incorrect? I tried using L'Hôpital's Rule on $\frac{g(x)}{x^2}$, but since we don't know what g'(x) is, it doesn't help much. So how does that term end up as 0?


问:
Stokes Divergence Question for Calculus 3
Let $F$ be the vector field $F=4xz\,i+xz^2\,j+x^7y^3\,k$. Let $S$ be the portion of the paraboloid $z=x^2+y^2$ below $z=9$, along with the disk $x^2+y^2\leq9$ in the plane $z=9$, all with outward orientation. Find $\iint_SF\cdot dS$.

How would I do this? It's one of my HW problems and I'm really struggling.

答:
Portion of the paraboloid $z = x^2 + y^2$ below $z = 9$, along with the disk $x^2 + y^2 \leq 9 \,$ in the plane $z = 9$. That tells you it is a closed surface. So it is easier to go ahead and apply Divergence Theorem (unless you have been asked to do it differently).

$F=4xz\,i+xz^2\,j+x^7y^3\,k$

Divergence of the vector field, Div $F = \nabla \cdot F = \frac{\partial}{\partial x} 4xz + \frac{\partial}{\partial y} xz^2 + \frac{\partial}{\partial z} x^7y^3m = \,?$

Using cylindrical coordinates $x = r \cos \theta, y = r \sin \theta, z = z$

So $dV = r \, dz \, dr \, d \theta$.

As you can see the bound is $r^2 \leq z \leq 9$, $0 \leq r \leq 3, 0 \leq \theta \leq 2 \pi$.

Now set up your integral $\iiint_R (\nabla \cdot F) \, dV$.


问:
Show that if $\mathfrak{p} \subset R$ is a prime ideal, then $S = R \setminus \mathfrak{p}$ is multiplicatively closed.
Here is the question that I want to answer:

Let $R$ be an integral domain with field of fractions $K,$ and let $S \subset R \setminus \{0\}$ be a multiplicatively closed subset.

Show that if $\mathfrak{p} \subset R$ is a prime ideal, then  $S = R \setminus \mathfrak{p}$ is multiplicatively closed.

My thoughts

I know that if $R$ is a commutative ring with unity then the proper ideal $\mathfrak{p}$ is prime iff $R/\mathfrak{p}$ is an integral domain. but how can this help me in my proof? the set $S$ is defined as the setminus not the quotient.

Could anyone help me in solving this please?

This is the definition I am using for multiplicatively closed set.

Definition:  A set $S \subset R \setminus \{0\}$ is multiplicatively closed if $x,y \in S \implies xy \in S.$

答: Hint: As with many statements like this, it's useful to proceed by way of contradiction. Assume that $x,y\in S$, but $xy\not\in S$. What does this mean about inclusions in $\mathfrak p$? Can you use that to show that this contradicts the definition of prime ideal?


问:
Why is the absolute integrability criterion "inverted" for local integrability with respect to improper integrability?
In the context of Riemann integrals, I've noticed that for improper integrability:

(1) $|f|$ is improperly  integrable $\implies f $ is improperly integrable

but when it comes to local integrability it is the other way around

(2) $f$ is locally integrable $\implies |f| $ is locally integrable

Why is that? what's wrong about the following proof:?

If $|f|$ is locally integrable
$|\int_a^b{f}|< \int_a^b{|f|}<+\infty $   $\implies f$  is locally  integrable

Can't this prove that (2)'s converse is true, that is that if $|f|$ is locally integrable $\implies f $ is locally integrable? I know however it can't be since as a counterexample if $g $ is the Dirichlet function , $|g|$ is locally integrable but $g $is not

答:
The converse to (2) is only true if you're talking about Lebesgue integration since measurability is invariant across taking absolute values. You're not completely correct in that the Dirichlet function, say $g$, would satisfy that $|g|$ is locally integrable, since $g=|g|$ as the Dirichlet function only takes positive values. If you define something like
$$\varphi(x) = \begin{cases}1 & x\in\mathbb{Q} \\
-1 & x\in\mathbb{R}\setminus\mathbb{Q},
\end{cases}$$
then $\varphi$ would give you the desired property in terms of local integrability. Further, regarding your concerns about relationship of (2) to (1), the hypothesis in (1) is actually: if $f$ is locally integrable, then $|f|$ is improper integrable implies $f$ is improper integrable, with the counterexample of the converse being $\frac{\sin x}{x}$. If you want to see a proof of why $\varphi$ is not locally integrable, let me know.

Proof of $\varphi$ not being locally integrable:

Let $[a,b]\subseteq \mathbb{R}$, we can compute $U(\varphi)=1$ and $L(\varphi)=-1$ because no matter how you partition the interval $[a,b]$, each sub-interval will have a rational point and an irrational point. Thus, for all partitions $P$ of $[a,b]$, we must have $U(f;P)=1$ and $L(f,P)=-1$. Then, by the definition of upper and lower Riemann sums, we have $\varphi$ is not Riemann integrable (this is the same proof as why Dirichlet function is not Riemann integrable).


问:
The set of $x \in C[a,b]$ such that $|x(t) - x_0| \le c \beta$ is closed in $C[a,b]$.
$
\newcommand{\set}[1]{ \left\{ #1 \right\} }
\newcommand{\setb}[2]{ \left\{ #1 \; \middle| \; #2 \right\} }
\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\ve}{\varepsilon}
$
Simplified Version of the Problem: Recall that we may define the following:

$$C[a,b] := \setb{ f : [a,b] \to \R }{ f \text{ is continuous on } [a,b] }$$

With the metric $d$ endowed by the supremum norm, this is a complete metric space. The metric, for clarity, is

$$d(f,g) := \max_{t \in [a,b]} \Big| f(t) - g(t) \Big|$$

Define a subset $B$ of $C[a,b]$ as follows:

$$B := \setb{ f \in C[a,b] }{ |f(t)-1| \le 2 }$$

I wish to prove that $B$ is closed.

More General Context/Application/Background: This comes up in the proof of Picard's existence theorem, presented in Kreyszig's Introduction to Functional Analysis with Applications, and is left as a problem to the reader in section $\S 5.3$, problem $6$. It is, however, much more general that the version I stated above. In particular (up to notational differences), we instead have the following:

Let $f(t,x)$ be continuous on the rectangle in $\R^2$ given by $R := [t_0 - a, t_0 + a] \times [x_0 - b, x_0 + b]$.

$f$ also has a Lipschitz condition on its second component: $|f(t,x) - f(t,v)| \le k|x-v|$ $\forall (t,x),(t,v) \in R$

$f$, moreover, is bounded: $|f(t,x)| \le c$ for a certain constant $c$.

Of particular concern is the initial value problem $x' = f(t,x)$ with $x(t_0) = x_0$. (Picard's theorem is concerned in this context of showing the solution to this exists, is unique, and lies on $J$ as below.)

Define $J := [t_0 - \beta, t_0 + \beta]$, for a constant $\beta$ such that $\beta < \min \{a, b/c, 1/k\}$

Define $B$ to be the set of $x \in C(J)$ such that $|x(t) - x_0| \le c \beta$.

In the proof of the theorem, one is meant to see that $B$ is a closed subset of $C(J)$. The completeness of $C(J)$ then gives completion to all of its closed subsets, namely $B$, which we later use to apply the Banach contraction principle: we define an operator $T : B \to B$, which happens to be a contraction, and then the principle gives us a unique fixed point, which is critical for the proof.

Of course, many of these details are not strictly related to showing that $B$ (in this more general context, even) is closed. This is in part why I developed the much simpler formulation at the start, to avoid overcomplicating matters with symbols and notation that don't make sense. So if you prefer working with the simpler version initially presented, or want the more general one I am working with, either works for me -- insights into one no doubt would play into the other.

Attempts So Far: For what is a comparatively complicated proof, proving $B$ is closed has evaded me for a bit; I guess because I'm not too used to working with metrics tied to functions? (Below, I'm working in the more general context the proof demands.)

So far my idea is this. Take $\set{f_n}_{n \in \N} \subseteq B \subseteq C[a,b]$ a sequence in $B$ converging to $f$; this would ensure that $f \in B'$ (definition of limit point), and we want to show that $f \in B$ in particular (a set is closed iff it contains its limit points). Note that since $C[a,b]$ is complete, then $f \in C[a,b]$ for sure (as $\set{f_n}$ is assumed convergent, thus is Cauchy in $C[a,b]$, and $C[a,b]$ is complete, so the limit is in $C[a,b]$).

Since $f_n \in B$ $\forall n \in \N$, then, $\forall t \in J$, it holds that

$$|f_n(t) - x_0 | \le c \beta$$

Consider the quantity $|f(t) - x_0|$. From the convergence $f_n \to f$ and continuity of metrics, it holds that

$$|f(t) - x_0| = \left| \lim_{n \to \infty} f_n(t) - x_0 \right| = \lim_{n \to \infty} |f_n(t) - x_0| \le c \beta$$

(where the inequality holds in the limit since, otherwise, there would be some point $M$ where $|f_n(t) - x_0| > c \beta$ $\forall n \ge M$, contradicting that $f_n \in B$ $\forall n \in \N$). From this, thus, $f$ has the defining property of being in $B$; thus, $f \in B$.

My Questions:

My questions are mostly straighforward: does this approach work? Are there any logical or rigorous failings? I just feel somewhat unconfident in it, so I'd like to have some second opinions (or better/correct explanations).

Thanks for any insights you can give!

答: Yes this is correct. In general if you have a set given by a "continuous" equality or (not strict ) inequality for example $B=\{ x \in X | F(x)≤C\} $ you can use a similar argument by assuming that you have a convergent sequence $(a_n)_n \rightarrow a $ and you can show that $F(a_n)≤C \implies F(a)≤C$ because $F$ is continuous and you can interchange $F$ and the limit.


问:
Determinant of Kronecker Product

Problem. Let $A$ be an $m\times n$ matrix and $B$ be an $n\times m$ matrix. Show that $$\det(A\otimes B)=\det(B\otimes A).$$

This formula apparently holds if $A$ and $B$ are square matrices, but here this assumption fails in general.

I am considering the general definition of determinant: $$\det(M)=\sum_{\sigma\in S_n}{(-1)^{\rm sgn(\sigma)}a_{1\sigma(1)}\cdots a_{n\sigma(n)}}.$$
We have $$\det(B\otimes A)=\det((B\otimes A)^T)=\det(B^T\otimes A^T).$$
The shape of $B^T$ is the same as $A$ and the shape of $A^T$ is the same as $B$. Intuitively, the summands in the two determinants are exactly the same, but this argument is vague and complicated. It would be better if anyone has some easier proofs.

答:
When $m\ne n$, both determinants are zero.

Suppose $m<n$. Then $Ax=0$ has a non-trivial solution $x$. Let $y\in\mathbb F^m\setminus0$. Then $x\otimes y$ and $y\otimes x$ are nonzero but $(A\otimes B)(x\otimes y)=(Ax)\otimes(By)=0$ and likewise $(B\otimes A)(y\otimes x)=0$. Therefore $A\otimes B$ and $B\otimes A$ are singular and their determinants are zero.

Title: Determinant of Kronecker Product


答:
If you reflect on this, you will start to see that this is true by definition of prime ideals: They are exactly those ideals into which you cannot multiply from outside. Whether you state it as
\begin{align*}
 xy ∈ \mathfrak p &\implies& x ∈ \mathfrak p &~~~\text{or}~~~ y ∈ \mathfrak p, &\quad \text{or as} \\
 xy ∈ R \setminus \mathfrak p &\impliedby&  x ∈ R \setminus \mathfrak p &~~\text{and}~~  y ∈ R \setminus \mathfrak p
\end{align*}
is just a matter of giving contrapositions of each other. The latter line says that $S = R \setminus \mathfrak p$ is mutliplicatively closed, which is what you want to prove.

By the way: You could in principle also kind of use the fact that $R' = R / \mathfrak p$ is an integral domain. This means that $S' = R' \setminus \{0\}$ is multiplicatively closed. So as $S = R \setminus \mathfrak p = \mathrm{pr}^{-1} (R' \setminus \{0\}) = \mathrm{pr}^{-1}(S')$ is the preimage of a multiplicatively closed set under a ring homomorphism (so in particular under a multiplicative homomorphism), $S$ itself is multiplicatively closed too. This exposes the core insight behind this to realizing that for a commutative ring $R'$
$$R' ~\text{is integral domain} \iff R'\setminus \{0\} ~\text{is multiplicatively closed},$$
but this line of reasoning is disproportionate to the puzzle at hand.

Body:
Here is the question that I want to answer:

Let $R$ be an integral domain with field of fractions $K,$ and let $S \subset R \setminus \{0\}$ be a multiplicatively closed subset.

Show that if $\mathfrak{p} \subset R$ is a prime ideal, then  $S = R \setminus \mathfrak{p}$ is multiplicatively closed.

My thoughts

I know that if $R$ is a commutative ring with unity then the proper ideal $\mathfrak{p}$ is prime iff $R/\mathfrak{p}$ is an integral domain. but how can this help me in my proof? the set $S$ is defined as the setminus not the quotient.

Could anyone help me in solving this please?

This is the definition I am using for multiplicatively closed set.

Definition:  A set $S \subset R \setminus \{0\}$ is multiplicatively closed if $x,y \in S \implies xy \in S.$


问:
The natural numbers are not open.
I am wondering if someone can let me know if my solution is correct. I want to show that the natural numbers are not open. My work so far,

To show that the natural numbers are not open we must find a neighborhood $V_\epsilon(a),\;a\in \mathbb{N}. $ Such that $V_\epsilon(a)\not\subset \mathbb{N}$. Take $a=1$ and $\epsilon = 2$, $V_2(1) = (-1,3)\not\subset\mathbb{N}$. Is this enough to show that the set is not open? Or do I have to be more general? Thanks!

答: You have to show that for the point 1 there does not exist ANY epsilon-neighbourhood. So basically try showing that $V_\epsilon (1) $ is not a subset of $\mathbb{N}$ for all $\epsilon>0$

Title: The natural numbers are not open.


问:
How can you be sure that an integral does not exist, if it has no indefinite integral?
Say you have the integral $\displaystyle\int_1^\infty{\frac{1}{x^{1+\frac{1}{x}}}}\;\mathrm{d}x$

This integral cannot be completed. Not that it goes to infinity, but it physically just cannot be completed. How can you realize this if you encounter it? How can you prove it?

答:
$$I=\int_{1}^{\infty}\frac{dx}{x^p}=\left .\frac{x^{1-p}}{1-p}\right|_{1}^{\infty}=\frac{1}{p-1}, ~if~ p>1,$$
because $0^{1-p}=0$ if only $p>1$, otherwise it is infinite. Hence the integral converges when $p>1$.


问:
Narrow convergence of probability measures with a not bounded function
Let $ (\nu_n)_{\mathbb{N}} $ be a sequecence of Borel probability measure on $ [0,1]\times\mathbb{R} $ such that $$\nu_n\rightarrow\nu_{\infty} \text{ narrowly},$$
i.e. for every bounded continuous function $ h\in\mathcal{C}_b([0,1]\times\mathbb{R})$
$$ \int_{[0,1]\times\mathbb{R}}h(x,y)\ d\nu_n(x,y)\rightarrow\int_{[0,1]\times\mathbb{R}}h(x,y)\ d\nu_{\infty}(x,y) $$
where $ \nu_{\infty} $ is also a probability measure.

I know that for each measure $\nu_n$ with $ n\in\mathbb{N}\cup\{\infty\} $ we have that
$$ \int_{[0,1]\times\mathbb{R}}y\ d\nu_n(x,y)<+\infty.$$
I would say that
$$ \int_{[0,1]\times\mathbb{R}}y\ d\nu_n(x,y)\overset{n\rightarrow\infty}{\longrightarrow}\int_{[0,1]\times\mathbb{R}}y\ d\nu_{\infty}.$$
It is true? I think that the tightness condition could help, but I failed to use it.

答: This is false. Let $m$ be Lebesgue measure on $[0,1]$. Let $X_n$ be a sequence of random variables converging almost surely to $0$ such that $EX_n$ does not tend to $0$. (Example: $X_n=nI_{(0,\frac 1n)}$ on $(0,1)$ with Lebesgue measure). Let $\nu_n=m \times \mu_n$ and $\nu=m \times \mu$ where $\mu_n$ is the law of $X_n$ and $\mu$ is the law of $0$ (i.e. $\mu =\delta_0$). Then you can check that $\nu_n \to \nu$ but $\int yd\nu_n =1$ for all $n$ and $\int yd\nu=0$.


问:
Triangle related problem

In the figure below if $tan \alpha=\frac{m}{n}$, where $m$ and $n$ are relatively prime, find the value of $m$ and $n$.

My approach is as follow

$\frac{{SinA}}{{14}} = \frac{{SinB}}{{15}} = \frac{{SinC}}{{13}}$

$\Delta APB$

$\angle PBA = B - \alpha $

$\angle APB = 180 - \left( {\alpha  + B - \alpha } \right) = 180 - B$

$\frac{{Sin\left( {180 - B} \right)}}{{13}} = \frac{{Sin\alpha }}{{BP}} \Rightarrow SinB = \frac{{13Sin\alpha }}{{BP}}$

$\Delta APC$

$\angle PAC = A - \alpha $

$\angle APC = 180 - \left( {\alpha  + A - \alpha } \right) = 180 - A$

$\frac{{Sin\left( {180 - A} \right)}}{{15}} = \frac{{Sin\alpha }}{{AP}} \Rightarrow SinA = \frac{{15Sin\alpha }}{{AP}}$

$\Delta BPC$

$\angle PCB = C - \alpha $

$\angle BPC = 180 - \left( {\alpha  + C - \alpha } \right) = 180 - C$

$\frac{{Sin\left( {180 - C} \right)}}{{13}} = \frac{{Sin\alpha }}{{PC}} \Rightarrow SinC = \frac{{13Sin\alpha }}{{PC}}$

$\frac{{SinA}}{{SinB}} = \frac{{\frac{{15Sin\alpha }}{{AP}}}}{{\frac{{13Sin\alpha }}{{BP}}}} = \frac{{14}}{{15}} \Rightarrow \frac{{{{15}^2}}}{{13 \times 14}} = \frac{{AP}}{{BP}}$

How do I approach from here?

答:
I am not solving the problem completely. Instead I will continue the solution from where the author has left.

Use sin property in $\Delta APB$, you will get the equation $\frac{sin(B-\alpha)}{sin(\alpha)} = \frac{AP}{BP}$.
Now, use $sin(B-\alpha) = sin(B)cos(\alpha) - sin(\alpha)cos(B)$ and substitute in the previous equation to get $sin(B)cot(\alpha) - cos(B) = \frac{AP}{BP}$.

Now, you can get $cos(B) = \frac{AB^2+BC^2-AC^2}{2*AB*BC}$ using the cosine formula in $\Delta ABC$ and $sin(B) = \sqrt{1-cos^2(B)}$.

Substitute the values of $cos(B)$, $sin(B)$ and $\frac{AP}{BP}$ in the above equation and get $cot(\alpha)$. Now, $tan(\alpha) = \frac{1}{cot(\alpha)}$.

Title: Triangle related problem


答:
P is first Brocard point of the triangle.

We have the relation $$\cot \alpha = \cot A + \cot B + \cot C$$

To quickly find these, we partition the triangle as below

so $$ \cot B = \dfrac{5}{12}, \cot C = \dfrac{3}{4} $$

and in any $\triangle$, $$ \cot A\cot B + \cot B\cot C +\cot C\cot A =1$$

Update :

Following @JeanMarie's link, we see $$ \cot \alpha = \cot A + \cot B + \cot C = \dfrac{a^2+b^2+c^2}{4\triangle} $$

Title: Triangle related problem

Body:
In the figure below if $tan \alpha=\frac{m}{n}$, where $m$ and $n$ are relatively prime, find the value of $m$ and $n$.

My approach is as follow

$\frac{{SinA}}{{14}} = \frac{{SinB}}{{15}} = \frac{{SinC}}{{13}}$

$\Delta APB$

$\angle PBA = B - \alpha $

$\angle APB = 180 - \left( {\alpha  + B - \alpha } \right) = 180 - B$

$\frac{{Sin\left( {180 - B} \right)}}{{13}} = \frac{{Sin\alpha }}{{BP}} \Rightarrow SinB = \frac{{13Sin\alpha }}{{BP}}$

$\Delta APC$

$\angle PAC = A - \alpha $

$\angle APC = 180 - \left( {\alpha  + A - \alpha } \right) = 180 - A$

$\frac{{Sin\left( {180 - A} \right)}}{{15}} = \frac{{Sin\alpha }}{{AP}} \Rightarrow SinA = \frac{{15Sin\alpha }}{{AP}}$

$\Delta BPC$

$\angle PCB = C - \alpha $

$\angle BPC = 180 - \left( {\alpha  + C - \alpha } \right) = 180 - C$

$\frac{{Sin\left( {180 - C} \right)}}{{13}} = \frac{{Sin\alpha }}{{PC}} \Rightarrow SinC = \frac{{13Sin\alpha }}{{PC}}$

$\frac{{SinA}}{{SinB}} = \frac{{\frac{{15Sin\alpha }}{{AP}}}}{{\frac{{13Sin\alpha }}{{BP}}}} = \frac{{14}}{{15}} \Rightarrow \frac{{{{15}^2}}}{{13 \times 14}} = \frac{{AP}}{{BP}}$

How do I approach from here?


问:
Conjecture about :$\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac94+\frac32 \cdot \frac{(a-b)^2}{ab+bc+ca}}$
It's a conjecture . I combine two inequalities : Stronger than Nesbitt inequality and Stronger than Nesbitt's inequality $\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}$

So now the conjecture :

Let $a,b,c>0$ such that $a\geq b \geq c$ then we have :

$$\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac94+\frac32 \cdot \frac{(a-b)^2}{ab+bc+ca}}$$

I cannot find any counter-example until now . To prove it I have tried the same method as here Prove this refinement of Nesbitt's inequality based on another without success .

My question :

Have you an element of proof or an counter-example ?

Thanks in advance !!

Max .

答:
There is a good reason that you can not prove:
$$\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac94+\frac32 \cdot \frac{(a-b)^2}{ab+bc+ca}}$$
It's wrong for $a=\frac53,b=1,c=\frac19.$
By the way, the following inequality is true:
$$\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac94+\frac{(a-b)^2}{ab+bc+ca}},$$
as verified by Bottema - Maple program.


问:
Archer shoots 2 times. He gets as many more goes as he hit the target. Find the probability of him hitting the target exactly k times.
Archer shoots 2 times. He gets as many more goes as he hit the target. Probability of him hitting the target is q=0.4.Find the probability of him hitting the target exactly k times.

My solution

Hi - archer hits i out of 2 in the first go

Ak - hit the target exactly k times

P(Ak) = P(Ak-2)P(H2) - P(Ak-1)P(H1)

I got a complicated recurrence equation so i am not sure if this is correct

Pk = 0.16 Pk-2  +  0.48 Pk-1

答:
Suppose there are two targets and the archer shoots at each target repeatedly until he misses. Then the probabilities of getting $k$ hits are easily seen to be equal to those of the original problem.

Using the revised formulation, the probability of getting $k$ hits is that of getting exactly one miss in $k+1$ attempts and then a second miss, for a final answer of
$$(k+1)(0.6)^2(0.4)^k$$
This is, in fact, a negative binomial distribution.


问:
Is there always going to be both a concave up and down?
Throughout YouTube, I've seen people solve concavity and inflection questions where they find both: concave up and concave down. I was wondering if there were any instances where there was only one of them? Like only a concave up but not a concave down and vice versa?

I have the following function:

$f\left( x\right) =4x^{3}+21x^{2}+36x-20$

According to my work from class, there was no concave up, only a concave down. I'm not sure if this is wrong and I had accidentally missed writing down the part involving the concave up, but if that's not the case, I'm confused as to why there's only a concave down. Could someone explain this to me?

Anyways, this is my work:

$f'\left( x\right) =12x^{2}+42x+36$

$f''(x)=24x+42$

$\begin{aligned}f''(x)=24x+42=0\\ -42=-42\\ \dfrac{24x}{24}=\dfrac{-42}{24}\end{aligned}$

$\begin{aligned}x=-1.75\end{aligned}$

By using the sign chart (via numberline) I've gathered the following:

Concave up: $\left( -1.75, \infty \right)$

Concave down: $\left( -\infty , -1.75\right)$

Is this incorrect?

答:
Your computation looks fine to me.

For the other question, try to look at the log function. It is always concave down.


问:
What is $\lim_{n \rightarrow \infty}\sum\limits_{k = 1}^n\frac{k}{n^2}$?
We have $$\dfrac{1+2+3+...+ \space n}{n^2}$$

What is the limit of this function as $n \rightarrow \infty$?

My idea:

$$\dfrac{1+2+3+...+ \space n}{n^2} = \dfrac{1}{n^2} + \dfrac{2}{n^2} + ... + \dfrac{n}{n^2} = 0$$

Is this correct?

答:
Your method is wrong. Consider the first values of the expression:

for $n=1$ you get $1$;

for $n=2$ you get $1/4+2/4=3/4$;

for $n=3$ yoy get $1/9+2/9+3/9=6/9=2/3$.

One could make the conjecture that these terms are always bigger than $1/2$, which can be proved by induction. Let
$$
f(n)=\frac{1}{n^2}\sum_{i=1}^n i
$$
We have $f(1)=1>1/2$. Suppose the assertion holds for $f(n)$ and consider
\begin{align}
f(n+1)&=\frac{1}{(n+1)^2}\sum_{i=1}^{n+1} i\\
&=\biggl(\frac{1}{(n+1)^2}\sum_{i=1}^{n} i\biggr)+\frac{n+1}{(n+1)^2}\\
&=\frac{n^2}{(n+1)^2}\biggl(\frac{1}{n^2}\sum_{i=1}^n i\biggr)+\frac{1}{n+1}\\
\text{(by induction hypothesis)}\qquad&>\frac{1}{2}\frac{n^2}{(n+1)^2}+\frac{1}{n+1}\\
&=\frac{1}{2}\frac{n^2+2n+2}{(n+1)^2}\\
&=\frac{1}{2}\biggl(\frac{n^2+2n+1}{(n+1)^2}+\frac{1}{(n+1)^2}\biggr)\\
&=\frac{1}{2}\biggl(1+\frac{1}{(n+1)^2}\biggr)\\
&>\frac{1}{2}
\end{align}
Therefore your conclusion that the limit is $0$ cannot be true.


问:
Question on proveing the extended Fermat's theorem on sums of two squares
Let $m$ be an odd positive integer. Show that $m$ can be
written as a sum of two squares $m = a^2 + b^2$ with $\gcd(a,b) = 1$ if and only if every
prime factor of $m$ is congruent to $1 (\text{mod}~4)$.

$\mathbf{My~Attempts:}$

Notice that if $m$ is an odd prime, then the statement holds by Fermat's theorem on sums of two squares.

So, let $m$ be the composite odd positive integer.

First prove if every prime factor of $m$ is congruent to $1~(\text{mod}\ 4)$ then $m = a^2 + b^2$ with $\gcd(a,b) = 1$.

Assume that every prime factor of $m$ is congruent to $1~(\text{mod}\ 4)$

Let $m = p_1 p_2 \cdots p_n$ be the prime factorization of $m$ and each $p_i$ are odd.

Then, by assumption, each $p_i \equiv 1 ~(\text{mod}~4)$ which by Fermat's theorem on sums of two squares, $p_i = a_i^2 + b_i^2$ for some $a_i, b_i \in \mathbb{N}$.

So, $m = (a_1^2 + b_1^2)(a_2^2 + b_2^2) \cdots (a_n^2 + b_n^2) = [(a_1 a_2 + b_1 b_2)^2 + (b_1 a_2 - a_1 b_2)^2](a_3^2 + b_3^2) \cdots (a_n^2 + b_n^2)$.

Let $x_1 = (a_1 a_2 + b_1 b_2)$ and $y_1 = (b_1 a_2 - a_1 b_2)$.

Then, we have $m = (x_1^2 + y_1^2)(a_3^2 + b_3^2) \cdots (a_n^2 + b_n^2)$.

Now repeat this process $n-2$ times and let each $x_i = (x_{i-1} a_{i+1} + y_{i-1} b_{i+1})$ and let each $y_i = (y_{i-1} a_{i+1} - x_{i-1} b_{i+1})$.

Then, we will have $m = (x_{n-1}^2 + y_{n-1}^2)$ where $x_{n-1} = (x_{n-2} a_n + y_{n-2} b_n)$ and $y_{n-1} = (y_{n-2} a_n - x_{n-2} b_n)$.

Where $x_{n-1}$ and $y_{n-1}$ are both positive integers.

Let $a = x_{n-1}$ and $b = y_{n-1}$.

So, we proved that $m$ can be written as a sum of two squares $m = a^2 + b^2$.

$\mathbf{Problems:}$

Now I stuck on how to prove that $\gcd(a,b) = 1$ in this case  !!
Also, I don't know how to prove the inverse of the statement where if $m = a^2 + b^2$ with $\gcd(a,b) = 1$ then every prime factor of $m$ is congruent to $1~(\text{mod}~4)$ !

答:
Here's a somewhat different approach. First, similar to what you did, the "if" part means each prime factor of $m$ is congruent to $1 \pmod{4}$. As shown in the answer to Sum of two squares and prime factorizations, Fermat's theorem on the sum of squares states each prime factor $p_i$ of $m$ can be written as the sum of squares. Also, for any $c, d, e, f \in \mathbb{R}$,

$$(c^2 + d^2)(e^2 + f^2) = (ce \pm df)^2 + (cf \mp de)^2 \tag{1}\label{eq1A}$$

shows whenever $2$ numbers can be written as a sum of squares, their product can be as well, in $2$ different ways. Using \eqref{eq1A} repeatedly with the previous result (starting at $1$) and for each $p_i \mid m$ means the final product, i.e., $m$, can be written as a sum of squares.

Regarding proving you can choose an $a$ and $b$ where $\gcd(a, b)$, the answer to Any product of primes in the form of 4n+1 is the sum of 2 relatively prime squares shows this, paraphrased below.

As shown in \eqref{eq1A}, the product of the $2$ sums of squares can be expressed in $2$ ways. Have $c^2 + d^2$, with $\gcd(c, d) = 1$, be a product of $1$ or more primes of the form $4n + 1$, and $e^2 + f^2$ be a prime of that form to be multiplied. Consider if the first form in \eqref{eq1A}, i.e., $(ce + df)^2 + (cf - de)^2$, is not valid, i.e., there's a prime $q$ which divides each term. This means

$$q \mid (ce + df)e + (cf - de)f = c(e^2 + f^2) \tag{2}\label{eq2A}$$

$$q \mid (ce + df)f - (cf - de)e = d(e^2 + f^2) \tag{3}\label{eq3A}$$

Since $q$ doesn't divide $c$ and $d$, then $q \mid e^2 + f^2 \implies q = e^2 + f^2$. If both solution types in \eqref{eq1A} are not valid, then $e^2 + f^2$ divides $ce - df$ as well as $ce + df$, and hence divides $2ce$ and $2df$. Since $e^2 + f^2$ doesn't divide $2e$ or $2f$, it must divide both $c$ and $d$, contrary to the hypothesis, meaning at least one of the $2$ forms must be valid. Thus, use the valid form, and repeat this procedure for each prime that is multiplied, to eventually get $m$.

For the "only if" part, similar to the answer to If $a \in \Bbb Z$ is the sum of two squares then $a$ can't be written in which of the following forms?, suppose there's a prime $p \equiv 3 \pmod{4}$ with $p \mid m$. If $p \mid a$, then $p \mid b$, and vice versa, but since $\gcd(a, b) = 1$, then $p$ can't divide either $a$ or $b$. Thus, $a$ has a multiplicative inverse, call it $a'$, modulo $p$. Let $r = \frac{p-1}{2}$ and note $r$ is odd. Also using Fermat's little theorem, this gives (note the argument below is basically equivalent to showing $-1$ is not a quadratic residue modulo $p$ if $p \equiv 3 \pmod{4}$)

$$\begin{equation}\begin{aligned}
a^2 + b^2 & \equiv 0 \pmod{p} \\
a^2(a')^2 + b^2(a')^2 & \equiv 0 \pmod{p} \\
1 + (ba')^2 & \equiv 0 \pmod{p} \\
(ba')^2 & \equiv -1 \pmod{p} \\
\left((ba')^2\right)^{r} & \equiv (-1)^r \pmod{p} \\
(ba')^{p-1} & \equiv -1 \pmod{p} \\
1 & \equiv -1 \pmod{p}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$

This, of course, is not possible, meaning the original assumption must be false. This confirms all prime factors of $m$ must be congruent to $1 \pmod{4}$.


问:
Prove this refinement of Nesbitt's inequality based on another
Let $a,b,c\in[1,2]$ such that $a,b$ are constants then prove  :

$$f(c)=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{b+a}\geq h(c)=(c-1)\frac{g(2)-g(1)}{2-1}+g(1)\geq g(c)=\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2}{ab+bc+ca}}$$

Yes it's a probable refinement of HN_NH's inequality/ Stronger than Nesbitt inequality

My refinement is based on two observations :

The function $g(c)$ is convex on $[1,2]$

2.The chord of a convex function is greater than the convex function.

To know if the LHS is good I have tried derivatives . It gives a quartic and it's very ugly so I can say that I have not the solution for the LHS but it seems to be true (numerical check).

Update 12/11/2020:

The function :

$$p(c)=f(c)-h(c)$$

Is convex on $[1,2]$ so there is a possibility to use Jensen's inequality but now I don't see any good issue .

Using Jensen's inequality we have :

$$p(c)+p(1)\geq 2p\left(\frac{1+c}{2}\right)$$

And :

$$p\left(\frac{1+c}{2}\right)+p(1)\geq 2p\left(\frac{3+c}{4}\right)$$

And :

$$p\left(\frac{3+c}{4}\right)+p(1)\geq 2p\left(\frac{3+c}{8}+\frac{1}{2}\right)$$

And so on...Playing with these inequalities we got the result I think !

Have you an idea to show the LHS(or confirm my update)?

Thanks in advance

Max.

答:
Just a comment, this inequality is true for all $a,b,c>0.$ By the way, the best inequality is:
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac94+{\frac {\sqrt {1158\sqrt {193}-3450}}{72}} \cdot \frac{(a-b)^2}{ab+bc+ca}}$$
I'm trying to prove your inequality. I will post it later.


问:
$U(n)$ is a subgroup of $SO(2n)$
How can I show that $U(n)$ is a subgroup of $SO(2n)$? I can see how we can identify $\mathbb{C}^n$ with $\mathbb{R}^{2n}$ by mapping $a+ib\mapsto (a,b)$, but after that I'm a bit confused. In particular, the determinant of a matrix in $U(n)$ has to be nonzero, whereas in $SO(2n)$ it has to be 1. How do we make this transition?</p>

答:
Let me write the answer, this is a good exercise.

All you have to do is to show the inclusion $U(n)\subset O(2n)$, which makes sense after identify $\mathbb{C}^n$ with $\mathbb{R}^{2n}$ by for instante $(z_1,\cdots,z_n) = (x_1+iy_1,\cdots,x_n+iy_n)\longleftrightarrow (x_1,\cdots,x_n,y_1,\cdots,y_n)$.

Consider then $U\in U(n)$, that is, $U \in Mat_{n\times n}(\mathbb{C})$ such that $UU^\dagger = U^\dagger U = 1$. Such matrices acts on $\mathbb{C}^n$ by $$(z_1,\cdots,z_n) \longmapsto (U_{1j}z_j,\cdots,U_{nj}z_j)$$

Using our identification, we will discover how $U(n)$ acts on $\mathbb{R}^{2n}$. To do so we decompose the matrices of $U(n)$ in real and imaginary parts: $$ U = U^R + iU^I $$ with $U^R,U^I \in Mat_{n\times n}(\mathbb{R})$ and

$$ UU^\dagger =  (U^R + iU^I)(U^R + iU^I)^\dagger = U^R (U^R)^T + U^I (U^I)^T + i\left( U^I(U^R)^T - U^R(U^I)^T \right)=1 $$

$$ U^\dagger U =  (U^R + iU^I)^\dagger(U^R + iU^I) =  (U^R)^TU^R +  (U^I)^TU^I + i\left( (U^R)^TU^I - (U^I)^TU^R \right)=1 $$

The condition follows as: $$U^R (U^R)^T + U^I (U^I)^T = (U^R)^TU^R +  (U^I)^TU^I = 1 \ \ (*)$$

$$ U^I(U^R)^T - U^R(U^I)^T = (U^R)^TU^I - (U^I)^TU^R = 0 \ \ (**)$$

With this notation:

$$ (z_1,\cdots,z_n) \longmapsto (U^R_{1j}z_j + iU^I_{1j}z_j,\cdots,U^R_{nj}z_j+iU^I_{nj}z_j) $$

writing in terms of $z_i=x_j+iy_j$  : $$ \left( (U^Rx)_1 - (U^Iy)_1 + i(U^Ix)_1+i(U^Ry)_1,\cdots, (U^Rx)_n - (U^Iy)_n + i(U^Ix)_n+(U^Ry)_n) \right) $$

Therefore, expliciting the identification: $$ (x_1,\cdots,x_n,y_1,\cdots,y_n)\longmapsto \left( (U^Rx)_1 - (U^Iy)_1, \cdots, (U^Rx)_n - (U^Iy)_n,(U^Ix)_1+(U^Ry)_1,\cdots, (U^Ix)_n+i(U^Ry)_n \right) $$

In matrix notation:

$$U = \begin{bmatrix}U^R& -U^I\\ U^I & U^R\end{bmatrix} \in Mat_{2n\times 2n}(\mathbb{R}) $$

To see that this matrix is on $SO(2n)$, we have to check the ortogonal condition: $UU^T=1$. This follows immediately by $(*)$ and $(**)$ :

$$ UU^T = \begin{bmatrix}U^R& -U^I\\ U^I & U^R\end{bmatrix}\begin{bmatrix}(U^R)^T& (U^I)^T\\ -(U^I)^T & (U^R)^T\end{bmatrix}  $$

$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \   = \begin{bmatrix}U^R(U^R)^T + U^I(U^I)^T& U^R(U^I)^T-U^I(U^R)^T\\ U^I(U^R)^T-U^R(U^I)^T & U^I(U^I)^T+U^R(U^R)^T\end{bmatrix}= \begin{bmatrix}1& 0\\ 0 & 1\end{bmatrix}$$

ps: Repeated indices always denotes sum over the indices $\left(a_jb_j=\sum_ja_jb_j\right)$

Title: $U(n)$ is a subgroup of $SO(2n)$


问:
Group Homomorphism mapping to the symmetric group on X elements
If we have a group G acting on a Set X define the group homomorphism$$ \varphi:G\to S_x $$ where$$\varphi(g)=\sigma_g ,\sigma_g(x)=g*x=x$$

I am trying to show that if this group homomorphism is surjective then $*$ is transitive. I'd like to see if I have my ideas/understanding clear and if there are points I am missing.

If the group homomorphism is surjective that means for some $g\in G$ we have $\varphi(g)$ that maps every to every permutation in $S_x$

Since $g*x$ is a permutation on $X$ then if we have an $x\in X$, $\varphi(g)$ will send it to itself or another element in X. Since $\varphi$ is surjective there is going to be a permutation  that sends a fixed $x$ to every $y\in X$. Which makes $*$ transitive.

Also do we need the group homormorphism to be surjetive for transitivity? If $S_x$ has n elements then there are n! permutations. I believe you need less than that to have this property hold true (atleast for $3\leqslant n$.)

答:
What you wrote is correct, though a bit unclear. You could make things more clear as follows:

Let $x,y\in X$. There is a permutation $\sigma\in S_X$ mapping $x$ to $y$ (one could take the permutation $(x\,\,\,\, y)$). Since $\varphi$ is surjective, there is $g\in G$ such that $\varphi(g)=\sigma$. Then we have $g*x=\varphi(g)\cdot x=(x\,\,\,\,y)\cdot x=y$ as desired.

For the second question, indeed surjectivity is not needed. In fact, it is enough for the image of $\varphi$ to act transitively on $X$. For example, if $X$ has $n$ elements, then the cyclic subgroup of $S_n$ generated by the cyclic permutation $(1\,2\,\dots\,n)$ (or, in fact, by any $n$-cycle) acts transitively on $X$, and this subgroup has only $n$ elements. Another example is the subgroup $A_n$ of even permutations, which acts transitively on $X$.


问:
Taylor series of $\sqrt{1+x}$ using sigma notation
I want help in writing Taylor series of $\sqrt{1+x}$ using sigma notation
I got till
$1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5x^4}{128}+\ldots$ and so on.
But I don't know what will come in sigma notation.</p>

答:
allow me to take one more tiny little step forward based on your excellent calculations.

$$
\sqrt{1+x}=\sum _{n=0}^{\infty } \frac{ 1}{(1-2 n)}
\frac{(2 n)!}{n! n!}\left(-\frac{x}{4}\right)^n
$$

or

$$
\sqrt{1+x}=\sum _{n=0}^{\infty } \frac{ C_{2n}^n}{(1-2 n)}
\left(-\frac{x}{4}\right)^n
$$


问:
Does $\sum_{k=1}^{\infty} (-1)^k \frac{x^{2k}}{5^k \sqrt k}$ converge?
Let $a_k=(-1)^k \frac{1}{5^k \sqrt k}$.

This is an example in my book, the author claims that according to the root test the limit does not exist, because $a_k=0$ for k odd (is this true?), but proceeds to show that it indeed has a convergence radius of $R=5$, with the argument that limsup for a bounded sequence always exists.

答:
Show that

$$|\frac{a_{k+1}x^{2k+2}}{a_kx^{2k}}|= \frac{1}{5} \cdot \frac{\sqrt{k}}{\sqrt{k+1}} \cdot x^2.$$

Hence

$$|\frac{a_{k+1}x^{2k+2}}{a_kx^{2k}}| \to \frac{x^2}{5}$$

as $k \to \infty$.

From

$$\frac{x^2}{5} <1 \iff |x| < \sqrt{5}$$

we get thst the radius of convergence is $R= \sqrt{5}$.


问:
show that harmonic series is divergent with $ \sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n} \geq\left(2^{k+1}-2^{k}\right) \frac{1}{2^{k+1}}=\frac{1}{2} $
I'm pretty stuck with this exercise. Hope somebody can help me:

show that
$$
\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n} \geq\left(2^{k+1}-2^{k}\right) \frac{1}{2^{k+1}}=\frac{1}{2}
$$
and use this to show that the harmonic series is divergent.

First I can't figure out how to show the inequality.
Second I'm not quite sure how to use the inequality to show divergence. Because it doesn't really help letting k go to infinity.

Writing out the series gave me this, but where do I go from here:
$$
\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n}=\frac{1}{2^{k}+1}+\frac{1}{2^{k}+2}+\cdots \frac{1}{2^{k}+2^{k}}=\frac{1}{2^{k}+1}+\frac{1}{2^{k}+2}+\cdots+\frac{1}{2^{k+1}}
$$

Edit:
So I understand how to show the inequality now. Thanks!
But I'm still lost with how to use the inequality for the divergence proof.

Body:
(I am summarizing in this answer the steps discussed with OP in the comments)

Let $k\geq 0$ be an integer.

First note that the sum $\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n}$ has $2^{k+1}-(2^{k}+1)+1= 2^{k+1}-2^k = 2^k$ terms, that is
$$\sum_{n=2^{k}+1}^{2^{k+1}} 1 = 2^k\tag{$*$}$$

Second, note that
$$\frac{1}{n}\geq \frac{1}{2^{k+1}},\qquad \forall n\, \text{ s.t. }\,2^{k}+1\leq n\leq 2^{k+1}\tag{$**$}$$

Combining the above observations, we have
$$\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n}\,\overset{(**)}\geq\,\sum_{n=2^{k}+1}^{2^{k+1}}\frac{1}{2^{k+1}}\,=\,\frac{1}{2^{k+1}}\sum_{n=2^{k}+1}^{2^{k+1}}1\,\overset{(*)}{=}\, \frac{2^k}{2^{k+1}}\,=\,\frac{1}{2}\tag{$*\!*\!*$}$$

Now, note that

$$\sum_{n=1}^{2^{k+1}}\frac{1}{n}\,\geq\, \sum_{n=2}^{2^{k+1}}\frac{1}{n}\, =\, \sum_{m=0}^k\, \sum_{n=2^{m}+1}^{2^{m+1}}\frac{1}{n} \, \overset{(***)}{\geq}\, \sum_{m=0}^k \frac{1}{2}\, = \, \frac{k+1}{2}$$

As the above holds for every $k\geq 0$, by letting $k\to \infty$, we conclude that the harmonic series diverges.


答:
$a_k$ is definitely not zero for $k$ odd, unless $x=0$. With the $x^{2k}$ inside:
$$\limsup_{k\to\infty}|a_k|^{1/k}=\limsup_{k\to\infty}\frac{x^{2k/k}}{(5^k)^{1/k}k^{1/2k}}=\frac{x^2}5$$
Thus the series converges when $x^2<5$, i.e. the radius of convergence is $\sqrt5$.

Title: Does $\sum_{k=1}^{\infty} (-1)^k \frac{x^{2k}}{5^k \sqrt k}$ converge?

Body:
Let $a_k=(-1)^k \frac{1}{5^k \sqrt k}$.

This is an example in my book, the author claims that according to the root test the limit does not exist, because $a_k=0$ for k odd (is this true?), but proceeds to show that it indeed has a convergence radius of $R=5$, with the argument that limsup for a bounded sequence always exists.


问:
Change of integration variables in multi-dimensional integral
I have the following two-dimensional integral at hand
\begin{equation}
\int_{0}^{\infty}dx_{1}\int_{0}^{x_{1}}dx_{2} F(x_{1}-x_{2},x_{2}),
\end{equation}
where $F(x,y): \mathbb{R}^{2} \rightarrow \mathbb{R}$ is an arbitrary scalar function on a two-dimensional space.

Intuitively, by imagining the integration domain it is clear that this should be equal to
\begin{equation}
\int_{0}^{\infty}dx_{1}'\int_{0}^{\infty}dx_{2}' F(x_{1}',x_{2}').
\end{equation}
How can I show this formally via a change of integration variables?

答: Since $F$ is integrated on$$\{(x_1,\,x_2)\in\Bbb R|0\le x_2\le x_1\}=\{(x_1,\,x_2)\in\Bbb R|x_1-x_2\ge0\land x_2\ge0\},$$the change of variables $x_1^\prime:=x_1-x_2,\,x_2^\prime:=x_2$ rewrites the integral as$$\int_0^\infty dx_1^\prime\int_0^\infty dx_2^\prime JF(x_1^\prime,\,x_2^\prime),$$with $J$ a Jacobian determinant we need to check is $1$. I'll leave that part to you.


问:
Why does the Denominator of the Denominator go to the Numerator?
Quite blindly I've learnt a basic rule about fractions:

The Denominator of the Denominator goes to the numerator.

I'm confused about it and I'll give an example as to why.
Imagine the following:

1/2/2

Now, if the denominator of the denominator went to the numerator this would be 2/2 which is 1

and that's the what everyone agrees upon.

But what I feel is that since 1/2 is 0.5 ,

1/2/2 = 0.5/2 = 0.25

So, I hope you can see why I feel confused.

I apologize if this question seems stupid

答:
You need to be careful with the order of operations:

$$
1/2/2 = (1/2)/2 = 1/4
$$
while
$$
1/(2/2) = 2/2 = 1
$$


问:
Convolution with the poisson kernel (what is)

I am trying to work on the last part of the above problem. I am not sure what a 'convolution with the poisson kernel' is and consequently, a way to express $g(x+i\epsilon)- g(x-i\epsilon)$, or solve the problem. I would appreciate some help.

答: Just a bit of algebra $$g\left(x+i\varepsilon\right)-g\left(x-i\varepsilon\right)=\frac{1}{2\pi i}\int_{-M}^{M}h\left(z\right)\left(\frac{1}{z-x-i\varepsilon}-\frac{1}{z-x+i\varepsilon}\right)dz$$ $$=\frac{\varepsilon}{\pi}\int_{-M}^{M}\frac{h\left(z\right)}{\left(z-x\right)^{2}+\varepsilon^{2}}dz=\int_{-\infty}^{+\infty}h\left(z\right)P_{\varepsilon}\left(x-z\right)dz=\left(h*P_{\varepsilon}\right)\left(x\right)$$ where $P_{\varepsilon}\left(x\right)$ is the Poisson kernel $P_{y}\left(x\right)=\frac{1}{\pi}\frac{y}{x^{2}+y^{2}}.$ Now it remanis to observe that $P_{\varepsilon}\left(y\right)$ is an approximate identity and so, since $h$ is continuous, $$\lim_{\varepsilon\rightarrow0^{+}}\left(h*P_{\varepsilon}\right)\left(x\right)=h(x).$$


问:
matrices product and angle
proof for equality of angles (cosines) between 2 matrices</p>

答:
Visibly, as you apply the matrix $A^T$ on the right to $u$ and $v$, it means that $u$ and $v$ are row vectors.

Instead of row vectors $uA^T$ and $vA^T$, it is better to work with corresponding column vectors  $(uA^T)^T=Au^T$ and $(vA^T)^T=Av^T$.

First of all: dot product

$$(Au^T).(Av^T)=(Au^T)^T(Av^T)=uA^TAv^T=uv^T=u.v \tag{1}$$

Besides

$$\|Au^T\|^2=(Au^T)^TAu^T=uA^TAu^T=uu^T=\|u^T\|^2 \ \implies \ \|Au^T\|=\|u^T\|\tag{2}$$

Therefore

$$\cos(Au^T,Av^T)=\dfrac{1}{\|Au^T\|\|Av^T\|}(Au^T).(Av^T)=$$

$$=\dfrac{1}{\|u^T\|\|v^T\|}(u^T).(v^T)=\cos(u^T,v^T)$$

Title: matrices product and angle


答:
Just convert such divisions into multiplication like this, to avoid confusion.

$$\dfrac{\dfrac{k}{m}}{\dfrac{p}{q}}=\dfrac{k}{m} \cdot \dfrac{q}{p}$$

Title: Why does the Denominator of the Denominator go to the Numerator?

Body:
Quite blindly I've learnt a basic rule about fractions:

The Denominator of the Denominator goes to the numerator.

I'm confused about it and I'll give an example as to why.
Imagine the following:

1/2/2

Now, if the denominator of the denominator went to the numerator this would be 2/2 which is 1

and that's the what everyone agrees upon.

But what I feel is that since 1/2 is 0.5 ,

1/2/2 = 0.5/2 = 0.25

So, I hope you can see why I feel confused.

I apologize if this question seems stupid


问:
Prove that $a_{n+2}=0.5(a_n+a_{n+1})$, $a_1=2, a_2=5$ has a limit, and find it
I have the following sequence: $a_{n+2}=0.5(a_n+a_{n+1})$, $a_1=2, a_2=5$, and I need to prove that $\lim\limits_{n\to \infty}a_n$ exists, and find it.

I don't know ho to prove that the limit exists, since the sequence neither decreasing nor increasing.

In addition, I don't know how to find it, since the equation I am getting is $L=0.5(L+L)$, whitch is true for all $L$.

答:
Here I leave a sketch. I may return and provide the details if someone asks for it when I am free.

Let $x_n = a_{2n+1}$ for each $n \ge 1$, and $y_n = a_{2n}$ for each $n \ge 1.$

Show that $(x_n)_{n \ge 1}$ and $(y_n)_{n \ge 1}$ are monotone and bounded and hence conclude that each of them converge to a finite limit. Let $l, l'$ be the limits of these two sequences respectively.

Now, use $a_{n+2} = \frac{a_n + a_{n+1}}{2}$ to show that $l = l'.$

Conclude from here that $(a_n)_{n \ge 1}$ converges.

Also you can find the value of the limits $l , l'$ and hence find the point where $(a_n)_{n \ge 1}$ converges.


问:
Maxima and minima of $\frac{x^2-3x+4}{x^2+3x+4}$ without calculus
How to find the minimum and max values of $y=\frac{x^2-3x+4}{x^2+3x+4}$ for all real values of $x$ without using calculus?

Perhaps it could be done graphically by noting the fact that the numerator and denominator are a pair of parabolas symmetric about the $x$ axis, but I do not know how to continue.

Thanks!

答:
Hint wehave $$\frac{x^2-3x+4}{x^2+3x+4}=y$$
$$x^2(y-1)+x(3y+3)+4y-4=0$$ set the discriminant greater or equal to zero


答:
Let the maximum of $f(x) = \frac{x^2-3x+4}{x^2+3x+4}$ be $m$. Then:

$$x^2-3x+4 = mx^2 + 3mx + 4m$$
$$(m-1)x^2 + (3m+3)x + (4m - 4) = 0$$

We want this equation to have only one real root (a double root), so:
$$\Delta = 0 \Rightarrow (3m+3)^2-4(m-1)(4m-4) = 0.$$

A similar process for the minimum ($n$) yields the same equation, as multiplying it by $-1$ does not change the values of $m$. Therefore, both the maximum and minimum values are given by this equation.

Title: Maxima and minima of $\frac{x^2-3x+4}{x^2+3x+4}$ without calculus

Body:
How to find the minimum and max values of $y=\frac{x^2-3x+4}{x^2+3x+4}$ for all real values of $x$ without using calculus?

Perhaps it could be done graphically by noting the fact that the numerator and denominator are a pair of parabolas symmetric about the $x$ axis, but I do not know how to continue.

Thanks!


答:
As mentioned in comments, the sequence may be shown to be Cauchy and hence convergent.

This is a second-order homogeneous linear recurrence. The characteristic equation is
$$2\lambda^2-\lambda-1=(2\lambda+1)(\lambda-1)=0$$
with roots $1$ and $-\frac12$. Therefore the general equation is $a_n=p+q(-1/2)^n$ where $p$ and $q$ are determined from the initial conditions; here $p=q=4$. Since $(-1/2)^n$ tends to zero as $n\to\infty$, the limit is $p=4$.

Title: Prove that $a_{n+2}=0.5(a_n+a_{n+1})$, $a_1=2, a_2=5$ has a limit, and find it

Body:
I have the following sequence: $a_{n+2}=0.5(a_n+a_{n+1})$, $a_1=2, a_2=5$, and I need to prove that $\lim\limits_{n\to \infty}a_n$ exists, and find it.

I don't know ho to prove that the limit exists, since the sequence neither decreasing nor increasing.

In addition, I don't know how to find it, since the equation I am getting is $L=0.5(L+L)$, whitch is true for all $L$.


答:
$a_1 = 2$, $a_2 = 5$

$$2a_3 = a_1 + a_2$$
$$2a_4 = a_2 + a_3$$
$$2a_5 = a_3 + a_4$$
$$\cdots$$
$$2a_{n - 1} = a_{n- 2} + a_{n - 3}$$
$$2a_n = a_{n- 1} + a_{n - 2}$$

Add all equations, you will get

$$2a_n  + a_{n - 1} = a_1 + 2a_2$$

Since you can show that the sequence is Cauchy and hence converges as I explained in the comment above, $a_n \to L$ and $a_{n - 1}\to L$ as $n \to \infty$

Hence, $$2L + L = a_1 + 2a_2$$
or $$L = \dfrac{2 + 2(5)}{3} = 4$$


问:
Extension of quasi-coherent sheaves
In the Hartshorne book (Algebraic Geometry), page 114, we have the following proposition:

Proposition 5.7. Let $X$ be a scheme. The kernel, cokernel, and image of any
morphism of quasi-coherent sheaves are quasi-coherent. Any extension of
quasi-coherent sheaves is quasi-coherent. If $X$ is noetherian, the same is
true for coherent sheaves.

My question is: what is an "extension of quasi-coherent sheaves"?

答: If $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is a short exact sequence, B is called an extension of $A$ by $C$.

Title: Extension of quasi-coherent sheaves


问:
Double Summation that has Same Upper Limit as Index: $\sum\limits_{i=1}^j\sum\limits_{j=1}^5 3ij$
This is the first time I encountered a double summation.
$$\sum_{i=1}^j\sum_{j=1}^5 3ij$$
I have solved the right summation but I don't know what to do with the 'j' upper limit of the left summation. Perhaps its order need to be changed first?

Body:
$$S=\sum_{i=1}^j\sum_{j=1}^5 3ij=\sum_{j=1}^5\sum_{i=1}^j 3ij= \sum_{j=1}^{5}3j \sum_{i=1}^{j} i=\sum_{j=1}^{5}[3j(j(j+1)/2].=\frac{1}{2}\sum_{j=1}^{5}[3j^3+3j^2] $$
$$\implies S=\frac{3}{2} \left(\frac{5(5+1)}{2}\right)^2+\frac{3}{2}\frac{5(5+1)(2.5+1)}{6}=420$$


问:
Unique linear transformation carrying projective hyperplane to projective hyperplane and point to point - a concise proof
Let $\mathbb P^n$ denote the projective $n$-space over an algebraically close field $k$, i.e. $\mathbb P^n$ is given by $(\mathbb A^{n+1}\setminus \{0\})/ \sim$ where $\mathbb A^{n+1}$ is the affine $(n+1)$-space and $\sim$ is the equivalence relation identifying points which are scalar multiples of one another, that is for two points $(a_0, \cdots , a_n)$ and $(b_0, \cdots , b_n)$ in $\mathbb A^{n+1}$,
$$(a_0, \cdots , a_n) \sim (b_0, \cdots , b_n) \iff \exists \hspace{1mm} \lambda \in k^\times \text{ s.t. } b_j = \lambda a_j \text{ for all }1 \leq j \leq n$$
By a hyperplane in $\mathbb P^n$, I shall mean the zero set of some linear homogenous polynomial $f \in k[x_0, \cdots , x_n]$, that is some polynomial of the form $f(x_0, \cdots , x_n) := \sum_{j=0}^n a_j x_j$ where $(a_0, \cdots , a_n) \in \mathbb P^n$.

I have seen the following result get used in a few contexts before, and although I can see intuitively why it must be true, I have been unable to find a rigorous argument justifying the same:

Fact(?) Let $H$ be a hyperplane and $P$ any point in $\mathbb P^n \setminus H$. Then there exists a linear transformation $A \in \text{GL}_{n+1}(k)$ such that $A(H)$ is the hyperplane $\{(x_0, \cdots , x_n) : x_0=0\}$ and $A(P) = (1, 0, \cdots , 0)$.

I am looking for a complete and concise proof of this result, which is clean if possible. I believe that one possible argument could rest on the following observations:

$H$ is uniquely determined by any $n$ points on it. So we now pick $n+1$ points $P_1, \cdots , P_n$ on $H$.

There exists a linear transformation sending $P$ to $(1, 0, \cdots,  0)$ and $P_j$ to $(0, \cdots , 0 , 1, 0, \cdots, 0) \in \mathbb P^n$ ($0$ in the $j$-th slot, here the $n+1$ slots are being called the $0$-th, $1$-st, ..., $n$-th slot slots) for each $1 \leq j \leq n$.

I have however been unable to make these clean and rigorous (I keep getting involved with too many linear equations) and am starting to doubt the accuracy of my intuition. I would really appreciate a complete argument for the above "Fact(?)" or a reference containing the same and if possible, suggestions on how to make my idea work.

Edit (Some Progress): Thanks to Roland's comment, I think I have made some progress:

Let $H$ be given by the equation $\sum_{j=0}^n a_j x_j = 0$. Then in $\mathbb A^{n+1}$, $H$ remains the same (nevertheless I will call it $H_0$ when viewed as a subset of $\mathbb A^{n+1}$) while $P := (p_0, \cdots , p_n)$ becomes the line $L_0 := \{(p_0 t, \cdots , p_n t) : t \in k\}$. I should first show that there is a matrix $A \in \text{GL}_{n+1}(k)$ such that $A(H_0) = H_1$ and $A(L_0)=L_1$, where $H_1 := \{(0, x_1, \cdots , x_n) : x_j \in k\} \subset \mathbb A^{n+1}$ and $L_1$ is the line $\{(t, 0, \cdots , 0) : t \in k\} \subset A^{n+1}$.

So now I can pick $n$ linearly independent points $A_j \in H_0$ ($1 \leq j \leq n$), which is possible since $H_0$ is an $n$-dimensional subspace of $\mathbb A^{n+1}$ and I get a linear transformation $A \in \text{GL}_{n+1}(k)$ which sends $A_j$ to $(0, \cdots , 0 , 1, 0, \cdots 0)$ (with $1$ in the $j$-th slot) for each $1 \leq j \leq n$. Thus $A$ sends $H_0$ to $H_1$. I still have to send $A(L_0)$ to $L_1$ so I need a linear transformation $T \in \text{GL}_{n+1}(k)$ which sends $A(L_0)$ (which is also a line through the origin) to $L_1$ and leaves $H_1$ invariant (as a set).

Finally, we let $T \in \text{GL}_{n+1}(k)$ be the linear transformation that sends $(p_0, \cdots , p_n) \in \mathbb A^{n+1}$ to $(1, 0, \cdots , 0)$ and fixes some basis of $H_1$ pointwise.

Upon getting this last linear transformation $T$, we note that $TA \in \text{GL}_{n+1}(k)$ sends $H_0$ to $H_1$ and $L_0$ to $L_1$ in $\mathbb A^{n+1}$. Therefore $TA$ should also do the required job, namely, send $H$ to $\{(0, x_1, \cdots , x_n)\} \subset \mathbb P^n$ and $P$ to $(1, 0, \cdots 0)$, thus completing the proof and making "Fact(?)"$ a fact.

My only follow-up question: Is this argument correct or are there are any gaps?

答:
A small note: $GL_{n+1}(k)$ acts differently for linear systems than for points. If $H$ is a hyperplane with equation $a^T \times x=0$, then the equation of $AH$ is $(aA^{-1})^T \times x=0$.

Now, your problem is the following: given nonzero vectors $a$ (row) and $x$ (column) with $ax \neq 0$, find an invertible matrix $A$ such that $aA^{-1}=(1,0,\ldots,0)$, $Ax=(r,0,\ldots,0)$ with $r \neq 0$.

Find a basis $(a_2,\ldots,a_{n+1})$ of the $n$-dimensional space of the row vectors orthogonal to $x$. Take $A_1$ the matrix the rows of which are $(a,a_2,\ldots,a_{n+1})$. Then by definition $aA_1^{-1}=(1,0,\ldots,0)$, and $x_1=A_1x$ has zero entries at indices $2 \leq i \leq n+1$, and is nonzero. So we are done.


问:
How to prove that () Δ () ≠ ( Δ ) for any 2 sets.
So the question is how can I prove that for any 2 Sets, A and B, the symmetric difference between the power set of A and the power set of B IS NOT equal to the powr set of A symmetric difference B.

() Δ () ≠ ( Δ )

I have been stuck on this question for a while and I can't find a way to prove it.

答:
The idea is that the empty set ($\varnothing$) lies in the RHS, but not in LHS. And thus the assertion is proved.

In details, let $A$ and $B$ be a set. Since any power set contains the empty set, we have that $\varnothing \in P(A),P(B),P(A\Delta B)$. By the definition of symmetric difference, we have $\varnothing \notin P(A)\Delta P(B)$. Hence, $P(A)\Delta P(B)\neq P(A\Delta B)$, as desired. QED


问:
Linear equivalence of Divisors on a Surface
Studying algebraic geometry, while the abstract theory is pretty clear to me I often feel puzzled in practice. Here I am trying to understand linear equivalence of divisors in some practical situations. Consider the following example.

Let $S\subset\Bbb{P}^3$ be a smooth surface of degree $d\geq3$ and let $\ell\subset S$ be a line. A plane containing $\ell$ cuts on $S$ the divisor
$$ H_0=\ell+C$$
where $C\subset S$ is a curve of degree $d-1$. Let $H$ be the divisor cut on $S$ by a generic plane. So $H$ cuts $\ell$ in one point. I am not sure about the following:

Question: why can we say that $H_0\sim H \ $ (linear equivalence) ?

This is then a nice example for showing that even the most simple divisor like $\ell$ can have negative self-intersection: if $H_0\sim H$ then
$$1=H\cdot\ell=H_0\cdot\ell=(\ell+C)\cdot\ell=\ell^2+C\cdot\ell=\ell^2+d-1 $$
Hence $\ell^2=2-d<0$.

答: (Turning my comment into an answer...) The point is that restriction of divisors respects linear equivalence (as follows quickly from the definition). Any two planes in the ambient space $\mathbf{P}^3$ are linearly equivalent divisors (easy exercise!), so the same remains true when we restrict them to the surface $S$.</p>


问:
The Question of Vector space ,Subspace and their Base
$V$ : Vector space ($\text{dim}V=n$)

$W_1, W_2$ : Subspace of $V$

$\text{dim}W_1 =1, \text{dim}W_2=2, W_1 \subset W_2$

If I let $\{ v_1 \}$ be basis of $W_1$, can I let $\{ v_1, v_2 \} \, (v_2 \in W_2)$ be basis of $W_2 $?

答:
Yes, as long as $v_2 \neq \alpha \cdot v_1, \forall \alpha \in \mathbb{K}.$

If $v_2 = \alpha \cdot v_1$ for some $\alpha \in \mathbb{K},$ then $\{v_1,v_2\}$ is not linearly independent, and it can’t be a base.


答: That all depend on what $v_2$ is. If $v_2 = \lambda v_1$, $\{v_1,v_2\}$ won't be a basis of $W_2$. That will only work if $v_1,v_2$ are linearly independent.

Title: The Question of Vector space ,Subspace and their Base

Body:
$V$ : Vector space ($\text{dim}V=n$)

$W_1, W_2$ : Subspace of $V$

$\text{dim}W_1 =1, \text{dim}W_2=2, W_1 \subset W_2$

If I let $\{ v_1 \}$ be basis of $W_1$, can I let $\{ v_1, v_2 \} \, (v_2 \in W_2)$ be basis of $W_2 $?


问:
Why does graph $y=(\ln y/\ln x) x-\ln y/\ln x (x^2/y)+x$ not include $x = y$?
Why does the graph $y=(\ln y/\ln x)x (-\ln y/\ln x (x^2/y))+x$ not include all positive $y$ and $x$ for which $y = x$? When I plotted this graph on desmos , the plot did not include the line $y = x$, even though the equation is solved for all $y = x$. This equation came up in the context of me trying to prove that the equations of the solutions to $x^y = y^x$ intersect at the point $(e,e)$

答:
If we substitute $y=x$ in the equation
$$y=-\frac{x^2 \ln y}{y \ln x}+\frac{\ln y}{\ln x}x+x$$
we get $x=-x+x+x\to 0=0$ undefined.

Thus the equation above is not equivalent to $$x^y=y^x\tag{1}$$

I strongly recommend to simply prove by direct inspection that $x=y$ is a set of solution of $(1)$, and use this answer to prove that $(e,e)$ is a self intersection.

Hope this helps


问:
How to factor $ {s - a_m} \over {s-b_n} $ to $ {1-{s \over a_m}} \over {1 - {s \over b_n}} $
I saw this polynomial or algebraic equation and can verified by plugging in some numbers and figuring out the constant for a specific sets of $ a_m $ and $ b_n $. But how can I derive the equation on the right from the left? I tried multiplying both sides by s, a or b but I can't come up with anything close to the term on the right.

${{(s-a_1)(s - a_2) \; ... \; (s-a_m)} \over {(s-b_1)(s-b_2) \; ... \; (s-b_n)}} = C_0 {({1-{s \over a_1})({1 - {s \over a_2}}) \; ... \;({1 - {s \over a_m}})} \over {({1 - {s \over b_1}})({1 - {s \over b_2}}) \; ... \; ({1 - {s \over b_n}})}}$

Body:
$${s-\alpha}=\alpha\left(\frac{s}{\alpha}-1\right)=(-\alpha)\left(1-\frac{s}{\alpha}\right),$$ provided $\alpha\ne0$. So

\begin{eqnarray}
\frac{(s-a_1)(s - a_2) \; ... \; (s-a_m)}{(s-b_1)(s-b_2) \; ... \; (s-b_n)}& =&\frac{(-a_1)(-a_2)\cdots(-a_m)(1-s/a_1)(1-s/a_2)\cdots(1-s/a_n)}{(-b_1)(-b_2)\cdots(-b_n)(1-s/b_1)(1-s/b_2)\cdots(1-s/b_m)}\\
&=&C_0 {({1-{s \over a_1})({1 - {s \over a_2}}) \; ... \;({1 - {s \over a_m}})} \over {({1 - {s \over b_1}})({1 - {s \over b_2}}) \; ... \; ({1 - {s \over b_n}})}}\end{eqnarray}
where $$C_0=(-1)^{m-n}\frac{a_1\cdots a_m}{b_1\cdots b_m},$$


答:
$$\dfrac{x^2-3x+4}{x^2+3x+4}=1-\dfrac{6x}{x^2+3x+4}=1-\dfrac6{x+\dfrac4x+3}$$

Now if $x>0, x+\dfrac4x\ge2\sqrt{x\cdot\dfrac4x}=4$

$\implies\dfrac1{x+\dfrac4x+3}\le\dfrac17\implies-\dfrac6{x+\dfrac4x+3}\ge-\dfrac67$

If $x<0, x=-y, y>0, x+\dfrac4x=-\left(y+\dfrac4y\right)$

Can you take it from here?


问:
A geometry problem related to circumcenter.
If a, b and c are the lengths of the sides of a triangle ABC and if $p_1$, $p_2$ and $p_3$ are the lengths of the perpendiculars drawn from the circumcenter onto the sides BC, CA and AB respectively, then show that

$$\dfrac{a}{p_1} + \dfrac{b}{p_2} + \dfrac{c}{p_3}= \dfrac{abc}{4p_1 p_2 p_3}$$

I have done some bashing but I am unable to arrive at the final result. This problem is of 10+2 level as well.

答:
We know

$ a=2R\sin A, b=2R\sin B, c=2R\sin C $

and

$ p_1=R\cos A, p_2=R\cos B, p_3=R\cos C $

This can be seen on a diagram of a triangle and its circumcircle. From the right triangles and little angle chasing we get these.


问:
Using goal seek in excel.
Hi I don't know if this is an appropriate question for this page but I thought it was worth a shot and if nothing else a learning experience.

So I am studying actuarial mathematics and I am currently working on problem involving calculating the yield of an investment.

I have been working through these questions alongside prerecorded lectures and have been fairly understanding the concept and how to use it however my lecturer constantly says that these answers are not that exact or precise but would be a lot better when worked out using Goal seek in excel.

However my problem is that he has never shown us goal seek or mentioned any further on the topic.

Hence I was wondering if anyone could explain this too me and/or how to use it as I'm getting answers to questions wrong due to the situation.

Thanks in advance.

Adding from comments:

The question I was given was: You invest £10,000 in a project and receive back £700.87 per annum payable half-yearly in arrears for the next 25 years. Calculate the yield on this investment.

So from this we worked out that the money coming back at the end of this question is £17521.75 and the average time for this question is 12.75 $(last time period-first time period) /2$ which is $(25.5)/(2)$

And from this we do the calculation

$(17521.75/10000)^(1/12.75)$

And we get 4.5% which my lecturer then states in his lecture that this isn't a very good answer as the real answer is 4.98% but doesn't proceed to tell us how to obtain this answer.

答:
Starting from @user2661923's answer, you need to solve for $v$ the equation
$$k= \frac{v \times [1 - v^{50}]}{1 - v}$$ and we know that $v$ is close to $1$.

Using series expansion, the rhs is
$$50+1275 (v-1)+20825 (v-1)^2+249900 (v-1)^3+\cdots$$ Using series reversion
$$v=1+t-\frac{49 }{3}t^2+\frac{3038}{9} t^3-\frac{1029049 }{135}t^4+\frac{72902494
   }{405}t^5+O\left(t^6\right)$$ where $t=\frac{k-50}{1275}$.

Using your number $k= \frac{10000}{350.435}$, this gives as an estimate
$v \sim 0.976071$ while the "exact" solution is $0.975900$.

Title: Using goal seek in excel.


问:
How to solve this trigonometric equation further?
Find the sum of the roots of the equation
$$
\sin (2\cos x - \sqrt{3}) = 0
$$

belonging to the segment

$$
A_m = [2m\pi; ~ \frac{\pi}{2} + (2m+2)\pi]
$$

where $m = -4$. Round the number to two decimal places if necessary.

my incomplete solution:

$$2\cos x - \sqrt 3 = k\pi$$

$$k = 0 \text{ fits}$$
$$k = 1$$

$$ 2\cos x = pi + \sqrt3 = 3.14 + 1.73> 2$$ - not suitable. For
larger $k$, it is even worse ...

$$k = -1$$
$$2\cos x = -3.14 + 1.73 = -1.41$$ Hmm Suitable unfortunately.
Smaller ones won't fit

This means that there are 2 options

$$cos x = 0$$ $$x = \pi / 2 + 2n\pi$$ The segment $$[-8\pi, \pi / 2 - 6\pi]$$ 2
roots hit $$-8\pi + \pi / 2$$ and $$-6\pi + \pi / 2$$

$$\cos x = -0.705$$ (coincidence with the root of 2 is random here) But
approximately you can say $$x = + -2\pi / 3 + 2n\pi$$

I have a problem with subsequent calculations, it lies in the fact that I do not understand how and what to calculate

I would be grateful if you could solve this problem to the end with an explanation

答:
Hint:

For real $x,$

$$-2-\sqrt3\le2\cos x-\sqrt3\le2-\sqrt3=\dfrac1{2+\sqrt3}$$

$\dfrac1{2+\sqrt3}<\pi\implies k<1$

Again, $2+\sqrt3<2\pi\implies k>-2$

So, what are the possible values of $k?$

Can you take it from here?


答:
$$2\cos x-\sqrt3=k\pi$$ is possible with $k=-1$ and $k=0$.

Now we have the solutions

$$x=2n\pi\pm\arccos\frac{\sqrt3}2$$ and $$x=2n\pi\pm\arccos\frac{\sqrt3-\pi}2.$$

As solutions are requested in the range $\left[-8\pi,-8\pi+\dfrac52\pi\right]$ you need to find suitable values of $n$. Start with $n=-4$ (which gives two solutions) and increase $n$ until the options become too large.

Title: How to solve this trigonometric equation further?

Body:
Find the sum of the roots of the equation
$$
\sin (2\cos x - \sqrt{3}) = 0
$$

belonging to the segment

$$
A_m = [2m\pi; ~ \frac{\pi}{2} + (2m+2)\pi]
$$

where $m = -4$. Round the number to two decimal places if necessary.

my incomplete solution:

$$2\cos x - \sqrt 3 = k\pi$$

$$k = 0 \text{ fits}$$
$$k = 1$$

$$ 2\cos x = pi + \sqrt3 = 3.14 + 1.73> 2$$ - not suitable. For
larger $k$, it is even worse ...

$$k = -1$$
$$2\cos x = -3.14 + 1.73 = -1.41$$ Hmm Suitable unfortunately.
Smaller ones won't fit

This means that there are 2 options

$$cos x = 0$$ $$x = \pi / 2 + 2n\pi$$ The segment $$[-8\pi, \pi / 2 - 6\pi]$$ 2
roots hit $$-8\pi + \pi / 2$$ and $$-6\pi + \pi / 2$$

$$\cos x = -0.705$$ (coincidence with the root of 2 is random here) But
approximately you can say $$x = + -2\pi / 3 + 2n\pi$$

I have a problem with subsequent calculations, it lies in the fact that I do not understand how and what to calculate

I would be grateful if you could solve this problem to the end with an explanation


问:
How to prove that this matrix is diagonalizable?
I am trying assignment questions of Linear algebra and couldn't solve this particular question regarding diagonalizabilty.

Let $n \times n$ complex matrix $A$ satisfies $A^k = I$ the $n \times n $ identity matrix, where $k$ is a positive integer $>1$ and let $1$ not be an eigenvalue of $A$. Then how to prove that A is necessarity Diagonalizable?

As $A^k=I$  and 1 is not an eigenvalue so $(A-I ) (A^{k-1}+...+ I)=0$ implies that $(A^{k-1}+...+ I)=0$ but I am unable to move foreward .

Can you please help?

答:
There is a Theorem which says that If any square Matrix A is Diagonizable then any positive power of A    ie A^k , k belong to Z+. A^k is also Diagonizable

BUT THE CONVERSE part is true Only if A is Invertible ie. If
We are given that A^k is Diagonizable and A is Invertible
Then A is Diagonizable
. You can see the proof here If $A$ is invertible and $A^n$ is diagonalizable, then $A$ is diagonalizable.

Here We are given that A^k = I
So A is Invertible and Identify is Always diagonizable So A is Diagonizable.

Hope This Would be Helpful for You


答:
The symmetric difference of two sets is the set of elements that are in both sets but not in their intersection and is given by:
$$ A Δ B = A \cup B \backslash A \cap B $$
The powerset of a set is the set of all elements of $S$ including the empty set and the set itself and is represented by $P(S)$.

The cardinality of $P(S)$ is given by $$|P(S)| = 2^{|S|}$$
We need to prove that $$() \space Δ \space () ≠ ( \space Δ \space )$$
$$P(A \cup B) \backslash P(A \cap B) ≠  P((A \cup B) \backslash (A \cap B)) $$

Let $X = A \cup B$ and $ Y = A \cap B$. We need to show that:
$$P(X) \space \backslash \space P(Y) ≠ P(X \backslash Y)$$
We can prove that two sets are not equal by showing that they have a different number of elements. Let $|X| =x, |Y|=y$.
$$|X \backslash Y| = x - y$$
$$|P(X) \space \backslash \space P(Y)| = 2^x - 2^y$$
Similarly, on the right hand side,
$$|P(X \backslash Y)| = 2^{x-y}$$

We can prove that $$2^x - 2^y ≠ 2^{x-y}$$ by using a counter example. Suppose $x=1, y=2$
$$2^1 - 2^2 = -3 ≠ 2^{1-2}=1/2$$

Title: How to prove that () Δ () ≠ ( Δ ) for any 2 sets.

Body:
So the question is how can I prove that for any 2 Sets, A and B, the symmetric difference between the power set of A and the power set of B IS NOT equal to the powr set of A symmetric difference B.

() Δ () ≠ ( Δ )

I have been stuck on this question for a while and I can't find a way to prove it.


问:
If the conorm of a linear transformation is positive, is the transformation an isomorphism?
We defined the conorm of a linear transformation as $\inf\{\frac{|Tv|}{|v|}:v\ne0\}$, where T is a linear map between norm spaces.

I have already proven that the conorm of an isomorphism is positive. However, I'm not sure about the converse.

I think that it should be true for finitely dimensional spaces, as it would imply that $\ker(T)=\{0\}$, and I know that for finitely dimensional spaces this happens iff T is an isomorphism.

Now, I don't really know much about spaces of infinite dimensions, so my question is whether there is a counterexample of that kind for which this would not be true?

Thanks for any help.

答:
On $\ell^{2}$ define $T(a_n)=(0,a_1,a_2,...)$. Then the infmum is $1$ but $T$ is not surjective so it is not an isomorphism of $\ell^{2}$ onto itself. However the following is true:

If the infimum is positive and $T$ is bounded then $T$ is an isomorphism onto its range (which may be a proper sub space).


问:
What exactly happens, when I do a bit rotation?
I'm sitting in my office and having difficulties to get to know that exactly happens, when I do a bit rotation of a binary number.

An example:
I have the binary number 1110. By doing bit rotation, I get 1101, 1011, 0111 ...
so I have 14 and get 7, 11, 13 and 14 again.

I can't get a rule out of it... can somebody help me?

Excuse my bad English, I'm just so excited about this problem.

答:
For reasons that are irrelevant, I was looking for the arithmetic or mathematical equivalance of a bitwise rotation.
Scavaging the internet, not much can be found. Even this post here is quite rare.

After looking at the difference answers, they didn't provide an adequate solution, but some of the answers here provided a nice starting point.

So I tried to make the most simplistic mathematical equation or representation for bitwise rotation (left and right) and here is what I came up with:

Left Rotation:

$
\normalsize{
\operatorname{Left}~\!(x,s,b)\equiv(x\cdot2^{s})~\bmod(2^{b}-1)
}
$

Example:

x = 5 (value) -> 0000 0101

s = 7 (steps or rotations)

b = 8 (bits)

$
\normalsize{
\operatorname{Left}~\!(5,7,8)\equiv(5\cdot2^{7})~\bmod(2^{8}-1)=130
}
$

result = 130 -> 1000 0010

What is happening?

Well moving the bits 1 position to the left in a radix 2 or binary is the same as multiplying by $2^1$.

Moving 7 position is the equivalence of multiplying by $2^7$

To make sure that the movement actually rotates within a certain amount of bits we are taking the remainder of the outcome based on the amounts of bits.

Right Rotation:

$
\normalsize{
\operatorname{Right}~\!(x,s,b)=\operatorname{Left}~\!(x,b-(s~\bmod b),b)
}
$

First I wanted to make right rotation equation independent from  the left rotation, but that would've made the equation much more complex. (because if the rotation surpasses the range or bits-container, it's much harder to redefine it.)

The left rotation equation is very simple, as it just multiplies the value and if the value is bigger than the 'mask', 'container' or better said value of $2^{b}-1$, it will take the remainder of it and thus our wanted result.

And if we rotate 1 position to the right in 8 bits, it is the same as rotating 7 times to the left. Hence we can use the Left rotation equation with a modified step argument or variable.

To make sure we have the correct amount of steps in case that steps or rotations is bigger than our bits. We take the remainder of it:

$
\normalsize{
b - (s ~\bmod b)
}
$

As this is isn't quite a popular thing, because if you would program it, you could simply use bitwise operations, but from a mathematical standpoint it is an interesting thing and thus why I decided to answer this ca. 9 year old question.

I hope you like it!


问:
How many ways to select 5 objects from 3 groups of 4 objects each?
Consider this problem:

A question paper on mathematics consists of twelve questions divided into three parts A, B
and C, each containing four questions. In how many ways can an examinee answer five
questions, selecting at least one from each part?

I solved this problem taking a case-by-case approach. The examinee could pick 1, 1 and 3 questions, or 2, 2 and 1 questions, from the three sections. Number of ways of picking these questions is ${4 \choose 1} \times {4 \choose 1} \times {4 \choose 3}$ and ${4 \choose 2} \times {4 \choose 2} \times {4 \choose 1}$.

We will now multiply the two numbers by $3!/2!=3$ to find the number of permutations. The correct answer is ${4 \choose 1} \times {4 \choose 1} \times {4 \choose 3} \times 3 + {4 \choose 2} \times {4 \choose 2} \times {4 \choose 1} \times 3 = 624$.

Obviously, this approach is not generalizable because it involves thinking about the possible distributions. If the number of questions in each section or the number of sections was large enough, it would be very time-consuming to employ this method. What would be an alternative approach that scales well?

答:
We could use the Inclusion-Exclusion Principle.

There are $\binom{12}{5}$ ways to select five of the twelve questions on the question paper.  From these, we must subtract those selections which do not contain at least one selection from each section.

There are three ways to exclude a section and $\binom{8}{5}$ ways to select five of the remaining eight questions.

Since it is not possible to exclude two of the sections and still select five questions from the remaining four questions, we are done.

$$\binom{12}{5} - \binom{3}{1}\binom{8}{5} = 624$$

Suppose the examinee only had to answer four of the twelve questions, but was still required to answer at least one question from each part.

Observe that the examinee must answer two questions from one part and one from each of the others, which can be done in
$$\binom{3}{1}\binom{4}{2}\binom{4}{1}\binom{4}{1} = 288$$
since there are three ways to choose the section from which two questions will be answered, $\binom{4}{2}$ ways to choose two questions from that section, and four ways to choose a single question from each of the other sections.

By the Inclusion-Exclusion Principle, there are
$$\binom{12}{4} - \binom{3}{1}\binom{8}{4} + \binom{3}{2}\binom{4}{4} = 288$$
admissible ways to select the questions since there are $\binom{12}{4}$ ways to select four questions, three ways to exclude one section and $\binom{8}{4}$ ways to select four of the remaining eight questions, and $\binom{3}{2}$ ways to exclude two sections and $\binom{4}{4}$ ways to answer all four of the remaining four questions.  The reason we must add the last term is that we subtracted each selection in which all the selected questions are drawn from one section twice, once for each way of designating one of the other two sections as the excluded section.

While the first method of solving this problem is simpler in this case, the Inclusion-Exclusion Principle can be more readily applied if the number of questions in each section ot the number of sections were large.


问:
Biharmonic functions on closed manifolds?
Let $(M,g)$ be a closed (compact, $\partial M=\varnothing$) smooth connected Riemannian manifold with Laplace–Beltrami operator $\Delta_g$. As it is well-known, every harmonic function ($u\in \ker \Delta_g$) on $M$ is constant (e.g. Hodge Isomorphism Theorem).

Question: What about biharmonic functions ($u\in \ker \Delta_g^2$)?
In particular, can we say that every (everywhere defined, at least continuous) (distributionally) biharmonic function on $M$ is constant?

A few remarks: of course, the answer is negative on non-compact manifolds (think polynomials on $\mathbb R^n$), as well as on bounded domains with appropriate boundary conditions. (See e.g. this question).

Equivalently, we may look for smooth solutions to the Poisson equation $\Delta_g u\equiv 1$. By multiplying on both sides by $u$ and integrating w.r.t. the Riemannian volume $\mu_g$ we get:
$$\int u \Delta_g u d\!\mu_g=\int |du|^2 d\!\mu_g=\int u d\!\mu_g$$
If $u$ has mean $0$, we immediately get that it is constant. If otherwise, we may assume that $\int u d\!\mu_g=1$, in which case the same holds for $|du|^2$. In particular if $u$ is a solution to the eikonal equation $|du|^2\equiv\mu_g(M)^{-1}$, then it is biharmonic. If we allow $du$ to have a singularity at some point $x_0$ in $M$, then this solution should be (?) the Green kernel for the bi-Laplacian pinned at $x_0$. If however $du\equiv 1$ everywhere, I do not see why a solution should exist.

答:
There is no solution to the Poisson equation $\Delta u  =1$ on a closed Riemannian manifold. Just use (weak) maximum principle: at the maximum $x_0$ of $u$ one has $\Delta _g u(x_0) \le 0$.

The same argument shows that $\Delta u = c$ has a solution only when $c = 0$, and in this case $u$ is harmonic and thus constant. So $\Delta^2 u = 0$ also implies that $u$ is constant.


问:
basic analysis: showing $e^{-1/x^2}$ is continuous at $0$
$f(x) = e^{-1/x^2}$

from definition I need to show $$\lim_{x\to 0} f(x) = f(0)$$ I used the substitution $x = 1/t$ to show that the limit $$\lim_{x\to 0} e^{-1/x^2} = 0$$ but what is $f(0)$ and how do I know that $0 = f(0)$

答:
As stated, the question can lead to a circular argument.

You don't define $f(0)$, so it is tempting to extend it by setting $f(0)=0$ (several answerers did it). But this choice is made to ensure that $f$ is continuous at $0$, so that... $f$ is continuous at $0$.

A better question is "evaluate $\lim_{x\to 0}e^{-1/x^2}$". You can easily establish this for example by squeezing in $[0,x]$.


问:
Inequality : definition interval
We have the following parameters :

$y \in [0,1]$

$z \in [0,1]$

and I have now to proof the following thing:

$(y+z-2yz) \in [0,1]$

$0 \leq y \leq 1$

$0 \leq z \leq 1$

Therefore  $\Rightarrow 0 \leq y + z \leq 2 \tag{1}$

Then, $0 \leq yz \leq 1$ $\Rightarrow 0 \leq 2yz \leq 2 $ $\Rightarrow -2 \leq - 2yz \leq 0 \tag{2}$

But $(1) + (2)$ :

$-2 \leq y + z - 2yz \leq 2 $
and not $0 \leq y + z - 2yz \leq 1 $.

答:
All of steps taken are valid. However, you are not interpreting the result in the correct way.

See for example: Let $x\in (0,1)$, so $0<x<1$, then we have $-5<x<5$, or $x\in(-5,5)$. Thus, the possible values of $x$ are in $(-5,5)$; However, note that it is not true that every value in $(-5,5)$ is actually attained. This is due to the fact that $-5$, $5$ are not the "optimal" bounds for $x$.

Comparing to your result, it is true that, by your deduction, $-2\leq y+z-2yz\leq 2$. However, your method of considering the bounds of each term individually is not "sharp" enough that it does not give the optimal bounds you're looking for, and hence does not prove your assertion. More precisely, the reason why your method is not sharp enough is the upper (and lower, resp.) bounds of $y+z$ and $-2yz$ are not attained at the same value of $(y,z)$. Hence, the sum of upper (and lower, resp.) bounds is not an actual attainable upper (and lower, resp.) bound.

The "sharper" approach that gives the optimal bound and also proves your assertion would take into account terms together as follows:

Let $y,z\in [0,1]$. It is easy to see that $y-\frac{1}{2}\in \left[-\frac{1}{2},\frac{1}{2}\right]$ and $1-2z\in [-1,1]$, thus we have that $\left(y-\frac{1}{2}\right)(1-2z) \in \left[-\frac{1}{2},\frac{1}{2}\right]$. Now consider the expression $$y+z-2yz=y+z-2yz-\frac{1}{2}+\frac{1}{2}\\= \left(y-\frac{1}{2}\right)(1-2z)+\frac{1}{2}.$$ Thus, the expression above lies in $\left[-\frac{1}{2}+\frac{1}{2},\frac{1}{2}+\frac{1}{2}\right]=[0,1]$, as desired.

Title: Inequality : definition interval


问:
How many partitions to divide a set of n elements into k subsets, mantaining order of the set?
For example I have a string 'bbbbb' (n = 5) and want to divide it in 3 (k = 3) groups. The possibilities are 6:

b|b|bbb

b|bb|bb

b|bbb|b

bb|b|bb

bb|bb|b

bbb|b|b

How can I find the number in advance?

答:
Here we are placing 2 partitions into 2 of 4 places. So it is $ 4 \choose 2 $$= 6 $ ways.

In general for string of length $n$, we have $k-1$ partitions into $n-1$ places. Then we will have $ {n-1} \choose {k-1}  $ ways.


答: Let $A=SJS^{-1}$ be the Jordan normal form then $A^{k}=(SJS^{-1})^k=(SJ^{k}S^{-1})=I$. Then multiplying by $S^{-1}$ on the left and $S$ on the right yields $J^k=I$. If $J$ had a Jordan block $J_i$ of size $n>1$ corresponding to the eigenvalue $\lambda_{i}$, then $J_i^k$ would have super diagonal entries $(2\lambda_{i})^{(k-1)}\neq 0$.

Title: How to prove that this matrix is diagonalizable?

Body:
I am trying assignment questions of Linear algebra and couldn't solve this particular question regarding diagonalizabilty.

Let $n \times n$ complex matrix $A$ satisfies $A^k = I$ the $n \times n $ identity matrix, where $k$ is a positive integer $>1$ and let $1$ not be an eigenvalue of $A$. Then how to prove that A is necessarity Diagonalizable?

As $A^k=I$  and 1 is not an eigenvalue so $(A-I ) (A^{k-1}+...+ I)=0$ implies that $(A^{k-1}+...+ I)=0$ but I am unable to move foreward .

Can you please help?


答: Yes, you can choose a basis of $W_2$ that contains the vector $v_1$. This is due to the assumption that $W_1$ is a subspace of $W_2$.


问:
Three circle touching each other and having a common tangent
Three circle with radius $r_1$, $r_2$ and $r_3$ where ($r_1$<$r_2$<$r_3$) touch each other externally . If they have a common tangent then the value of $\sqrt{\frac{r_1}{r_2}}+\sqrt{\frac{r_1}{r_3}}$=__________

My approach is illustrated in the diagram

How will I proceed from here

答:
Length of common tangent (CT) of two touching circlesis  $L=2\sqrt{r_1r_2}$.
CT of 1 and 3 is $L_{13}=2\sqrt{r_1r_3}$, CT of 1 and 2 is $L_{12}=
\sqrt{r_1r_2}$, CT of 2 and 3 is $L_{23}=2\sqrt{r_2r_3}.$

Next, $$L_{23}=L_{12}+L_{13}\implies 2 \sqrt{r_2r_3}=2\sqrt{r_1r_3}+2 \sqrt{r_1r_2}$$
Hence we get $$\sqrt{\frac{r_1}{r_2}}+\sqrt{\frac{r_1}{r_3}}=1$$


答:
You could have assumed less. The condition "1 is not an eigenvalue of $A$" is unnecessary.

Recall that a complex matrix $A$ is diagonizable if and only if its minimal polynomial has no multiple roots. If $A^k = I$, then the minimal polynomial of $A$, which we denote by $f(x)$, necessarily divides $x^k - 1$. We conclude that $f(x)$ cannot have multiple roots since $x^k - 1$ has $k$ distinct roots in $\mathbb{C}$. Hence $A$ is diagonizable.

Alternatively, we may use Jordan Canonical Form to see the diagonizability. Suppose on the contrary $A$ is not diagonalizable, then there must exist some nontrivial Jordan block $B$ of the Jordan Canonical Form in the following form:
$$\begin{pmatrix}
\lambda & 1       & 0       & \cdots  & 0 \\
0       & \lambda & 1       & \cdots  & 0 \\
\vdots  & \vdots  & \vdots& \ddots  & \vdots \\
0       & 0       & 0        & \lambda & 1       \\
0       & 0       & 0       & 0       & \lambda \end{pmatrix}$$
which satisfies $B^k = I$. This is impossible by calculating, for example, the $(1,2)$-entry of $B^k$, since $\lambda \neq 0$.


问:
T-test vs Z-test

It is known that the diameter of a particular type of automobile tyre follows a normal distribution with mean and variance being $600$ millimetres and $3$ millimetres respectively. Ten samples are collected from a large population of manufactured tyres as :

It is believed that the manufactured lot has a diameter different from $600$ mm. Which hypotheses test can be used here? What conclusion can be drawn on $H_0$.

Select one:

T-test, Failed to reject $H_0$

Z-test, Failed to reject $H_0$

Z-test, Reject $H_0$

F-test, Reject $H_0$

Note: I'm confused between T-test and Z-test. When one should go for T-tset and when for Z test. In this particular senario which one is best and why?

答:
You are given the true mean and variance of the tire diameter.  As such, you would use a $Z$-test because the test statistic does not need you to estimate the population variance from the sample.

The $F$-test is not appropriate because that is a test of variance, but the hypothesis concerns the mean of the sample.  The $T$-test can be used, in the sense that the testing procedure is statistically valid, but is not the test of first choice here, because it is not as powerful as a $Z$-test when you know the variance of the population.

Title: T-test vs Z-test