问:
Find what values of $n$ give $\varphi(n) = 10$

For what values of $n$ do we get $\varphi(n) = 10$?

Here, $\varphi$ is the Euler Totient Function. Now, just by looking at it, I can see that this happens when $n = 11$. Also, my friend told me that it happens when $n = 22$, but both of these are lucky guesses, or educated guesses. I haven't actually worked it out, and I don't know if there are any more.

How would I go about answering this question?

答:
The second is not a lucky guess. If $n$ is odd, then $\varphi(2n)=\varphi(2)\varphi(n)=\varphi(n)$.

There is clearly only one prime $n$ such that $\varphi(n)=10$. And it has the automatic companion $22$. To search for other non-primes, recall that $\varphi$ is multiplicative, meaning that if $a$ and $b$ are relatively prime, we have $\varphi(ab)=\varphi(a)\varphi(b)$.

Let us look at the possibilities $\varphi(a)=2$, $\varphi(b)=5$. The equation $\varphi(b)=5$ has no solutions, since by symmetry $\varphi(n)$ is even if $n\gt 2$.

Remark: Consider $n=2k$ where $k$ is odd. If $k=1$, the equation $\varphi(n)=n$ has the solutions $3$, $6$, and $4$.

If $k\gt 1$, and $2k+1$ is not prime, then we cannot find an $x$ such that $\varphi(a)=2k$. If $2k+1$ is prime, the equation $\varphi(x)=2k$ has precisely two solutions, $2k+1$ and $4k+2$. The argument is basically the same as in the case $k=5$ above.


问:
Show for the centroid $M$ of a triangle $ABC$ that $\vec{OM}=\frac13(\vec{OA}+\vec{OB}+\vec{OC})$
Show for the centroid $M$ of a triangle $ABC$ that $$\vec{OM}=\dfrac13\left(\vec{OA}+\vec{OB}+\vec{OC}\right)$$
where $O$ is an arbitrary point.

I haven't studied position vectors and am very new to vectors so I would like a simple solution. Any help would be appreciated. I don't even know how to start. Maybe we can use the fact that the centroid divides each median in a ratio of $2:1$? Thank you!

答:
If the coordinate of vertices A, B and C are $(x_1, y_1),( x_2, y_2)$ and $(x_3, y_3)$ then coordinate of centroid is $M(\frac{x_1+x_2+x_3)}3,\frac{y_1+y_2+y_3)}3)$.

The vector form is:

$\vec{OM}=(x_M i+y_Mj)=(\frac{x_1+x_2+x_3)}3 i+\frac{y_1+y_2+y_3)}3 j)$

$A(x_1,y_1)$ means vector $\vec{OA}=(x_1 i+y_1 j)$ , so is about B and C so we can write:

$(x_M i+ y_M j)=\frac 13[(x_1 i+y_1j)+(x_2 i+y_2 j)+(x_3 i+y_3 j)]$

Or in vector form:

$\vec{OM}=\frac 13(\vec{OA}+\vec{OB}+\vec{OC})$

Now if you transform coordinate to an arbitrary point like $O(x_O. y_O)$ the coordinates  of vertices will be:

$(x_1-x_O, y_1-y_O)$ which is vector $\vec{OA}=(x_1-x_O)i+(y_1-y_O)j$ ,$(x_2-x_o, y_2-y_o)$ which is vector $\vec{OB}=(x_2-x_O)i+(y_2-y_O)j$  and $(x_3-x_O, y_3-y_O)$ which is vector $\vec{OC}=(x_3-x_O)i+(y_3-y_O)j$ .Hence the coordinate of centroid is:

$M(\frac{(x_1-x_o)+(x_2-x_o)+(x_3-x_o)}3),(\frac{(y_1-y_o)+(y_2-y_o)+(y_3-y_o)}3)$

Or :

$\vec{OM}=(\frac{(x_1-x_o)+(x_2-x_o)+(x_3-x_o)}3)i+(\frac{(y_1-y_o)+(y_2-y_o)+(y_3-y_o)}3 )j$

The vector representation of this is:

$\vec{OM}=\frac{\vec{OA}+\vec{OB}+\vec{OC}}3$

ContentLicense: CC BY-SA 4.0


问:
Terminologie of y1/y0
If I set a list of points: {2,3,5}

"Derivative" will be {3-2=1, 5-3=2}

How about {3/2=1.5, 5/3~=1.666}.  What is the name of this ?

答:
To nuance and slightly correct @Mark S.'s answer, there is more depth to your question than it seems.

In fact, you can truly consider the sequence of differences of a sequence (the sequence defined by the "difference equation") to be what we should call the "discrete derivative". In this context, recurrent sequences are actually 'discrete differential equations' and series are 'discrete integrals', $2^x$ is the discrete equivalent of $e^x$, etc. All of these operations are linked to the operations of the arithmetic mean.

However, there also exist a bunch of derivatives and integrals linked to the operations of the geometric mean. This is generally referred to as "non-newtonian calculus". See https://en.wikipedia.org/wiki/Product_integral or https://www.calculushowto.com/non-newtonian-calculus/ for example. In this context, the name you are looking for is "discrete multiplicative derivative", corresponding to $u^{[1]}_n = \frac{u_{n+1}}{u_n}$. I have personally encountered uses of it in statistics, in some methods of analysis of time series (though the authors did not seem to know about non-newtonian calculus in general).

You can also take a look at this very interesting table: https://en.wikipedia.org/wiki/List_of_derivatives_and_integrals_in_alternative_calculi

And here, for an article which makes use of these techniques:
https://d1wqtxts1xzle7.cloudfront.net/46356975/khatami.pdf?1465464788=&response-content-disposition=inline%3B+filename%3DAn_Analytical_Method_for_Some_Nonlinear.pdf&Expires=1612625066&Signature=c~tREtS5~4f3yRRWrVyRQeLpnq46rgFfgCYQk2DFssvCNB13YcOlvgqB-8gxXfkCsWfpOqQJL2mYwZpThx9-qgurW9vlZR39jSBhasmZdJQJcOUmtvBaO4o4i6W0yBpkEpeACrOiNaUPMz3ECJWQ~5Mi2AuthIXHKgQUiv-MSUjwj9GDE6S~aXkHRuXAFRiW4gJvyU1~8~ynH3umlwRpvYUoc67lIamDquyupmsgcHzYihCXOrtwgo8DHMja4kaozRKvHLpIXeDO0QKQAfI3KOGEZE9rAt1WjdJYqERWhC6~uQh0Z66dOizd1my6nfjmr96V8QQoxtQhEJh5exY5UQ__&Key-Pair-Id=APKAJLOHF5GGSLRBV4ZA

Title: Terminologie of y1/y0

Body:
To nuance and slightly correct @Mark S.'s answer, there is more depth to your question than it seems.

In fact, you can truly consider the sequence of differences of a sequence (the sequence defined by the "difference equation") to be what we should call the "discrete derivative". In this context, recurrent sequences are actually 'discrete differential equations' and series are 'discrete integrals', $2^x$ is the discrete equivalent of $e^x$, etc. All of these operations are linked to the operations of the arithmetic mean.

However, there also exist a bunch of derivatives and integrals linked to the operations of the geometric mean. This is generally referred to as "non-newtonian calculus". See https://en.wikipedia.org/wiki/Product_integral or https://www.calculushowto.com/non-newtonian-calculus/ for example. In this context, the name you are looking for is "discrete multiplicative derivative", corresponding to $u^{[1]}_n = \frac{u_{n+1}}{u_n}$. I have personally encountered uses of it in statistics, in some methods of analysis of time series (though the authors did not seem to know about non-newtonian calculus in general).

You can also take a look at this very interesting table: https://en.wikipedia.org/wiki/List_of_derivatives_and_integrals_in_alternative_calculi

And here, for an article which makes use of these techniques:
<a href="https://d1wqtxts1xzle7.cloudfront.net/46356975/khatami.pdf?1465464788=&response-content-disposition=inline%3B+filename%3DAn_Analytical_Method_for_Some_Nonlinear.pdf&Expires=1612625066&Signature=c%7EtREtS5%7E4f3yRRWrVyRQeLpnq46rgFfgCYQk2DFssvCNB13YcOlvgqB-8gxXfkCsWfpOqQJL2mYwZpThx9-qgurW9vlZR39jSBhasmZdJQJcOUmtvBaO4o4i6W0yBpkEpeACrOiNaUPMz3ECJWQ%7E5Mi2AuthIXHKgQUiv-MSUjwj9GDE6S%7EaXkHRuXAFRiW4gJvyU1%7E8%7EynH3umlwRpvYUoc67lIamDquyupmsgcHzYihCXOrtwgo8DHMja4kaozRKvHLpIXeDO0QKQAfI3KOGEZE9rAt1WjdJYqERWhC6%7EuQh0Z66dOizd1my6nfjmr96V8QQoxtQhEJh5exY5UQ__&Key-Pair-Id=APKAJLOHF5GGSLRBV4ZA" rel="nofollow noreferrer">https://d1wqtxts1xzle7.cloudfront.net/46356975/khatami.pdf?1465464788=&response-content-disposition=inline%3B+filename%3DAn_Analytical_Method_for_Some_Nonlinear.pdf&Expires=1612625066&Signature=c~tREtS5~4f3yRRWrVyRQeLpnq46rgFfgCYQk2DFssvCNB13YcOlvgqB-8gxXfkCsWfpOqQJL2mYwZpThx9-qgurW9vlZR39jSBhasmZdJQJcOUmtvBaO4o4i6W0yBpkEpeACrOiNaUPMz3ECJWQ~5Mi2AuthIXHKgQUiv-MSUjwj9GDE6S~aXkHRuXAFRiW4gJvyU1~8~ynH3umlwRpvYUoc67lIamDquyupmsgcHzYihCXOrtwgo8DHMja4kaozRKvHLpIXeDO0QKQAfI3KOGEZE9rAt1WjdJYqERWhC6~uQh0Z66dOizd1my6nfjmr96V8QQoxtQhEJh5exY5UQ__&Key-Pair-Id=APKAJLOHF5GGSLRBV4ZA


问:
$\text{E}[xW^2 + yW + z] = x \text{E}[W^2] + y \text{E}[W] + z$
I am currently helping a friend with a math problem, but I am currently stuck with it. The question asked to prove the following identity, where $x$, $y$, $z$  are constants:

$$\text{E}[xW^2 + yW + z] = x \text{E}[W^2] + y \text{E}[W] + z$$

May I ask, what is the topic that this falls under? I took an interest in statistics, but I'm not sure what topic in statistics this falls under.

答: Recall that if $W$ is a random variable on a probability space $(\Omega, \mathcal{B}, \mathbb{P}),$ then the expected value of $W$ is defined as $\mathbb{E}[W] := \int_\Omega W d \mathbb{P}$ (a Lebesgue integral). Since this integral is linear, it immediately follows that $\mathbb{E}[x W^2 + y W + z] = x \mathbb{E}[W^2] + y \mathbb{E}[W] + z$ (maybe the only thing that you could elaborate on is the fact that $\mathbb{E}[1] = 1,$ but this is trivial as well, because $\int_\Omega 1 \,\,d\mathbb{P} = 1 \cdot \mathbb{P}(\Omega) = 1,$ since $\mathbb{P}$ is probabilistic). Hope this clears the air.


问:
Quadrature for numerically evaluating Cauchy integral formula using unit circle as closed contour
I am currently trying to evaluate the derivatives of a function $F(z)$ in z=0, which is only known numerically on the unit circle ("$UC$") in the complex plane. My question is this:

given $z=\mathrm{e}^{-iwT}$ on $UC$ and Cauchy's integral formula, we get

\begin{equation}
F^{(n)}(z=0) = \frac{1}{2\pi i}\int_{UC} \frac{F(z)}{z^{n+1}}\mathrm{d}z = -\frac{\omega}{2\pi}\int_0^{2\pi/\omega}F(\mathrm{e}^{-i\omega T})\mathrm{e}^{in\omega T} \mathrm{d}T\\ =-\frac{1}{2\pi}\int_0^{2\pi}F(\mathrm{e}^{-i\tilde{T}})\mathrm{e}^{in\tilde{T}} \mathrm{d}\tilde{T}.
\end{equation}
Does anybody know of any quadrature schemes fitting for evaluating the above expression without too many sampling points? Im thinking there might be something equivalent to e.g. Gauss-Laguerre quadrature, but with another weight-function? Any suggestions would be appreciated.

For anybody curious about what it is for, the derivatives of $F$ in $z=0$ are actually Greens functions of a perturbed quantum system in the adiabatic limit. The order of derivative is related to how many photons are absorped/emitted. $F$ is actually matrix valued and will consume a lot of RAM, which is why i would like to keep the sampling points to a minimum. $F$ is formally a Laurent-series on the form $F(z,E) = \sum _{n=-\infty}^{\infty}z^n G_n(E)$, where the $G_n$'s are various Greens functions of the system.

答: The trapezoidal rule would be ideal here, because the function is periodic (as well as smooth) and this rule converges almost exponentially fast for such functions.</p>


问:
Bijective mapping defined by modulo operation

Q. If $m$ and $n$ are two positive integers and $f:A\to A,$ where $A=\{0,1,2,3,...,m-1\},$ is defined by $f(x)= nx\pmod m.$ Find the condition on $n$ and $m$ which ensures that $f$ is one-to-one and onto.

My attempt: Here $f^{-1}$ must exist if $f$ is a bijection. Now as $f$ maps $x$ to $nx$ under modulo $m$ so $f^{-1}$ must map $x$ to $n^{-1}x$ under modulo $m.$ But in $\mathbb Z_m$(or here $A$), $n^{-1}$ will exist iff $\gcd(m,n)=1.$ So I think the required condition should be that $m$ and $n$ are relatively prime. (I don't know what the actual answer is!)

I have just started with number theory and discrete mathematics this semester. So I don't have much knowledge to deal with such type of questions involving moduler arithmetic. So kindly guide me saying that whether my approach is correct or not. If not, please suggest me a way to reach the answer to this question or if it is correct, then can we reach the same conclusion using a different approach? I would love to know about your insights. Thanks in advance...

答:
Your conjecture is valid.

Observe that a function $g:X\to X$ on a finite set is one-to-one exactly when it's onto: indeed, $g$ collapses two elements of $X$ if and only if its range contains less than $|X|$ elements.

If $d$ is a common divisor of $n$ and $m$, then $d$ also divides $f(x)=nx \mod m$, so in particular, if $d>1$, e.g. $1$ is not in the range of $f$.

On the other hand, if $m,n$ are relatively prime, then $f$ is one-to-one, because then we have
$$nx\equiv ny\pmod m\iff m\,|\,n(x-y)\iff m\,|\,(x-y)\iff x\equiv y\pmod m\,.$$


问:
How to verify the orthogonal projection formula?

Let $B = \{\vec{b}_1, \vec{b}_2, \vec{b}_3\}$ a orthogonal basis $\in V^3$.
  Verify that $\forall\,\vec{u} \in V^3$,
  $$\vec{u} = \left(\frac{\vec{u}\cdot\vec{b}_1}{\vec{b}_1\cdot\vec{b}_1}\right)\vec{b}_1 + \left(\frac{\vec{u}\cdot\vec{b}_2}{\vec{b}_2\cdot\vec{b}_2}\right)\vec{b}_2 + \left(\frac{\vec{u}\cdot\vec{b}_3}{\vec{b}_3\cdot\vec{b}_3}\right)\vec{b}_3$$

I know the concept and how to use it, but I have no ideia how to verify it.

Can anyone please help me to check it?

Thanks in advance

答:
Let me restate your problem as I can make sense of it:

Let $B = \{\vec{b}_1, \vec{b}_2, \vec{b}_3\}$ a orthogonal basis $\in V$, a 3-dimensional vector space. Verify that $\forall\,\vec{u} \in $,
$$\vec{u} = \left(\frac{\vec{u}\cdot\vec{b}_1}{\vec{b}_1\cdot\vec{b}_1}\right)\vec{b}_1 + \left(\frac{\vec{u}\cdot\vec{b}_2}{\vec{b}_2\cdot\vec{b}_2}\right)\vec{b}_2 + \left(\frac{\vec{u}\cdot\vec{b}_3}{\vec{b}_3\cdot\vec{b}_3}\right)\vec{b}_3$$

If that is what you meant, then what follows may help you:

Given that $B = \{b_1,b_2, b_3\}$ is a basis, we can write $u$  as a linear combination of the $b_i$'s:
$$u=\lambda_1 b_1+\lambda_2 b_2+\lambda_3 b_3$$
Now we should find tha $\lambda_i$'s. Since this basis is an orthogonal one, $$b_1 \cdot b_2=b_1 \cdot b_3=b_2 \cdot b_3=0$$
So we can find the $\lambda_i$ by computing the inner product of $u$ and $b_i$. For example,
$$u \cdot b_1=(\lambda_1 b_1+\lambda_2 b_2+\lambda_3 b_3)\cdot b_1=\lambda_1b_1\cdot b_1+\lambda_2b_2\cdot b_1+\lambda_3b_3\cdot b_1=\lambda_1b_1\cdot b_1$$
So $$\lambda_1=\frac{u \cdot b_1}{b_1 \cdot b_1}$$

ContentLicense: CC BY-SA 3.0


问:
What math do I start with as an adult learner?
I am hilariously bad at math. Even in High school I didn't reach past geometry, I just couldn't care less when I was in HS and didn't pay attention at all. When I did pay attention I was lost, frustrated and painfully confused. I'm 25 now with almost zero math skills. I am super interested in picking a STEM major but I feel like I need to start from square one as I am not even confident in my multiplication skills. What type of math do I start with and in what order? What exercises do I need to practice to become better at math?</p>

答:
Assuming you reached the point of studying functions, Thomas' Calculus by Hass, Heil, and Weir (now in it's 14th edition) is an excellent, modern textbook. It is full of great examples and exercises, which is critical for a beginner. This should help you achieve a strong foundation for further studies.

For reference, here is the first page of my text:


问:
Irrationality of ${\pi}$ and $e^{x/y}$
I attempt to prove that ${\pi}$ is an irrational number. For this, I use the beautiful continued fraction given by Brouncker who rewrote Wallis' formula as a continued fraction, which Wallis and later Euler (1775) proved to be equivalent.

The continued fraction given by Brouncker is
$$\frac{4}{\pi} =
1 + \cfrac{1}{3 + \cfrac{4}{5 + \cfrac{9}{7 + \dots}}}$$

Now we know that every infinite continued fraction is irrational. Does it mean that since RHS has an infinite continued fraction the LHS $\frac{4}{\pi}$ must also be irrational? And hence ${\pi}$ must be an irrational number?

EDIT 1: From this post General Continued Fractions and Irrationality If, in the continued fraction
$$\cfrac{b_1}{a_1 + \cfrac{b_2}{a_2 + \cfrac{b_3}{a_3 + \dots}}}$$
the values $a_{i}, b_{i}$ are all positive integers, and if we have $a_i \geq b_i$ for all $i$ greater than some $n$, then the value of the continued fraction is irrational(Is this a necessary and sufficient condition? I am in doubt since the golder relation $\phi$ has $a_i \geq b_i=1$ for all $n\geq1$). As the condition is not satisfied the proof is actually wrong.

EDIT 2:
How to prove the following continued fraction of  $e^{x/y}$

$${\displaystyle e^{x/y}=1+{\cfrac {2x}{2y-x+{\cfrac {x^{2}}{6y+{\cfrac {x^{2}}{10y+{\cfrac {x^{2}}{14y+{\cfrac {x^{2}}{18y+\ddots }}}}}}}}}}}$$

Since $a_i \geq b_i$ for all $i \geq 1$. By the condition of irrationality of generalized continued fraction, proving this directly proves that $e^{x/y}$ will be an irrational number!

答:
Regarding the second part of the question.

I think this might be a solution.

The Continued Fraction Expansion of the hyperbolic tanh function discovered by Gauss is

$$\tanh z = \frac{z}{1 + \frac{z^2}{3 + \frac{z^2}{5 + \frac{z^2}{...}}}} \\$$

We also know that the hyperbolic tanh function is related to the exponential function with the following formula

$$\tanh z=\frac{e^z-e^{-z}}{e^z+e^{-z}}$$

Putting $\frac{x}{y}$ in place of $z$ in the previous equation we get
$$\frac{e^{\frac{x}{y}}-e^{-{\frac{x}{y}}}}{e^{\frac{x}{y}}+e^{-{\frac{x}{y}}}}= \frac{{\frac{x}{y}}}{1 + \frac{{\frac{x}{y}}^2}{3 + \frac{{\frac{x}{y}}^2}{5 + \frac{{\frac{x}{y}}^2}{...}}}} \\$$

This continued fraction can be simplified into

$$\frac{e^{\frac{x}{y}}-e^{-{\frac{x}{y}}}}{e^{\frac{x}{y}}+e^{-{\frac{x}{y}}}}= \frac{x}{y + \frac{x^2}{3y + \frac{x^2}{5y + \frac{x^2}{...}}}} \\$$

This equation can be further be simplified as

$$1+\frac{2}{e^{\frac{2x}{y}}-1}=y + \frac{x^2}{3y + \frac{x^2}{5y + \frac{x^2}{...}}} \\$$

$$\frac{e^{\frac{2x}{y}}-1}{2}=\frac{1}{(\frac{y}{x}-1) + \frac{x}{3y + \frac{x^2}{5y + \frac{x^2}{...}}}}$$

From this equation after some algebraic manipulation, we finally get the continued fraction expansion of $e^{x/y}$ as

$${\displaystyle e^{x/y}=1+{\frac {2x}{2y-x+{\frac {x^{2}}{6y+{\frac {x^{2}}{10y+{\frac {x^{2}}{14y+{\frac {x^{2}}{18y+\ddots }}}}}}}}}}}$$

This is an infinite generalized continued fraction. We will now state the necessary and sufficient condition for the continued fraction proved by Lagrange given in Corollary 3, on page 495, in chapter XXXIV on "General Continued Fractions" of Chrystal's Algebra Vol.II to converge into an irrational number.

\begin{theorem}The necessary and sufficient condition that the continued fraction
$$\frac{b_1}{a_1 + \frac{b_2}{a_2 + \frac{b_3}{a_3 + \dots}}}$$

is irrational is that the values $a_{i}, b_{i}$ are all positive integers, and if we have $|a_i| > |b_i|$ for all $i$ greater than some $n$
\end{theorem}

In the continued fraction of $e^{x/y}-1$ as we have derived $a_{i}, b_{i}$ are equals to $2(2i-1),x^2$ except $i=1$.Therefore we have $|a_i| > |b_i|$ for all $i>\frac{\frac{x^2}{2}+1}{2}$. Hence we have proved that $e^{x/y}-1$ is irrational which in turn means $e^{x/y}$ is irrational.

Title: Irrationality of ${\pi}$ and $e^{x/y}$


问:
Function satisfying a symmetric two-variable inequality
Does anyone know if there is a measurable function $f: {\mathbb R} \to {\mathbb R}$
such that

$$
|x+y|-|x-y| \geq f(x)f(y)
$$

for any $x,y \in {\mathbb R}$.

Motivation : if $f$ exists, one can immediately deduce an easy
answer to that other question.

答:
Let $f$ be a function (no further assumptions, not even measurabilty) $\mathbb R\to \mathbb R$ with
$$\tag1f(x)f(y)\le |x+y|-|x-y|\quad\text{for al }x,y\in\mathbb R.$$
Assume there is $y>0$ with $f(y)>0$. Then for $x\ge y$ we have $f(x)\le \frac{2y}{f(y)}$ and for $0\le x<y$, $f(x)\le \frac{2x}{f(y)}\le \frac{2y}{f(y)}$. Hence $f$ is bounded from above over the nonnegative reals.
Since the functions $x\mapsto -f(x)$ and  $x\mapsto f(-x)$ also satisfy $(1)$, we conclude that $f$ is bounded, say $|f(x)|\le M$ for all $x$.
But if we plug $x=M^2+1>0$ and $y=-x$ into $(1)$, we get $$-M^2\le f(M^2+1)f(-M^2-1)\le-2M^2-2,$$ and hence $M^2\le-2$, a contradiction.

We conclude that no function satisfying $(1)$ exists.


问:
Understand use of absolute values in proof for Quotient Rule for convergent sequences
My question is about the following proof:

Lemma: If $\lim_{n \to \infty} a_n = l$ and $l > 0$, then there is an integer $N$ such that $a_n > \frac{1}{2}l$ for all $n > N$.

Quotient Rule for convergent sequences

If $\lim_{n \to \infty} a_n = l$ and $\lim_{n \to \infty} b_n = m$, then $\lim_{n \to \infty} \frac{a_n}{b_n} = \frac{l}{m}$, for $m \ne 0$.

Proof: For $m > 0$.

Write the required expression in terms of the null sequences $a_n - l$ and $b_n - m$.

\begin{equation*}
 \frac{a_n}{b_n} - \frac{l}{m} = \frac{m(a_n -l) - l (b_n - m)}{b_n m}
 \end{equation*}

Note that $m(a_n -l) - l (b_n - m)$ is a null sequence, but the denominator $b_n m$ can take the value of $0$.
However by the lemma, we know that for some integer $N$ we have $b_n > \frac{1}{2}m$, for all $n > N$, so the terms of $(b_n)$ are eventually positive.

Thus, for all $n > N$, we find

\begin{align*}
 \bigg| \frac{a_n}{b_n} - \frac{l}{m} \bigg| = \frac{|m(a_n -l) - l (b_n - m)|}{b_n m} \le \frac{|m(a_n -l) - l (b_n - m)|}{ \frac{1}{2} m^2 }
 \end{align*}

Since the RHS defines a null sequence, it follows by the Squeeze Theorem for null sequences that $\big( \frac{a_n}{b_n} - \frac{l}{m} \big)$ is null, as required.

In general I follow how the proof is structured, but I get lost with the last inequality.

Using the lemma, we know there must exist some $N$ such that the denominator $b_n m$ will be eventually positive. So we don't have to worry about division by $0$ for all $n > N$.

What I don't get: But why do they introduce absolute value signs in the last inequality? Couldn't we just continue with $\frac{a_n}{b_n} - \frac{l}{m} = \frac{m(a_n -l) - l (b_n - m)}{b_n m}$,  as for all $n > N$, we avoided division by $0$.

答:
First of all, I really wouldn't call this result a Theorem, but that's not really the point of your question. The reason why the absolute value of the aforementioned quantity appears in the proof of your proposition is precisely the way convergence is set up in $\mathbb{R}.$ To elaborate on that, recall that convergence is a topological concept and, given $(X, \tau)$ a topological space, we say that a sequence $(x_n)_{n \in \mathbb{N}} \subset X$ is convergent to $\lambda \in X$ if for every open set $U \in \tau$ such that $\lambda \in U,$ there is $n_U \in \mathbb{N}$ so that $x_n \in U$ for all $n \geq n_U.$

In the particular case of a metric space $(X, d)$, we have a very natural way to topologise $X,$ namely we consider the topology generated by the family of balls $B(x, r) := \{ y \in x: d(x, y) < r \},$ where $x \in X$ and $r > 0.$ On $\mathbb{R}$ we have exactly the topology generated by the standard metric, so putting all of these facts together explains why we need to consider the absolute value of the quantity you mentioned, not just the quantity itself.

I hope I understood your question correctly and that you will find this helpful.


问:
Is the given statement true?
Does there exist a continous function from $\mathbb{R}$ to $\mathbb{R}$ such that $f(\mathbb{R})= \mathbb{Z}$ (the set of integers)?

If yes then please provide with an example. Any help would be greatly appreciated.

答:
If $(X,T)$ is a connected topological space and $f$ a continous and surjective function between $(X,T)$ and $(Y,S)$, then $(Y,S)$ is connected.
There is no such function because $\mathbb{R}$ is connected and $\mathbb{Z}$ is not. The continous function would be surjective over $\mathbb{Z}$, making $\mathbb{Z}$ connected (in the usual topology restricted to $\mathbb{Z}$).

Title: Is the given statement true?


问:
Why is $a^{-1} \cdot a \equiv 1 \text{ mod } m$ a lemma in universal hashing?
I have been given the following lemma in a online lecture on universal hashing:

Lemma: Let m be a prime. For any $a \in \{ 1, \dots, m-1 \} $ there exists a unique inverse $a^{-1}$ such that

$a^{-1} \cdot a \equiv 1 \text{ mod } m$

($\mathbb{Z}_{m}$ is a field)

Example. m = 7

a
1
2
3
4
5
6

$a^{-1}$
1
4
5
2
3
6

So to understand this question, I have found out that the $a^{-1}$ is actually what we call the Modular Multiplicative Inverse. So i went to some online calculator of that and inserted for example 2 and 7 as modulu. But what i don't understand is, if we replace the modular muliplicative inverse in this equation we get for example:
$$ 4 * 2 \equiv 1 \text{ mod } 7$$
$$ 5 * 3 \equiv 1 \text{ mod } 7$$
$$ 6 * 6 \equiv 1 \text{ mod } 7$$

Which is not true. The only way this is true is that
$$
(4 * 2) \text{ mod } 7 \equiv 1 \text{ mod } 7
$$

But I am not sure if that's what they mean and why they wouldn't write that out?

答:
When we write $4 \cdot 2 \equiv 1 \mod 7$ the $\mod 7$ (usually placed in parentheses) is telling us what equivalence is used.  The statement is true as written because $4 \cdot 2$ and $1$ belong to the same equivalence class $\mod 7$

If I start by saying I am working in the field $\Bbb {Z/7Z}$ it becomes correct to write $4 \cdot 2 = 1$


问:
asymptotical stability
If my system has the equilibirum point at $(x_1,0,0)$ where $x_1\in  \mathbb{R}$. Why from this information I can say that the system is not asymptotically stable?

答:
Assume it were. Then e.g. the equilibrium point $(0, 0, 0)$ is asymptotically stable. So there exists some $\varepsilon > 0$ such that $\displaystyle \lim_{t \rightarrow \infty} \varphi(t) = (0, 0, 0)$ for all solutions $\varphi$ with $\varphi(0) \in B_\varepsilon((0, 0, 0))$.

Now choose $x \in \mathbb{R} \setminus \lbrace 0 \rbrace$ such that $(x, 0, 0) \in B_\varepsilon((0, 0, 0))$. But then there exists $\delta > 0$ such that all solutions $\varphi$ with $\varphi(0) \in B_\delta((x, 0, 0))$ converge to $(x, 0, 0)$ as $t \rightarrow \infty$. But by construction, $ U:= B_\delta((x, 0, 0)) \cap B_\varepsilon((0, 0, 0))$ is nonempty. So solutions $\varphi(0) \in U$ converge to $(x, 0, 0)$ and $(0, 0, 0)$ as $t \rightarrow \infty$. But this makes no sense at all.

Title: asymptotical stability


问:
Using finite difference for a simple second order BVP
I'm trying to solve this BVP using the finite difference method
$$u_{xx}(x) +sin(x) = 0$$
with $u(0)=u(2\pi)=0$. When I solve it on python, I get the correct curve shape I expect, a sinusoidal one, but the magnitude is much lower than what I expect. I'd expect a maximum and minimum of $1$ and $-1$, but they come out very small, like ~$0.04$.

Here's my code

```<code>import numpy as np
import math
import matplotlib.pyplot as plt

N = 100 #number of nodes
L = 2.0*math.pi #length of x-axis
x = np.linspace(0.0,L,N)
h = L/N #step length
u = np.zeros(N) #solution vector

u[0] = 0.0
u[N-1] = 0.0

#solve
for j in range(10):
    for i in range(1,N-1):
        u[i] = (u[i-1]+u[i+1]+h**2*math.sin(x[i]))/2.0

plt.plot(x,u)
</code>```

and here's the curve

So what causes the incorrect magnitude? I should get a regular $sin(x)$ curve

Body:
10 iterations is far too few for Jacobi method (or in general any linear method) to converge with any reasonable accuracy.

In fact, a quick experiment printing result every 200 iterations gives
$$
\begin{array}{c||c|c}
\text{iteration}&\text{max error }(+)&\text{max error }(-)\\\hline
200 & 0.461371 & -0.455683 \\\hline
400 & 0.223509 & -0.208337 \\\hline
600 & 0.116791 & -0.098056 \\\hline
800 & 0.068118 & -0.049773 \\\hline
1000 & 0.045499 & -0.029056 \\\hline
1200 & 0.034605 & -0.020533 \\\hline
1400 & 0.029090 & -0.017293 \\\hline
1600 & 0.026095 & -0.016330 \\\hline
1800 & 0.024342 & -0.016295 \\\hline
2000 & 0.023215 & -0.016601 \\\hline
&\dots&\\\hline
10000 & 0.019570 & -0.019567 \\\hline
&\dots&\\\hline
20000 & 0.019568 & -0.019568
\end{array}
$$
which shows how slowly Jacobi converges.


问:
Intuition of the meaning of homology groups
I am studying homology groups and I am looking to try and develop, if possible, a little more intuition about what they actually mean. I've only been studying homology for a short while, so if possible I would prefer it if this could be kept relatively simple, but I imagine it is entirely possible there is no real answer to my query anyway.

As I said above, I want to gain a little deeper understanding of what the n-th homology group actually means: I can happily calculate away using Mayer-Vietoris but it doesn't really give me a great deal of intuition about what the n-th homology group actually means. For example, with homotopy groups, the fundamental group is in some sense a description of how loops behave on the object in question, and it is obvious to me why that is what it is for say, the torus or the circle. However, I have no idea what, if anything, I am actually saying about a triangulable object when I talk about it having 0-th homology group this or 1st homology group that.

The best I have been able to find online or in my limited book selection is the brief description "intuitively, the zeroth homology group counts how many disjoint pieces make up the shape and gives that many copies of $\Bbb Z$, while the other homology groups count different types of holes". What 'different types of holes' are there, roughly speaking? I appreciate that it may often be completely non-obvious what the low-order homology groups are for some complicated construction, but perhaps in simpler examples it might be more explicable. Are there (simple) cases where I could say, just from looking something like e.g. the torus, what its zero-th or first or second etc. homology group was based on the nature of the object? I guess in the zero-th case it is, as my source (http://teamikaria.com/hddb/wiki/Homology_groups) above says, related to the number of disjoint pieces. Can we delve deeper than this for the other homology groups?

Any book/website suggestions would be welcomed (preferably websites as I am nowhere near a library!) - I have Hatcher but not a great deal else, and I haven't gleaned as much as I wish to from that alone. Of course I know that there is a great deal we don't know about homology groups even today, so I don't expect some magical all-encompassing answer, but any thoughts you could provide would be appreciated. I hope this question is appropriate for SE Mathematics, apologies if not! -M

答:
Let's restrict ourselves to orientable spaces that are homotopic to CW complexes. In low dimensions, there is a very intuitive way to think of homology groups. Basically, the rank of the $n$-th dimensional homology group is the number of $n$-dimensional “holes” the space has. As you stated in your example, for $H_0$, this is counting connected components. Moving to $H_1$, we are counting literal holes. The torus has $H_1\cong \mathbb Z \oplus \mathbb Z$ since it has two holes, one inside and one outside.

You can think of a 2-dimensional hole as an empty volume. The best analogy I’ve heard is to think of your space as an inflatable object. The rank of the second homology group is the number of different plugs you’d need to blow air into to inflate it. The torus has one empty volume, so you’d only need one plug to inflate it. If you take the wedge of two 2-spheres, you’d need two different plug to inflate it, one for each empty volume, so it has rank 2.

As is usual in topology, we now wave our hands and say “it works the same for higher dimensions.”


问:
2T(n/2) +n by induction
I try to proof by induction that:

$$ T(n)= 2 T(n/2)+n \quad n>2,\quad T(2)=2,\quad n = 2^{k}$$

is
$$ n*lg_2(n) $$

How can I do this?

Thanks

Steps that i went throw:

==Base Case==
$$T(2) = 2, \quad 2\cdot lg(2) = 2, \quad$$
My base case is verified

==Induction Step ==

Inductive hypothesis

$$ 2T((n+1)/2) +(n+1) = (n+1)\cdot lg_2(n+1)$$

I'm stuck here...

答:
The reason you are confused is that (if I understand your problem correctly) $T(n)$ is defined only for $n = 2^k$, i.e. only when $n$ is a power of two.

Often when you do induction, you assume the statement is true for $n$, and then prove it is true for $n + 1$.  But here, $T$ isn't even defined for $n + 1$ because it is only defined for powers of two.  So you're going to have to be more clever.

You don't have to worry about $n + 1$ because it's not a power of two.  What you do have to worry about is all the powers of two.  So the way you would prove this by induction is, assuming it is true for $n$, prove it is true for the next power of two.

Letting $n = 2^k$, the next power of two is $2^{k+1}$.  Therefore, you're going to want to assume that
$$
T(n) = n \log_2 ( n ) \quad \textbf{for } \textbf{n = } \textbf{2}^\textbf{k}
$$
and prove that
$$
T(n) = n \log_2 (n) \quad \textbf{for } \textbf{n = } \textbf{2}^\textbf{k + 1}.
$$

Title: 2T(n/2) +n by induction


问:
Transforming equation in terms of new independent variables.
Transform the equations taking $u$ and $v$ for new independent variables and $w$ for a new function:

$$z_{x x}-2 z_{x y}+\left(1+\frac{y}{x}\right) z_{y} y = 0$$
\begin{align}
u&=x\\
v&=x+y\\
w&=x+y+z
\end{align}

In this example I can't understand, how $z$ is depends on $x,y$, or $u$ and $v$. So how should I differentiate it in a right way?

答:
You want to transform an equation that depends on $z$ into an equation that depends on $w$ and you know how $z$ and $w$ relate to each other.

$$z_{x x}-2 z_{x y}+\left(1+\frac{y}{x}\right) z_{y} y = 0$$

If $z = w - x - y$, it is possible to state that $z_{xx} = w_{xx}$, $z_{xy} = w_{xy}$ and $z_{y} = w_{y} - 1$.

$$w_{x x}-2 w_{x y}+\left(1+\frac{y}{x}\right) (w_{y} - 1) y = 0$$

Now we have to focus on changing the variables.

First, solving the system $\{u = x, v = x + y\}$ for $\{x,y\}$ leads to $\{x = u, y = v - u\}$.

Second, following the chain rule, we know that

$$\begin{matrix}
w,w_u,w_v & -\rightarrow          & u & -\rightarrow          & x\\
          & \diagdown \rightarrow & v & -\rightarrow          & x\\
          &                        &   & \diagdown\rightarrow & y\\
\end{matrix}$$

$$\left\{\begin{matrix}
w_{x} = w_{u} u_{x} + w_{v} v_{x} = w_{u} + w_{v}\\
w_{y} = w_{v} v_{y} = w_{v}
\end{matrix}\right.$$

$$\left\{\begin{matrix}
w_{ux} = w_{uu} u_{x} + w_{uv} v_{x} = w_{uu} + w_{uv}\\
w_{uy} = w_{uv} v_{y} = w_{uv}
\end{matrix}\right.$$

$$\left\{\begin{matrix}
w_{vx} = w_{vu} u_{x} + w_{vv} v_{x} = w_{vu} + w_{vv}\\
w_{vy} = w_{vv} v_{y} = w_{vv}
\end{matrix}\right.$$

$$\left\{\begin{matrix}
w_{xx} = w_{ux} + w_{vx} = w_{uu} + w_{uv} + w_{vu} + w_{vv}\\
w_{xy} = w_{uy} + w_{vy} = w_{uv} + w_{vv}
\end{matrix}\right.$$

Finally,

$$w_{x x}-2 w_{x y}+\left(1+\frac{y}{x}\right) (w_{y} - 1) y = 0$$

$$w_{uu} + w_{uv} + w_{vu} + w_{vv}-2w_{uv} -2 w_{vv}+\left(1+\frac{v-u}{u}\right) (w_{v} - 1) (v-u) = 0$$


问:
Is this proposition True? $[(∀x)A(x) → (∀x)B(x)] → [(∀x)(A(x) → B(x))]$
Is this proposition True? $[(∀x)A(x) → (∀x)B(x)] → [(∀x)(A(x) → B(x)]$

I cannot think if there is a difference between the first implication and the last one. Therefore, I think the proposition is true, although I got someone saying that the proposition is false, again, for the proposition to be false, there must be a difference between $(∀x)A(x) → (∀x)B(x)$ and$(∀x)(A(x) → B(x))$ and I just can't think in one.

This is one of the solution that I tried, but now I think that this is wrong, because if $(∀x)A(x)$ is false in the first implication, then so It's in the last one.

答:
I think, the proposition is false.

Let $S := \{1,2,3,4,5\}$ and $f : S \to \mathbb{R}$ be a function. If $\forall x \in S: f(x) > 0$, than surely $\forall x \in S : \sum_{i=1}^{x} f(i) > 0$. In this case, $\forall x \in S: f(x) > 0 \Rightarrow \sum_{i=1}^{x} f(i) > 0$ does not necessarily hold: If $f(2) = 1$ and $f(1) = -2$, than $f(2) > 0$, but $\sum_{i=1}^{2} f(i) < 0$.


问:
What does $K^{m \times m}$ mean?
$K^{m \times m}$ is unclear to me. I know what $K^{m}$ means. I know what $K_{m}$ means.
My guess ist that $K^{m \times m}$ is the set of all quadratic matrices. May someone confirm that?

答:
Based on your comments, I would infer that $K^{m \times m}$ means 'the set of all $m \times m$ matrices over the field$^{[1]}$ $K$'.

[1] https://en.wikipedia.org/wiki/Field_(mathematics)

Title: What does $K^{m \times m}$ mean?


答: The exact solution is $u(x)=\sin x$, so you have good reason to be suspicious of your result. As a side note, it is much better to use implicit methods to solve this kind of BVP: The numerical solution $u=(u_1, \cdots, u_n)$ is the solution of a tridiagonal linear system. However, your fixed point iteration should work... You have probably not used enough iterations.

Title: Using finite difference for a simple second order BVP

Body:
I'm trying to solve this BVP using the finite difference method
$$u_{xx}(x) +sin(x) = 0$$
with $u(0)=u(2\pi)=0$. When I solve it on python, I get the correct curve shape I expect, a sinusoidal one, but the magnitude is much lower than what I expect. I'd expect a maximum and minimum of $1$ and $-1$, but they come out very small, like ~$0.04$.

Here's my code

```<code>import numpy as np
import math
import matplotlib.pyplot as plt

N = 100 #number of nodes
L = 2.0*math.pi #length of x-axis
x = np.linspace(0.0,L,N)
h = L/N #step length
u = np.zeros(N) #solution vector

u[0] = 0.0
u[N-1] = 0.0

#solve
for j in range(10):
    for i in range(1,N-1):
        u[i] = (u[i-1]+u[i+1]+h**2*math.sin(x[i]))/2.0

plt.plot(x,u)
</code>```

and here's the curve

So what causes the incorrect magnitude? I should get a regular $sin(x)$ curve


答:
I cannot think if there is a difference between the first implication and the last one.

There is a difference.  The first proposition says 'If everything is an $A$, then everything is a $B$'. The second proposition says that 'everything that is an $A$ is a $B$' ... which we typically phrase as 'All $A$'s are $B$'s'.

Now, if everything in the domain is an $A$, then by the first proposition everything would also be a $B$, and hence everything would be both an $A$ and a $B$, and it would therefore also be true, as stated by the second proposition, that every $A$ is a $B$. So yes, oif everything is an $A$, then both propositions are true, and the whole statement that is given to you would be true as well.

However, what if only some of the things in the domain are an $A$?  Then the first proposition does not 'apply'... that is, it would not be true that everything is an $A$, and hence you cannot conclude that everything is a $B$ .. let alone that anything is a $B$ at all.

Maybe somewhat counterintuitively, in this case (where some things are an $A$ but others things are not an $A$), the first proposition would still be true: whenever the antecedent ('if' part) of a conditional is false, then the whole conditional is true (indeed, since the 'if' part does not 'apply', the proposition is not making any claims about everything being a $B$, and hence the proposition cannot be said to make a false statement).

However, if we also assume that there are in fact no $B$'s at all in the world, this would also mean that the second proposition is false, since those things that are an $A$ would not be a $B$.

And so, with a world where some things are $A$, but not everything is an $A$, but where there are no $B$'s at all, the first proposition is true, but the second one is false. Hence, the whole statement would be false.

This tells us that the statement given to you can be true or false depending on the world/domain in which you evaluate it. Indeed, when the title of your question asks whether the statement is true or false, we just can't tell.  However, asking whether some logical statement is true or false is in fact a rather unusual question, and I wonder whether this was really the question you had to answer. More typically, we would ask: is some statement logically true (or: is it necessarily true), meaning: is it true no matter what the domain is like? (roughly: 'is it always true?')  And if that is the question (I bet it is), then the answer is no ... because as I showed above, we can find a world (sometimes called a counterexample) where the statement is false. So, it is not a necessary or logical truth.

Title: Is this proposition True? $[(∀x)A(x) → (∀x)B(x)] → [(∀x)(A(x) → B(x))]$

Body:
Is this proposition True? $[(∀x)A(x) → (∀x)B(x)] → [(∀x)(A(x) → B(x)]$

I cannot think if there is a difference between the first implication and the last one. Therefore, I think the proposition is true, although I got someone saying that the proposition is false, again, for the proposition to be false, there must be a difference between $(∀x)A(x) → (∀x)B(x)$ and$(∀x)(A(x) → B(x))$ and I just can't think in one.

This is one of the solution that I tried, but now I think that this is wrong, because if $(∀x)A(x)$ is false in the first implication, then so It's in the last one.


问:
Find what values of $n$ give $\varphi(n) = 10$

For what values of $n$ do we get $\varphi(n) = 10$?

Here, $\varphi$ is the Euler Totient Function. Now, just by looking at it, I can see that this happens when $n = 11$. Also, my friend told me that it happens when $n = 22$, but both of these are lucky guesses, or educated guesses. I haven't actually worked it out, and I don't know if there are any more.

How would I go about answering this question?

答:
Suppose $\varphi(n)=10$.  If $p \mid n$ is prime then $p-1$ divides $10$.  Thus $p$ is one of $2,3,11$.

If $3 \mid n$, it does so with multiplicity $1$.  But then there would exist $m \in \mathbb{N}$ such that $\varphi(m)=5$, and this quickly leads to a contradiction (e.g. note that such values are always even).

Thus $n$ fits the form $2^a\cdot 11^b$, and we claim that $b=1$.  If $b>1$ we have $11 \mid \varphi(n)$, a contradiction.  As well, $b=0$ gives $\varphi(n)$ a power of $2$.  Thus $n=2^a \cdot 11$, and it's easy to see from here that $n=11,22$ are the only solutions.


问:
Show that $f(x,y) = \frac{x^2y^2}{x^2+y^2}$ is (totally) differentiable
I want to prove that $f(x,y) = \frac{x^2y^2}{x^2+y^2} , (f(0,0) = 0)$ is (totally) differentiable at $ \begin{pmatrix}  0 \\ 0 \end{pmatrix}$. I want to use the criterion that a function which is continuously partially differentiable is differentiable. So I compute $\frac{\partial f}{\partial x}=\frac{2xy^4}{(x^2+y^2)^2}$ (The partial derivative $\frac{\partial f}{\partial y}$ works analogously). Going back to the definition of partial deriviative with respect to the $x$ direction, we see that $\frac{\partial f}{\partial x}(0,0)=0$.

So it remains to show that $\lim_{(x,y) \to (0,0)} \frac{\partial f}{\partial x}(x,y) = 0.$ I use the $\epsilon-\delta$ criterion: Let $\epsilon >0 $. Let $\delta := \frac{1}{2} \epsilon$. So let $d((x,y),(0,0)) = \sqrt{x^2+y^2}< \delta$, say $\sqrt{x^2+y^2} = \delta'<\delta.$ This implies $|x|\leq\delta'<\delta, |y|\leq\delta'<\delta$. Therefore: $|\frac{\partial f}{\partial x}(x,y)| \leq \frac{2(\delta')^5}{(\delta')^4} = 2\delta'<2\delta=\epsilon$, as desired.

Is this analysis correct?

答:
Your idea is fine but $0<\sqrt{x^2+y^2}<\delta$ doesn't imply
$$\frac{1}{(x^2+y^2)^2}<\frac{1}{\delta^4} $$
The correct inequality is actually in the opposite direction. Furthermore, you have to show
$$\left|\frac{2xy^4}{(x^2+y^2)^2}\right|<\varepsilon $$
(The absolute value is important here.) I would instead use $|xy|\leq x^2+y^2$ and $y^2\leq x^2+y^2$ so
$$\left|\frac{2xy^4}{(x^2+y^2)^2}\right|\leq \frac{2|y|^3}{x^2+y^2}\leq 2|y| $$
Then, $|y|\leq \sqrt{x^2+y^2}<\varepsilon/2$, and you can finish from here. So, your choice of $\delta$ works here.


问:
Ratio of two areas in a triangle
Suppose you have the following triangle $ABC$:

with the following properties:
$$|AB|=4\cdot |AA'|,\>\>\>|AC|=4\cdot |CC'|\>\>\>|BC|=4\cdot |BB'|$$

I have to find the ratio of the total area of the triangle and the red area. I tried a lot of algebraic manipulations with the lengths of the sides, but couldnt solve it that way.

I really need hints for this one. Thanks!

Title: Ratio of two areas in a triangle

Body:
Evaluate the area ratio

$$\frac{[ABB'']}{[ABC']} = \frac{[ABB']}{[ABB'C']}
= \frac{\frac14[ABC]}{[ABC]-[B'C'C]}= \frac{\frac14[ABC]}{[ABC]-\frac14\frac34[ABC]}
=\frac4{13} $$
which leads to
$\frac{[ABB'']}{[ABC]} =\frac{[ABB'']}{\frac43[ABC']}  =\frac3{13}$.
Likewise, $\frac{[BCC'']}{[ABC]} =\frac{[CAA'']}{[ABC]} =\frac3{13}$, Then
$$\frac{[A''B''C'']}{[ABC]} = \frac{[ABC]-[ABB'']-[BCC'']-[CAA'']}{[ABC]} = 1-3\cdot\frac3{13}=\frac4{13}
$$


问:
How to determine if a system of equations has a solution using the rank of a matrix
Consider this problem:

Given are the matrix $A$ and the vector $\mathbf{b}$ with $ \alpha $  and $\beta  \in \Bbb R $;

$$

State for which alpha and beta the linear system of equations $$ A \cdot x = \mathbf b $$ has none, one or infinite number of solutins.

Now what I tried to do is set up a system of equations. That looks like this

$$ \begin{matrix}
0 & 1 & 2 | 5\\
1 & 3 & 4 | 6 \\
2 & 4 & a | b
\end{matrix} $$

Now the goal is to get $0 \ 0 \ a$ in the last line so what I did i took the $-2$ second line and added it to the 3rd. So we get

$$ \begin{matrix}
0 & 1 & 2 | 5\\
1 & 3 & 4 | 6 \\
0 & 2 & -8+a |-12+ b
\end{matrix} $$

Than I had to get rid of the so I took $-2 \cdot 1 + 3$ and I got.

$$ \begin{matrix}
0 & 1 & 2 | 5\\
1 & 3 & 4 | 6 \\
0 & 0 & -4+a |-22+ b
\end{matrix} $$

Now I've look at the cases when when the system has no solutions. That is when $\operatorname{rank} (A) < \operatorname{rank}(A|b)$. Now I'm not sure how I can  confirm the rank of a matrix, I would assume the rank is 3 because we have 3 lines. So if the rank of $A$ is 3 than the rank of $(A|b)$ has to be 2 or lower. Now if I want to get a rank 2, $(A|b)$ do I have to set  $\alpha=0$?

Because than we would have all $0$ in the last line thus the rank would be $2$. Some insight would be great. Thank you!

PS: If the solution to my system is false, please let me know.

答:
Suppose that $\alpha=4$. Then you have the matrix$$\left[\begin{array}{ccc|c}0&1&2&5\\1&3&4&6\\0&0&0&b-22\end{array}\right].\tag1$$The matrix to the left of the vertical bar has rank $2$ (the rank is the number of linearly independent lines, whereas the rank of $(1)$ depends upon $b$: it is $3$ if $b\ne22$, but it is $2$ if $b=22$. So, if $b\ne22$, the system has no solutions, but it has solutions if $b=22$.

And if $\alpha\ne4$ both matrices have rank $3$, and therefor the system has solutions.


答:
The notation $K^{m\times n}$ usually means the set of all $m\times n$ matrices over a commutative ring $K$. It is also denoted by $M_{m,n}(K)$. Both notations are (practically) standard.

When $m=n$, the matrices has $m$ rows and $m$ columns. Hence the notation means the set of all square matrices of order $m$ over $K$. It is also denoted by $M_m(K)$.

When $n=1$, $K^{m\times n}$ is identified with $K^m$, the $m$-fold Cartesian product of $K$. It is also viewed as the set of all column vectors of with $m$ entries in $K$.

(In contrast, the use of the notation $K_m$ to mean all $m$-tuples of elements in $K$ is neither standard nor common in linear algebra literature.)

Title: What does $K^{m \times m}$ mean?

Body:
$K^{m \times m}$ is unclear to me. I know what $K^{m}$ means. I know what $K_{m}$ means.
My guess ist that $K^{m \times m}$ is the set of all quadratic matrices. May someone confirm that?


问:
Probability problem regarding 3rd and 4th pick
Say, there are $3$ bags and there are $6,5 \text{ and }4$ balls in those bags respectively. Now if $4$ balls were taken randomly without replacement, What is the probability that 3rd and 4th balls are taken from the same bag?

My Try:
$$=\dfrac{15}{15}\times\dfrac{14}{14}\times\dfrac{13}{13}\times(\dfrac{3}{12}+\dfrac{2}{12}+\dfrac{1}{12})=\dfrac{1}{2}$$

Is there any problem ?

答:
The problem with your solution is that it cannot be followed. You should explain how did you come to this expression. In any case the result is not correct as the following simple calculation shows.

As clarified in the comment the problem is equivalent to that of drawing of balls of three colors from a bag, which can be solved as follows.

There are altogether:
$$
N=15\cdot14
$$
ways to fill the third and fourth positons.

Fixing the color of the balls on these positions one obtains the following numbers of corresponding combinations:
$$
N_1=6\cdot5,\quad N_2=5\cdot4,\quad N_3=4\cdot3,
$$
so that the probability in question is:
$$
\frac{N_1+N_2+N_3}{N}=\frac{31}{105}.
$$


问:
Is there a program that can take matrices of variables and compute the inverse?
I have a control theory / optimal control problem that is formulated like this:

$A(x)\dot x = f(x,u)$

Where A(x) is a matrix containing state variables. And I want to isolate $\dot x$ by doing this:

$\dot x = A(x)^{-1}f(x,u)$

So that I have state rates equal to states and inputs. The issue is inverting A(x), it's a 12x12 matrix with probably 100 or more instances of states. Is there a program that will automatically output the inverse matrix or, even better, the entirety of:

$A(x)^{-1}f(x,u)$

I currently know f(x,u) and A(x). Could someone give me the name of a free program or the type of program I'm looking for?

答:
The matter is that if A has so many variables, even with  very simple functions of them, any CAS (even the most powerful one) either will refuse to work or will output such a tremendously complicated output that it will be useless.

That unless the matrix is highly symmetrical and simple, that it can be simply LU decomposed, or has simple blocks and /or eigen-values, eigen-vectors etc.


问:
What is the remainder when $4^{96}$ is divided by 6
What is the remainder when $4^{96}$ is divided by 6

My approach :
When 4 is divided by 6 gives 4 remainder, when $4^2$ is divided by 6 gives remainder 4 ,... is it for the complete series.. that means when $4^{96}$ is divided by 6 gives 4 as remainder or is there any short cut or other answer of this question.

Please suggest if there is any sort of short cut to such problems.

Thanks.

答:
In all cases,

n-odd$\rightarrow 2^n\equiv 2 \mod6$

n-even$\rightarrow 2^n\equiv 4\mod6$

$\implies 2^{96}\equiv 4\mod 6$


答:
So you haven't written down the discrete equation that you are trying to solve, which is a bit troubling. I think the discrete equation you want to use is probably

$$\frac{u_{n+1}-2u_n+u_{n-1}}{h^2}=-\sin(x_n)$$

for $n=1,2,\dots,N-1$ with $u_0=u_N=0$. This can be done with just a linear system solver easily enough, but if you want to use an iterative method, you can do, say, the Jacobi method. The iteration for the Jacobi method is

$$u_n^{(k+1)}=\frac{h^2 \sin(x_n) + u_{n-1}^{(k)} + u_{n+1}^{(k)}}{2}$$

which is not quite the same as what you implemented. What you did is actually Gauss-Seidel:

$$u_n^{(k+1)}=\frac{h^2 \sin(x_n) + u_{n-1}^{(k+1)} + u_{n+1}^{(k)}}{2}$$

which is usually a little bit faster than Jacobi.

But this will converge extremely slowly in this context. Indeed in the worst case the eigenvalues of the Jacobi iteration matrix with the largest magnitude are larger than 0.995 in absolute value, so Jacobi would need about $\frac{\log(\epsilon)}{\log(0.995)} \approx 200 \log(1/\epsilon)$ iterations to multiply the error by a factor of $\epsilon$. Gauss-Seidel is a little faster (I have forgotten exactly how much faster). The error that you committed in your initial guess is exactly the worst case scenario for Jacobi as well.


问:
Conjecture: all primes $p>5$ have smaller primes $q_1,q_2$ s.t. $p \pmod {q_1} = p \pmod {q_2}$
I assert

$\forall p \in \mathbb P\text{ where }p>5 : \exists q_1,q_2\in\mathbb P\text{ such that }q_1<q_2<p$ and $$p\!\!\!\!\pmod {q_1} = p\!\!\!\!\pmod {q_2}$$

E.g. $7 \equiv 1 \pmod {2}\text{ and } 7\equiv 1 \pmod {3}$ so, $7\pmod{2}=7\pmod{3}$

I am almost certain this is true. In fact, I think there's a minimum number of such duplicates which increases without bound, e.g. you'll find at least two distinct duplicate pairs $(q_1,q_2)$ for all primes $\geq 31$.

I am wondering whether anyone sees an easy proof for this, or failing that, a good heuristic argument for why it's true. (Also, is that legal use of mod notation?)

Addendum (Shubhrajit Bhattacharya)

For any prime $p$, define $\Phi(p)$ to be the number of unordered pairs $\{q_1,q_2\}$ of primes smaller than $p$ such that $$p\!\!\!\!\pmod{q_1}=p\!\!\!\!\pmod{q_2}$$ It has been clear that $\Phi(p)\geq1$ for all primes $p\geq 5$. Now, a reasonable question is that how $\Phi(p)$ grows as $p\rightarrow\infty$?

A not so useful formula for $\Phi(p)$. Let $\omega(n)$ denotes the number of distinct prime factors of $n$. Let $$\delta(a,p)=\begin{cases}0\quad\text{if $(p-a)$ is a prime power}\\\binom{\omega(p-a)}{2}\quad\text{otherwise}\end{cases}$$

For any prime $q<p$, we must have that $q\leq p-2$ as $p\geq5$. Then $p\pmod{q}\leq q-1\leq p-3$. Then the remainder can be at most $p-3$ when we divide $p$ by a prime $q<p$. Now, if $p-a$ is not a power of prime, any pair of two distinct prime factors of $p-a$ counts in $\Phi(p)$. Now, $\binom{\omega(p-a)}{2}$ such unordered pairs are there. When $p-a$ is a prime power, we certainly can't get two distinct primes modulo which $p$ is $a$. So, we get that
$$\boxed{\Phi(p)=\sum_{m=1}^{p-3}\delta(m,p)}$$ for all $p\geq5$

答:
Yes this is true. This is equivalent to saying that one of $p-1, p-2$ is divisible by at least two distinct primes.

Case 1: $p-1$ is not a power of two. Then $p-1$ is even and has an odd prime factor, and $p \equiv 1 \pmod 2$ and $p \equiv 1 \pmod {\text{"odd prime factor"}}$.

Case 2: $p-1$ is a power of two. Then $p$ is a Fermat prime, i.e. $p = 2^{2^n} +1$.

Then $p-2 = 2^{2^n} - 1 = (2^{2^{n-1}}-1)(2^{2^{n-1}}+1)$. Just pick an odd prime factor from each factor and we have $p \equiv 2 \pmod {\text{"them"}}$. Here observe that $\gcd\{2^{2^{n-1}}-1,2^{2^{n-1}}+1\} = 1$ and $p> 5$.


问:
Spheres in different dimension are not homotopy equivalent
Is there a way to prove that $\textbf{S}^n$ and $\textbf{S}^m$ are not homotopy equivalent if $n\neq m$ without using the machinery of homology or higher homotopy groups?</p>

答:
I can prove, without mentioning groups, that if spheres $S^m$ and $S^n$ with $m<n$ were homotopy equivalent, then they'd be contractible, but I don't immediately see a group-free proof that spheres aren't contractible.

Suppose $f:S^m\to S^n$ and $g:S^n\to S^m$ were inverses of each other up to homotopy. By cellular or simplicial approximation, you can deform $f$ so that it's not surjective (as $m<n$), and then you can deform it to a constant by pushing away from a point not in the image of $f$.  So $f$ is homotopic to a constant; therefore so are $f\circ g$ and $g\circ f$, which are homotopic to the identity maps of the two spheres.  Therefore, these spheres are contractible,


答:
A very rough idea:

If $p$ is prime then $p-1$ has unique factorization in the form $p-1=2^k\cdot p_1^{n_1}\cdot p_2^{n_2}\cdots p_m^{c_m}$. If such $p_i$ exists then we have $p\equiv 1\mod 2$ and $p\equiv 1\mod p_i$. Therefore we consider $p-1=2^k$.

Here $p=2^k+1$. Since $p$ is prime then $p$ must be a Fermat Number, i.e. $p=2^{2^a}+1$ for some natural number $a$. So $p=2^{2^a}+1=(2^{2^a}-1)+2=(2^{2^{a-1}}-1)(2^{2^{a-1}}+1)+2$. Note that $2^{2^{a-1}}-1$ and $2^{2^{a-1}}+1$ would be odd, then they must contain distinct prime factors, which cannot be $2$. For such prime $p_1$ in $2^{2^{a-1}}-1$ and $p_2$ in $2^{2^{a-1}}+1$, we have the same remainder.

Title: Conjecture: all primes $p>5$ have smaller primes $q_1,q_2$ s.t. $p \pmod {q_1} = p \pmod {q_2}$

Body:
I assert

$\forall p \in \mathbb P\text{ where }p>5 : \exists q_1,q_2\in\mathbb P\text{ such that }q_1<q_2<p$ and $$p\!\!\!\!\pmod {q_1} = p\!\!\!\!\pmod {q_2}$$

E.g. $7 \equiv 1 \pmod {2}\text{ and } 7\equiv 1 \pmod {3}$ so, $7\pmod{2}=7\pmod{3}$

I am almost certain this is true. In fact, I think there's a minimum number of such duplicates which increases without bound, e.g. you'll find at least two distinct duplicate pairs $(q_1,q_2)$ for all primes $\geq 31$.

I am wondering whether anyone sees an easy proof for this, or failing that, a good heuristic argument for why it's true. (Also, is that legal use of mod notation?)

Addendum (Shubhrajit Bhattacharya)

For any prime $p$, define $\Phi(p)$ to be the number of unordered pairs $\{q_1,q_2\}$ of primes smaller than $p$ such that $$p\!\!\!\!\pmod{q_1}=p\!\!\!\!\pmod{q_2}$$ It has been clear that $\Phi(p)\geq1$ for all primes $p\geq 5$. Now, a reasonable question is that how $\Phi(p)$ grows as $p\rightarrow\infty$?

A not so useful formula for $\Phi(p)$. Let $\omega(n)$ denotes the number of distinct prime factors of $n$. Let $$\delta(a,p)=\begin{cases}0\quad\text{if $(p-a)$ is a prime power}\\\binom{\omega(p-a)}{2}\quad\text{otherwise}\end{cases}$$

For any prime $q<p$, we must have that $q\leq p-2$ as $p\geq5$. Then $p\pmod{q}\leq q-1\leq p-3$. Then the remainder can be at most $p-3$ when we divide $p$ by a prime $q<p$. Now, if $p-a$ is not a power of prime, any pair of two distinct prime factors of $p-a$ counts in $\Phi(p)$. Now, $\binom{\omega(p-a)}{2}$ such unordered pairs are there. When $p-a$ is a prime power, we certainly can't get two distinct primes modulo which $p$ is $a$. So, we get that
$$\boxed{\Phi(p)=\sum_{m=1}^{p-3}\delta(m,p)}$$ for all $p\geq5$


答:
If $p-1$ is not a power of two, then there is a prime $q\neq 2$ dividing $p-1$. Also, $p-1$ is even. Then we have $p \equiv 1 \pmod 2$ and $p \equiv 1 \pmod q$, as desired.

If $p-1$ is a power of two, then consider the number $p-2$. If $p-2$ is divisible by two distinct primes $r$ and $s$ (note that $r,s>2$ as $p-2$ is odd), then $p \equiv 2 \pmod r$ and $p\equiv 2\pmod s$, as desired.

We are left with the case when $p-1$ is a power of two and $p-2$ is a power of a prime. We shall prove that this case is impossible. Write $p-1=2^k$ and $p-2=m^n$ where $m$ is prime. Note that $2^k-m^n=1$, which cannot happen if $n\ge 2$ due to Mihailescu theorem. Therefore $p-2$ is prime. But then, since $p-2$ and $p$ are primes, $p-1$ is divisible by three, which is a contradiction as $p-1$ is a power of two.


问:
Solving the system $\sqrt{x/y}+\sqrt{y/z}+\sqrt{z/x}=\sqrt{y/x}+\sqrt{z/y}+\sqrt{x/z}=3$, $\sqrt{xyz}=1$
I am having problems solving this particular system of equations (SOE). The only thing I see that could be useful is the first and second equations are reciprocals of each other. I just don't know how I can use that information to help solve the SOE.

$$
\left\{
\begin{array}{c}
\displaystyle\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}}=3 \\
\displaystyle\sqrt{\frac{y\:}{x\:}}+\sqrt{\frac{z\:}{y\:}}+\sqrt{\frac{x\:}{z\:}}=3 \\
\displaystyle\sqrt{xyz}=1
\end{array}
\right.
$$

Is there a way to solve this algebraically other than solving by inspection?
Thank you for any help/assistance you can provide!

答: Unless you are looking for a generalization with RHS other than $3$ and $1$, this system of equations is solvable by inspection where $$x=y=z=1$$


问:
Revisit "What is the rank of $Q+Q^T-Q\circ I$ if $Q = qq^T$"
Based on my previous problem: What is the rank of $Q+Q^T-Q\circ I$ if $Q = qq^T$

$Q = qq^T$, $q\in \mathbb{R}^n$. Suppose $q^Tq=1$.

$A\circ B:$ Hadmard (elementwise) product

$Q\circ I$ just takes the diagonal entries of $Q$.

My question: Is $Q+Q^T-Q\circ I$ still positive semidefinite?

I guess the answer is not fixed by Gershgorin circle theory since the diagonal terms become small. Not sure if I am correct.

答:
It is clearly not always positive semidefinite. E.g. when $q^T=(x,y)$, we have
$$
qq^T+qq^T-(qq^T)\circ I=\pmatrix{x^2&2xy\\ 2xy&y^2},
$$
whose determinant $-3x^2y^2$ is nonnegative only when $x=0$ or $y=0$.


问:
Fat tail threshold
How 'fat' do tails have to be to be give a distribution undefined variance? Is there an inequality for the kurtosis of the distribution which determines this? No matter how fat the tails, they must converge to the axis as the distribution must enclose a finite area of 1 (kind of reminds me of Gabriels horn...)</p>

答:
For one straightforward example look at Student's t distributions with 2 and 3 degrees of freedom, respectively. The variance is not defined for DF=2, but is defined for DF=3. (See Wikipedia on t distributions.)

The PDF of Student's t distribution with $\nu$ degrees of freedom has
density function of the form $f(x)=K(1 + \frac{x^2}{\nu})^{-(\nu + 1)/2},$ where $K$ is the constant that makes $f(x)$ integrate to $1$ over the real line.

For $\nu = 1,$ this is a Cauchy distribution which has a median, but no mean.
That is, $\mu=\int_0^\infty xf(x)\,dx$ is an improper integral; multiplication of the PDF by $x$ makes the tails of the integrand 'too fat' for a properly convergent integral.

For $\nu = 2,$ the mean is $0,$ but multiplying the PDF by $x^2$ leads to
an improper integral, so that the second moments do not exist.

For $\nu > 2,$ both $\mu = 0$ and $\sigma^2 = \frac{\nu}{\nu-2}$ exist.

For somewhat similar examples with convergent and divergent infinite
series for discrete probability distributions, you may want to look at
zeta distributions based on the Riemann zeta function (linked from there).

Title: Fat tail threshold


答:
Are you familiar with the semantic tableaux method? that way it is easy to find if there's a couterexample.

Consider an element of the domain $a$ that is not $A$, so $\lnot Aa$, what would then be the truth value of the formula $\forall x Ax \to \forall x Bx$?

Now take another element $c$ such that: $Ac$ and $\lnot Bc$.

What does that tell us about the truth value of the consequent of your formula?


问:
Group acting properly on $\Bbb R^2$ and restriction of a covering map
Let $G$ be free group with countable-infinitely many generators acting on $\Bbb R^2$ properly, i.e., for each point $x\in \Bbb R^2$ we have an open neighborhood $V_x$ of $x$ such that $g\cdot V_x\cap V_x=\varnothing$ for all non-trivial element $g\in G$.

Let $H\subsetneqq G$ be an infinite cyclic subgroup, so we have a covering map $p\colon \Bbb R^2\to \Bbb R^2/H$ as $H$ also acts properly on $\Bbb R^2$. Let $g\in G\backslash H$ and $\ell \colon\Bbb R\to \Bbb R^2$ be a proper map, i.e., the inverse image w.r.t. $\ell$ of a compact set is compact.

Is it true that the restriction of the covering map $p\colon \Bbb R^2\to \Bbb R^2/H$
on the subset $g\cdot \text{im}(\ell)$ of $\Bbb R^2$ is a proper map?

So, if $K\subseteq_\text{compact} \Bbb R^2/H$ is contained in an admissible open set of the covering $p$, then we have $p^{-1}(K)=\bigcup_{h\in H}h\cdot C$ for some compact set $C\subseteq_\text{compact} \Bbb R^2$. Now, one needs to show there are only finitely many $h\in H$ for which $hC\cap g\cdot \text{im}(\ell)\not=\varnothing$.

$\bullet$ My guess is that we need to also assume $G$ acts cocompactly on $\Bbb R^2$ i.e. there is a compact subset $S$ of $\Bbb R^2$ with $G\cdot S=\Bbb R^2$. But if I assume so, then by Lemma 1.17(b) $G$ has to be finitely generated.

答:
What you ask for is not true.

For a counterexample, start with $\ell' : \mathbb R \to \mathbb R^2$ which is an "axis" for $h$, meaning a proper embedding which is invariant under $h$, such that the restriction of $h$ to $\text{im}(\ell')$ is a translation homeomorphism.

Notice: the map $p : \mathbb R^2 \to \mathbb R^2 / H$ restricted to $\text{im}(\ell')$ is not proper, because $p(\text{im}(\ell'))$ is a circle and hence is compact, but the inverse image of this circle under $p \bigm| \text{im}(\ell')$ is all of $\text{im}(\ell')$ which is homeomorphic to $\mathbb R$ and hence is not compact.

Now define $\ell = g^{-1} \circ \ell' : \mathbb R \to \mathbb R^2$. It follows that $p$ is not proper on $g \cdot \text{im}(\ell) = \text{im}(\ell')$.


问:
Inequality : $ab^2+bc^2+cd^2+da^2 \leq 4$
I have a excercise like that: let $a;b;c \geq 0$ and $a+b+c=3$, prove: $ab^2+bc^2+ca^2 \leq 4$

We can assume $b=mid\{a;b;c\}$, then $(a-b)(b-c) \geq 0$, by that way, the problem is solved.

But I have a question: what should I do if the problem is for 4 variables, eg:

Let $a;b;c;d \geq 0$ and $a+b+c+d=3$, prove: $ab^2+bc^2+cd^2+da^2 \leq 4$

Thank you.

答:
If we fix $b$ and $d$ and write $a=x, c=s-x$ where $s=3-b-d\ge 0$ then

$$f(x) = ab^2+bc^2+cd^2+da^2 = b^2 x+b(s-x)^2+d^2(s-x)+dx^2, x \in [0,s]$$
is quadratic with leading coefficient $b+d \ge 0$ so its maximum is obtained at either $x=0 (a=0) $ or $x=s (c=0)$.

Similarly if we fix $a$ and $c$, then setting either $b=0$ or $d=0$ will increase $ab^2+bc^2+cd^2+da^2$. Now WLOG we can assume $c=d=0$. Then we need to prove

$$ab^2 \le 4 \text { where } a+b=3$$

which is obvious via AM-GM:

$$\frac 14 ab^2 = a \cdot \frac b2 \cdot \frac b2 \le \left( \frac{a+\frac b2 + \frac b2}{3} \right)^3 = 1$$

Remark 1: We can easily generalize to: if $a_i\ge 0, \sum_{i=1}^n a_i = 3$ then $S=\sum_{i=1}^n a_i a_{i+1}^2 \le 4$. Note that $a_{n+1}$ is defined as $a_1$.

For any non-adjacent $a_i$, $a_j$ we can show that if we fix everything else, then $S$ obtains maximum at either $a_i=0$ or $a_j=0$ so we are left with two adjacent $a_k, a_{k+1}\ge 0$ while every other terms are zeros. Then we apply AM-GM and we are done.

Remark 2: It's easy to see that your inequality becomes equality if and only if $(a_1, \cdots, a_n)=(1,2,0, \cdots,0)$ or one of its cyclic permutations.


问:
Apparent discrepancy in the calculated value of integrals using two different metods
The integral to be evaluated -

$I=\lambda\mu\int_{0}^{\infty}\int_{0}^{\infty}min(x,y)e^{-\lambda x}e^{-\mu y}\, dx\,dy$

Now one approach can be using the fact that this integral can be treated as the expectation value of the exponential variable $min(x,y)$ whose rate constant is $\lambda+\mu.$(This is known from the property that the minimum of two independent exponential variables is itself an exponential variable .)

The other approach that I used was evaluating this integral analytically. Now the way this could be done is as follows-

Divide the first quadrant using the $x=y$ line such that $min(x,y)=y$ in the lower half and $min(x,y)=x$ in the upper half .

Now , the lower half integral -
$\lambda\mu\int_{0}^{\infty}\int_{0}^{x}ye^{-\mu y}e^{-\lambda x}\,dy\, dx$
comes out to be $\frac{1}{\lambda+\mu}$

Similarly the upper half one would also be $\frac{1}{\lambda+\mu}$

and thus the total integral comes out to be  $\frac{2}{\lambda+\mu}$.

Where am I going wrong?

答:
$$\lambda\mu\int_{0}^{\infty}\int_{0}^{x}ye^{-\mu y}e^{-\lambda x}\,dy\, dx=\lambda\mu\int_{0}^{\infty}e^{-\lambda x}\int_{0}^{x}ye^{-\mu y}\,dy\, dx$$
$$=\lambda\mu\int_{0}^{\infty}e^{-\lambda x}\frac{1-e^{-\mu x}(1+\mu x)}{\mu^2} dx=\frac{\mu}{(\lambda+\mu)^2}$$

The other integral is

$$\lambda\mu\int_{0}^{\infty}\int_{x}^{\infty}xe^{-\mu y}e^{-\lambda x}\,dy\, dx=\frac{\lambda}{(\lambda+\mu)^2}$$


问:
$3$ paylines that can sum up simultaneously, slot machine. RTP(return to player)
How to calculate RTP (Slot Machine Return) if,

The number of symbols is $3$, i.e. black, white, red

Each symbol appears equally, there may be e.g. $9$ red ones

symbols are independent, on each of the $9$ fields it is possible to draw $1$/$3$ of a given symbol

Winning means drawing $3$ symbols of the same level

There are $3$ paylines simultaneously, that is Top, Middle and Bottom.

Lines won can add up

Of course, there can also be $1$ or $2$ win lines at once

The winnings table is not important, it is important that the RTP is around $95 \%$

答:
In each line, there are $3^3=27$ possibilities and $3$ of them are winners, so the probability of winning is $\frac3{27}=\frac19$, and the probability of losing is $\frac89$.

The probability of winning on all $3$ lines is $\left(\frac19\right)^3=\frac1{729}$.

The probability of winning on exactly $2$ lines is $3\left(\frac19\right)^2\frac89=\frac8{243}$, because there are $3$ ways to choose the losing line.

The probability of wining on exactly one line is $3\left(\frac89\right)^2\frac19=\frac{64}{243}$, because there are $3$ ways to choose the winning line.

The probability of losing on all lines is $\left(\frac89\right)^3=\frac{512}{729}$

As a sanity check,
$$\frac1{729}+\frac8{243}+\frac{64}{243}+\frac{512}{729}=\\
\frac1{729}+\frac{24}{729}+\frac{192}{729}+\frac{512}{729}=\\
\frac{729}{729}=1$$


问:
What are the algebraic operations of the following vector space?
I'm working on this exercise:

Let $\{e_1,..., e_n\}$ be a basis for a complex vector space $X$. Find a basis for $X$ regarded as a real vector space. What is the dimension of $X$ in either case?

My question is: What are (explicitly) the algebraic operations of the 'new' real vector space defined from that complex? I understand one of its basis is $\{e_1,..., e_n, ie_1,..., ie_n\}$ for $\{1, i\}$ is a basis of $\mathbb{C}$; but I don't see what 'do' its operations...

答: The addition on $X$ is the addition that you already add there. And the product of a real number $\lambda$ by a vector $x\in X$ is $\lambda x$, where you see $\lambda$ as a complex number. In other words, you restrict to the reals the multiplication by a complex scalar.


问:
Show $A_q(n,n-1)=q$, if $n > \binom{q+1}{2}$
See title: Show $A_q(n,n-1)=q$, if $n>\binom{q+1}{2}$, where $A_q(n,n-1)$ is the maximum number of words of length $n$ a code can have, over a $q$-ary alphabet, with minimum Hamming distance at least $n-1$. I try to count, but my counting is very rusty and I cannot get the desired result. Thanks!

答: Let $C$ be a set of $q+1$ words, and suppose by way of contradiction that each pair of words agree in at most one position. There are $\binom{q+1}2$ pairs of words, and $n$ positions. Since $n>\binom{q+1}2$, there would exist a position such that no two codes agree in that position. Can you take it from here?


问:
Split multiplicative reduction question
Let $E/\mathbb{Q}$ be an elliptic curve and $E_{d}$ be the quadratic twist of $E$ by a squarefree integer $d$. Let $\ell$ be a prime of multiplicative reduction for $E$. If $(d, \ell) = 1$, then $\ell$ is a prime of multiplicative reduction for $E_{d}$. I read that $\ell$ is of split multiplicative reduction for $E$ if and only if $\ell$ is of split multiplicative reduction for $E_{d}$. Are there any sources for this?</p>

答:
The statement is false.

An elliptic curve $E$ with multiplicative reduction at $\ell$ has split reduction if and only if $-c_6$ is a square in $\mathbb{F}_\ell$. Now suppose that $\gcd(d,\ell)=1$, so that $E'=E_d$, the quadratic twist by $d$, also has multiplicative reduction. The corresponding coefficient for $E'=E_d$ is given by $c_6'= d^3 c_6$. Hence, $E'$ has split reduction if and only if $-c_6' = -d^3c_6$ is a square in $\mathbb{F}_\ell$. In particular, if $d$ is not a square mod $\ell$, then $d^3$ is not a square either, and $-c_6'$ is a square if and only if $-c_6$ is not a square.

For instance, let $E:y^2=x^3-x^2+35$. Then $E$ has split multiplicative reduction at $p=5$. Now:

Let $d=11$. Since $11\equiv 1 \bmod 5$ is a square, the quadratic twist by $11$ also has split multiplicative reduction at $5$.

Let $d=13$. Since $13\equiv 3 \bmod 5$ is not a square, the quadratic twist by $13$ has non-split multiplicative reduction at $5$.

Title: Split multiplicative reduction question


问:
Why is $\int f(x) dx = \int_0^1 |f\ge t|dt$, where $0\le f\le 1$?

Why is $$\int f(x) dx = \int_0^1 |f\ge t|dt$$
true? Apparently, it comes from Fubini's theorem - which allows us to write the integral of $f$ as the integral of the length of its level set? Also note that $\sup f = 1$, $f \ge 0$ and $f$ is integrable.

Could someone help me understand where this comes from, and what's going on? Intuition would be helpful too.

Given a function $f$, I understand that $\{x:f(x) = t\}$ is one of its level sets. Is that right? I don't see how this relates with the above equality though.

Reference: Pg. 31, Notes.

答: This is called the layer-cake representation of the integral. In this case, I assume that $0 \le f(x) \le 1$ for all $x$. Then you can write $$\lvert \{ f \ge t \} \rvert = \int_{\{f \ge t\}} 1  \, dx = \int \mathbb I_{\{f \ge t\}}(x) dx,$$ where $\mathbb I_{\{f\ge t\}}$ is the indicator function of the set $\{ x: f(x) \ge t\}$. Thus $$\int_0^1 \lvert \{ f \ge t \} \rvert dt = \int^1_0 \int \mathbb I_{\{f \ge t\}} (x) dx dt = \int\int^1_0\mathbb I_{\{f \ge t\}} (x) dt dx,$$ where the last step follows from Fubini's theorem. Now notice that $$\mathbb I_{\{ f \ge t\}} (x) = \left \{ \begin{matrix} 1, & t \le f(x), \\ 0, & t > f(x). \end{matrix}\right.$$ Thus the inner integral simplifies and we get $$\int_0^1 \lvert \{ f \ge t \} \rvert dt = \int \int^{f(x)}_0 dt dx = \int f(x) dx.$$


问:
Find $n$ such that $\lim_{x \to 0}x^n\frac{\phi_2(x)}{\phi_1(x)}=0$
Let $\phi_1(x)$ and $\phi_2(x)$ are two linearly indpendend solution of

$$2x^2y''-(x+x^2)y'+(x^2-2)y=0$$

such that $\phi_1(0)=0$
, then find the  smallest positive integer $n$ such that  $\displaystyle \lim_{x \to 0}x^n\frac{\phi_2(x)}{\phi_1(x)}=0$

i have tried lots of examples but not getting anywhere, and i have no idea how to deal this second order linear homogeneous differential equation. Please provide me a hind so that i can solve this further, or is there any short trick to solve this?

答:
$$2x^2y''-(x+x^2)y'+(x^2-2)y=0$$
The solutions of this ODE involves confluent hypergeometric functions. This "brute force" approach would be too arduous.

We can look for approximate solutions valid close to $x=0$, which are solutions of
$$2x^2y''-xy'-2y=0$$
because $(x+x^2)\simeq x$ and $(x^2-2)\simeq -2$ at small $|x|$. Solving this "simplified" ODE leads to :

$$y=c_1x^2+c_2x^{-1/2}\quad\implies\quad \Phi_1\simeq x^2\quad\text{and}\quad \Phi_2\simeq x^{-1/2}$$
$$x^n\frac{\Phi_2}{\Phi_1}=x^{n-2-\frac12}=x^{n-\frac52}$$
The smallest $n$ in order to have $\lim_{x \to 0}x^n\frac{\phi_2(x)}{\phi_1(x)}=0$ is
$$n=3$$


问:
Prove that if $n$ is sufficently large, there is a prime gap $G(p_k, p_{k+1})$ with $p_k \leq n$ and $p_{k+1} - p_k > \frac{1}{2}\ln(n)$

Let $p_k ( k \geq 1)$ be an enumeration of all the positive primes with $p_1 = 2$ and $p_k < p_{k+1}$ for all $k \geq 1$. Prove that if $n$ is sufficiently large, then there is a prime gap $G(p_k, p_{k+1})$ with $p_k \leq n$ and $p_{k+1} - p_k > \frac{1}{2}\ln(n)$.

HINT: Let $m$ be the largest integer with $p_m \leq n$. Consider $\sum_{k=1}^m (p_{k+1} - p_k)$ and apply the Prime Number Theorem to $\pi(x)$.

$\pi(x)$ is defined as the number of positive primes $\leq x$.

The Prime Number Theorem states

$$\lim_{n \to \infty} \frac{\pi(n)}{n / \ln(n)} = 1.$$

I'm not sure how to go about doing this. I don't get how the $\frac{1}{2}$ would come into play. Can someone help me please?

答:
So let's apply the hint given. By the definition, $m=\pi(n).$ If each gap $p_{k+1}-p_k\le \frac{1}{2}\ln n$ then the sum $p_m-p_1=\sum_{k=1}^{m-1}(p_{k+1}-p_k)\le m\cdot  \frac{1}{2}\ln n=\frac{1}{2}\pi(n)\cdot \ln n.$ Hence,
$p_m\le 2+\frac{1}{2}\pi(n)\cdot \ln n.$ Now, by the prime number theorem, starting from $n=n_0$ the quantity  $\frac{\pi(n)\cdot \ln n}{n}<(1+\varepsilon),$ or $\pi(n)\cdot \ln n\le (1+\varepsilon)n.$ This can be used to estimate $p_m$ for $n\ge n_0:$ $p_m\le 2+\frac{1}{2}(1+\varepsilon)n.$ This means, that between
$2+\frac{1}{2}(1+\varepsilon)n$ and $n$ there is no primes This contradicts some  stronger version of Bertand's postulate.</p>


问:
Proof regarding double integral
We need to prove that:
$$\int_{0}^{a}\int_{0}^{x}\frac{f'(y)dydx}{\sqrt{(a-x)(x-y)}}=\pi(f(a)-f(0))$$
where, $a$ is a positive constant.

I couldn't see anything by looking at the LHS so I switched the order of integration as:
$$\int_{0}^{a}\int_{0}^{x}\frac{f'(y)dydx}{\sqrt{(a-x)(x-y)}}=\int_{0}^{a}\int_{0}^{y}\frac{f'(y)dxdy}{\sqrt{(a-x)(x-y)}}$$

After calculating the first integral I was left with:
$$\int_{0}^{a}(\frac{-\pi}{2}+\arcsin(\frac{a+y}{a-y}))f'(y)dy$$
but I can't figure out how to calculate the second part of this integral to get the result.

Any help would be appreciated.

答:
It seems that the easiest way is to change the order of integration:
$J(a)=\int_{0}^{a}\int_{0}^{x}\frac{f'(y)dydx}{\sqrt{(a-x)(x-y)}}=\int_{0}^{a}f'(y)dy\int_{y}^{a}\frac{dx}{\sqrt{(a-x)(x-y)}}$

Then the integral type $I(a,b)=\int_{a}^{b}\frac{dx}{\sqrt{(b-x)(x-a)}}$ ($b>a$) after the change $x=(b-a)t+a$ becomes

$I(a,b)=(b-a)\int_{0}^{1}\frac{dt}{\sqrt{(b-a)t}\sqrt{(b-a)(1-t)}}=B(\frac{1}{2};\frac{1}{2})=\Gamma^2(\frac{1}{2})=\pi$

$J(a)=\pi\int_{0}^{a}f'(y)dy=\pi(f(a)-f(0))$

Title: Proof regarding double integral


答:
One thing I'd say is that this clearly is not always the case - in order to answer this you need to know that $f$ is non-negative and $\sup f = 1$, as is given in the notes you linked. As for your identity, let me work from right to left. We have
$$|f \geq t| = \int_{\mathbb{R}^n} \chi_{\{f(x) \geq t\}},$$
which is just a basic property of the Lebesgue measure - the 'length' of the set is the integral of its characteristic function w.r.t the Lebesgue measure. Now we substitute this in to your original identity and apply Fubini to exchange the order of integration to obtain
$$\int_{0}^{1}|f \geq t| d t = \int_{\mathbb{R}^n} \int_{0}^{1} \chi_{\{f(x) \geq t\}} dt dx.$$
Here, in the integral in the middle $x$ is fixed and t varies. So on $0 \leq t \leq f(x)$ the characteristic function will equal $1$, and on $f(x) \leq t \leq 1$ it will equal $0$, where here we're using the fact $f$ takes values in $[0,1]$. Thus,
$$\int_0^1 \chi_{\{f(x) \geq t\}} dt = f(x),$$
and we obtain the identity you're after. I think when they say 'level sets' in your reference, they're referring to the set $\{x \in \mathbb{R^n} \; | \; f(x) \geq t\}$ which I suppose ought to be called a superlevel set.

Title: Why is $\int f(x) dx = \int_0^1 |f\ge t|dt$, where $0\le f\le 1$?

Body:
Why is $$\int f(x) dx = \int_0^1 |f\ge t|dt$$
true? Apparently, it comes from Fubini's theorem - which allows us to write the integral of $f$ as the integral of the length of its level set? Also note that $\sup f = 1$, $f \ge 0$ and $f$ is integrable.

Could someone help me understand where this comes from, and what's going on? Intuition would be helpful too.

Given a function $f$, I understand that $\{x:f(x) = t\}$ is one of its level sets. Is that right? I don't see how this relates with the above equality though.

Reference: Pg. 31, Notes.


问:
Quaternion group associativity
Consider the eight objects $\pm 1, \pm i, \pm j, \pm k$ with multiplication rules:

$ij=k,jk=i,ki=j,ji=-k,kj=-i,ik=-j,i^2=j^2=k^2=-1$,

where the minus signs behave as expected and $1$ and $-1$ multiply as expected. Show that these objects form a group containing exactly one involution.

Well, it is easy to determine closure from the definition. $1$ is clearly the identity, and the inverses can also be determined $(i,-i),(j,-j),(k,-k)$, $-1$ with itself, and $1$ with itself. The only involution is $-1$. Is there an easy way to check associativity of this group? There are too many possible combinations $(ab)c=a(bc)$ to check directly. ($8^3$ possible combinations)

答:
One idea that saves some work, but still might make your eyes cross is to realize that negatives work as expected, so you don't need to test associativity with 1, -1, or any of the negative i,j,k. That leaves only $\{i,j,k\}^3$ which is 27 tests, each requiring 4 very easy multiplications, 108 easy lookups. If you notice that $i \mapsto j \mapsto k$ is an automorphism, then you can say that $a=i$ in $a(bc) \stackrel{?}{=} (ab)c$ reducing it to 9 checks, 36 easy multiplications.

However, a method with more fringe benefits is to find matrices that multiply as expected. It is easiest if you know complex numbers, since $i$ and $j$ are both like $\sqrt{-1}$, but they don't commute.

Hopefully you can discover the matrices by yourself, but maybe it is tricky without a lot of experience. Here are matrices that work ($\sqrt{-1}$ is any fixed square root of $-1$ in a field of characteristic not 2, say $\mathbb{C}$ for instance; take 1 to be the identity matrix, and if $a$ corresponds to the matrix $A$, then $-a$ corresponds to $-A$, so that negatives work just like expected).

 $$i = \begin{bmatrix} \sqrt{-1} & 0 \\ 0 & -\sqrt{-1} \end{bmatrix}, \qquad j = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \qquad k = \begin{bmatrix} 0 & \sqrt{-1} \\ \sqrt{-1} & 0 \end{bmatrix}$$

Title: Quaternion group associativity


问:
A simple problem on Finding factors
There's a question wherein we have to find the number of factors of $480$ of the form $8n+4$ where $n \geq 0$.
Now, the prime factorisation of $480$ gives $2^5 \times 3 \times 5$.

This is to be of the form $8n+4$, i.e. $4(2n+1)= 4x$ (where $x$ is some odd number).
If we equate $4kx$ with $32 \times 3 \times 5$ (k is $\frac{480}{x}$) we obtain the expression $kx = 8 \times 3 \times 5$.

And as $x$ is an odd number, $8$ can't be one of the prime factors of $x$. So $x$ can either be $3, 5$ or $15$. The corresponding values of $k$ would be $160, 96$ and $32$.
Therefore, there can be three such values.

But trial and error gave me four values- $4, 12, 20$ and $60$.
Where am I going wrong?

答:
A full answer would be as follows. We require:-
$$4(2n+1)=2^a3^b5^c.$$
Therefore $a$ is precisely $2$ and we have four possibilities given by $$(b,c)\in (0,0),(0,1),(1,0),(1,1).$$
That gives you the four numbers obtained by trial and error.

Title: A simple problem on Finding factors


问:
Checking holomorphic domain
Let $z=x+iy$ and $f(z) = 3 x y^2 + i y^3$ let $\frac{\partial}{\overline{\partial z}} = \frac{1}{2} (\frac{\partial}{\partial x} + i \frac{\partial}{\partial y})$

Computing the Cauchy–Riemann equations of $f$ leads to $\frac{\partial}{\overline{\partial z}}f(z)=3ixy =0 \iff x=0 \text{ or }y=0$

Can we conclude that $f$ is holomorphic on the real and the imaginary line ?

I'm new to complex analysis I thought the domain where a function is holomorphic must be an open set.

Thank you for your help

答: Indeed. And therefore there is no non-empty open set $\subset\Bbb C$ such that the restriction of $f$ to $A$ is holomorphic.

Title: Checking holomorphic domain


问:
Grid coordinates $\Longleftrightarrow$ Triangle numbering
Pictures say more than a thousand words.

Grid coordinates $\Longleftrightarrow$ Triangle numbering:

Poorman's option. I have tabulated, not "formulated" the values in the second picture.
It has been done for a grid that is such that the horizontal / vertical legs of our isosceles right triangle contains $n=9$ gridpoints.
Of course, we seek to generalize this for abitrary $n$.

What I need first is a formula or a short algorithm (preferrably a one-liner) that converts Grid coordinates into Triangle numbering.
Call this function $\,\operatorname{nr}(i,j)\,$, giving as shown in the pictures:

```
nr(1,1) = 1; nr(2,1) = 2; nr(1,2) = 3; nr(3,1) = 4; nr(2,2) = 5; nr(1,3) = 6; and so on
```

And also the other way around, formulas or short algorithms that converts Triangle numbering into Grid coordinates.
Call them $\,\operatorname{nr2i}(n)\,$ and $\,\operatorname{nr2j}(n)\,$. Examples as in the pictures:

```
nr2i(1) = 1; nr2j(1) = 1; nr2i(2) = 2; nr2j(2) = 1; nr2i(3) = 1; nr2j(3) = 2; and so on
```

A hint may be that arithmetic sequences are involved with $\,\operatorname{nr}(i,j)$: $1+2+3+\cdots +n=n(n+1)/2\,$.
And the sum $\,(i+j)\,$ is a constant for one green $\color{green}{\bf \mbox{\\}}$ "level".

EDIT.

Implementation of the answer given by
cr001 has been done as follows.
My favorite programming language is Pascal.

```
function nr(i,j : integer) : integer;
begin
  nr := (((i+j-1)*(i+j)) div 2)-(i-1);
end;<br>
function nr2i(n : integer) : integer;
var
  h : integer;
begin
  h := Trunc(1/2+sqrt(2*n))+1;
  nr2i := (((h-1)*h) div 2)-(n-1);
end;<br>
function nr2j(n : integer) : integer;
var
  h : integer;
begin
  h := Trunc(1/2+sqrt(2*n))+1;
  nr2j := h - nr2i(n);
end;
```

Testing involves the fact that composing a function with its inverse reproduces identity:

```
n = nr(nr2i(n),nr2j(n));
i = nr2i(nr(i,j)); j = nr2j(nr(i,j));
```

答:
One thing you immediately notice is that $$nr(1,y)={y+1\choose 2}$$

Then you also know $$nr(x+1,y-1)=nr(x,y)-1$$

Therefore the result is

$$nr(x,y)={x+y\choose 2}-(x-1)$$

Edit:

For the other way around, first find the closest bigger triangular number.

Suppose you have $n=nr(x,y)$ and you know $n$.

Then $$x+y=\lceil{-1+\sqrt{8n+1}\over 2}\rceil+1$$

(This comes from solving ${u^2+u\over 2}=n$ where $u=x+y$)

Once you find $x+y$ you calculate $$x={x+y\choose 2}-n+1$$ and $$y=(x+y)-x$$.

To summarize in one formula, you have

$$nr2i(n)={\lceil{-1+\sqrt{8n+1}\over 2}\rceil+1\choose 2} - n + 1$$

$$nr2j(n)=(\lceil{-1+\sqrt{8n+1}\over 2}\rceil + 1) - ({\lceil{-1+\sqrt{8n+1}\over 2}\rceil+1\choose 2} - n + 1)$$


问:
How to solve this simultaneous equation of $3$ variables.
I've stuck in this equation system. No clue how to start ?

$$\begin{eqnarray}
x+y+z &=&a+b+c\tag{1} \\
ax+by+cz &=&a^{2}+b^{2}+c^{2}\tag{2} \\
ax^{2}+by^{2}+cz^{2} &=&a^{3}+b^{3}+c^{3}\tag{3}
\end{eqnarray}$$

Find the value of $x,y,z$ is in the form of $a,b$ and $c$.
I want to know steps of solution.

答:
I'm supposed to solve it using simultaneous eqn method like
  elimination,substitution,cross-multiplication .NO knowledge of
  matrices

By inspection we see that $(x,y,z)=(a,b,c)$ is a solution of the given system

$$
\begin{array}{l}
\text{Eq. 1}\qquad x+y+z=a+b+c \\
\text{Eq. 2}\qquad ax+by+cz=a^{2}+b^{2}+c^{2} \\
\text{Eq. 3}\qquad ax^{2}+by^{2}+cz^{2}=a^{3}+b^{3}+c^{3}.
\end{array}
 \tag{0}$$

The other solution can be found as follows. Solve Eq. 1 for $z$. Multiply original Eq. 1 by $a$, subtract Eq. 2 and solve for $z$. This results in
$$
\begin{array}{l}
z=a+b+c-x-y,
\end{array}
 \tag{1}$$

$$
\begin{array}{l}
z=\frac{b-a}{a-c}y+\frac{ab+ac-b^{2}-c^{2}}{a-c} .
\end{array}
 \tag{2}$$

Equate the right hand sides of $(1)$ and $(2)$

$$
\begin{array}{l}
\frac{b-a}{a-c}y+\frac{ab+ac-b^{2}-c^{2}}{a-c}=a+b+c-x-y,
\end{array}
 \tag{3}$$

and solve for $x$

$$
x=\frac{c-b}{a-c}y+\frac{-ac+b^{2}+a^{2}-bc}{a-c}.
\tag{4}$$

Substitute $x,z$ in $(0)$, Eq. 3, using $(4)$ and $(2)$

$$
a\left( \frac{c-b}{a-c}y+\frac{-ac+b^{2}+a^{2}-bc}{a-c}\right)
^{2}+by^{2}+c\left( \frac{b-a}{a-c}y+\frac{ab+ac-b^{2}-c^{2}}{a-c}\right)
^{2}=a^{3}+b^{3}+c^{3}.
\tag{5}$$

Solving for $y$ we get$^1$ the solution $y=b$ and the solution

$$y=\frac{B}{D},\tag{6}$$

where

$$
B=-2a^{3}c+2a^{3}b-a^{2}b^{2}-a^{2}bc+4a^{2}c^{2}-2acb^{2}-2ac^{3}+ab^{3}-abc^{2}+2bc^{3}+b^{3}c-c^{2}b^{2}$$
$$
D=a^{2}c+ac^{2}-6abc+a^{2}b+bc^{2}+ab^{2}+b^{2}c
$$

Finally substitute $y=b$ and $y=B/D$ in $(4)$ and $(2)$. We get the solutions $(x,z)=(a,c)$, and

$$(x,z)=\left(\frac{A}{D},\frac{C}{D}\right),\tag{7}$$

where

$$
A=a^{3}c+a^{3}b-2a^{2}bc-a^{2}b^{2}-a^{2}c^{2}+2ac^{3}-abc^{2}+2ab^{3}-acb^{2}-2bc^{3}-2b^{3}c+4c^{2}b^{2}
$$

$$
C=2a^{3}c-2a^{3}b+4a^{2}b^{2}-a^{2}c^{2}-a^{2}bc-acb^{2}-2ab^{3}+ac^{3}-2abc^{2}+bc^{3}+2b^{3}c-c^{2}b^{2}.
$$

So the two solutions of $(0)$ are: $$(x,y,z)=(a,b,c)\qquad\text{and }\qquad(x,y,z)=\left(\frac{A}{D},\frac{B}{D},\frac{C}{D}\right).$$

--

$^1$ Eq. $(5)$ is equivalent to

$$\begin{equation*}
\left( cb^{2}+c^{2}b+ac^{2}+ca^{2}+ab^{2}-6acb+ba^{2}\right) (y-b)\left( y-\frac{B}{D}\right) =0.
\end{equation*}$$


问:
Partial sum of Combination Induction Problem
How do I prove the following?

$\binom{n}{1} + 2\binom{n}{2}+ 3\binom{n}{3}+...+n\binom{n}{n}=n2^{n-1}$

$\forall n\epsilon \mathbb{N}$

Base Case:

$\binom{1}{1} = (1)2^{1-1}$

$1=1\;$, Which is True.

With $\;n=k$ let us assume that

$\binom{k}{1} + 2\binom{k}{2}+ 3\binom{k}{3}+...+k\binom{k}{k}=k2^{k-1}\;$
is true

then, for n=k+1 we have

$\binom{k+1}{1} + 2\binom{k+1}{2}+ 3\binom{k+1}{3}+...+k\binom{k+1}{k+!}=(k+1)2^{k}\;$

We know that

$_{}^{k}\textrm{C}_{r}=\frac{k!}{r!(k-r)!}$

then our inductive hypothesis can be seen as:

$\frac{k!}{k!(k-1)!} +2\frac{k!}{k!(k-2)!}+ 3\frac{k!}{k!(k-3)!} ...+k\frac{k!}{k!(k-k)!} = k2^{k-1}$

and what we need to prove should look like:

$\frac{(k+1)!}{(k+1)!(k)!} +2\frac{(k+1)!}{(k+1)!(k-1)!}+ 3\frac{(k+1)!}{(k+1)!(k-2)!} ...+(k+1)\frac{(k+1)!}{(k+1)!(k+1-k-1)!} = (k+1)2^{k}$

Is this the way to proceed? How would you proceed? Please show me the way, I am stuck.

答: I wouldn't actually go down the induction route. We know from the binomial theorem that $(1+x)^n=1+\binom n1 x+\binom n2 x^2+\cdots +\binom nn 1$. Notice that $(1+1)^n=2^n$, where we have set $x=1$. Now differentiate this - we get $x+2\binom n2 x+3\binom n3 x^2+\cdots$. Can you see how to finish this proof?


问:
Prove that $E_{\alpha} \setminus \bigcup\limits_{\beta \neq \alpha} E_{\beta} \neq \varnothing,$ for all $\alpha \in I.$
Let $\Lambda = \left \{E_{\alpha}\ |\ \alpha \in I \right \}$ be an uncountable family of infinite subsets of $\Bbb N$ such that intersection of any two subsets in $\Lambda$ has a finite intersection. Show that $E_{\alpha} \setminus \bigcup\limits_{\beta \neq \alpha} E_{\beta} \neq \varnothing,$ for any $\alpha \in I.$

If possible let there be an $\alpha \in I$ such that $E_{\alpha} \setminus \bigcup\limits_{\beta \neq \alpha} E_{\beta} = \varnothing$ i.e. $E_{\alpha} \subseteq \bigcup\limits_{\beta \neq \alpha} E_{\beta}.$ That means there is a countable subset $J$ of $I \setminus \left \{\alpha \right \}$ such that $E_{\alpha} \subseteq \bigcup\limits_{\beta \in J} E_{\beta}.$ From here how do I arrive at a contradiction? Any help in this regard will be appreciated.

Thanks in advance.

答:
The statement is not true.

Let $I_0,I_1$ be two disjoint subsets of $I$ such that $|I_0|=|I_1|=\aleph_0$, so we can write them as $\alpha_n^0$ and $\alpha_n^1$ for the $n$th set in $I_0$ and $I_1$ respectively. Now redefine your family:

If $\alpha\notin I_0\cup I_1$, $E_\alpha$ stays the same.

If $\alpha=\alpha_n^i$ for either $i\in\{0,1\}$, $E_\alpha$ is replaced by $E_\alpha\cup\{n\}$.

It is now easy to see that $\bigcup_{\alpha\in I_0}E_\alpha=\Bbb N$, and the same for $I_1$. So if $\alpha\in I$, then either $\alpha\notin I_0$ or $\alpha\notin I_1$. But in either case, $\bigcup_{\beta\neq\alpha}E_\beta=\Bbb N$, so $E_\alpha\setminus\bigcup_{\beta\neq\alpha}E_\beta=\varnothing$.


问:
Integrate in Mathematica takes forever
I'm trying to calculate the length of a curve from a polynomial in Mathematica and for some reason Integrate doesn't complete, it just runs until I abort the execution.

The polynom:

```<code>58.1735 - 36.4175 x + 7.10945 x^2 - 0.538311 x^3 + 0.0138115 x^4
</code>```

Is a part of an integration like this:

where "koordfitprim" is the polynom above (i.e. what is integrated is in fact the square root of 1 + the above polynom^2).

Any ideas why this wouldn't execute?

答:
You should try to use NIntegrate (instead of Integrate). Integrate is for symbolic integration. NIntegrate for numerical integration. So try</p>

```<code>NIntegrate[Sqrt[1+koordfitprim^2],{x,3,18}]
</code>```

Title: Integrate in Mathematica takes forever

Body:
You should try to use NIntegrate (instead of Integrate). Integrate is for symbolic integration. NIntegrate for numerical integration. So try

NIntegrate[Sqrt[1+koordfitprim^2],{x,3,18}]


问:
Fermat primes and the equation $x^2+y^2 = 10^n x+y$
This is related to the question by Naroza which was ably answered by E. Wong. In a nutshell, it seeks to find more examples of the curiosities,

$$12^2+33^2 = 1233$$

$$88^2+33^2 = 8833$$

or, in general, to solve,

$$x^2+y^2 = 10^n x+y$$

for integer $x,y$ (naturally, with size constraints on $x,y$). Wong gave,

$$a=4,\;\;b=10^{4m},\;\;x = \tfrac{a}{17}(ab-1),\;\;y = \tfrac{a}{17}(a+b)\tag{1}$$

where $m=4k+3$. For example, for $k=0$, we get,

$$ \color{red}{941176470588}^2 + \color{blue}{235294117648}^2 = \color{red}{941176470588}\color{blue}{235294117648}$$

and its partner as,

$$ {\color{red}{588}\color{blue}{2352}\color{red}{9412}}^2 + \color{blue}{235294117648}^2 = \color{red}{588}\color{blue}{2352}\color{red}{9412}\color{blue}{235294117648}$$

Note that,

$$ \color{red}{588}^2 + \color{blue}{2352}^2 = \color{red}{588}\color{blue}{2352}$$

$$ \color{red}{9412}^2 + \color{blue}{2352}^2 = \color{red}{9412}\color{blue}{2352}$$

Anyway, I noticed the Fermat prime 17 and, after a little experimentation, found,

$$a=16,\;\;b=10^{64m},\;\;x = \tfrac{a}{257}(ab-1),\;\;y = \tfrac{a}{257}(a+b)\tag{2}$$

$$a=256,\;\;b=10^{16384m},\;\;x = \tfrac{a}{65537}(ab-1),\;\;y = \tfrac{a}{65537}(a+b)\tag{3}$$

Question: Are the $x,y$ of (2), (3) also integers for all positive integer $m=4k+3$?

(I've tested it for the first few $k$ and it is true. Other Fermat numbers don't work. It seems the divisor must be prime > 5.)

答:
Looking at just the first one,

$\frac{a}{257}(ab-1)$ is an integer, if and only if

$(a^2+1)|(a^2b-a)=a(ab-1)$, if and only if

$(a^2+1)|ab-1$ (since $a,a^2+1$ are relatively prime), if and only if

$ab\equiv 1\pmod{a^2+1}$

By Fermat's little theorem, since $a^2+1$ is prime in this case, $10^{a^2}=10^{256}\equiv 1\pmod{257}$.

Hence, $ab(10^{256})\equiv 1\pmod{a^2+1}$.  Set $c=b10^{256}=10^{64(4k+3)}10^{256}=10^{64(4(k+1)+3)}$.  This can be repeated, to get $10^{64(4(k+2)+3)}$, etc, hence any $m=4k+3$.


问:
Limit of fourier series
I want to compute the limit $f(x)$ of

$$
f(n,x)=\sum_{k=1}^n\frac{1}{k^2+1}\sin(kx)
$$

and didn't succeed. I tried with the means used to compute the >>easy<< limits of trigonometric series like those listed in (e.g.) Bronstein, however, the above series is not in Bronstein. I plotted the graph with sagemath (see my post on ask.sagemath.org: https://ask.sagemath.org/question/55596/limit-of-fourier-series/). At least this suggests that f(x) exists and is not a polynomial.

Is it possible to compute the above limit and if so, how?

答:
Strating from @user1337's answer, what is interesting is that for small values of $x$, we can write
$$f(\infty,x)=\sum_{n=0}^\infty \frac{a_n-\Big[\frac{H_{i}+H_{-i}}{2} +\log(x)\Big]}{(2n+1)!} x^{2n+1}$$
where the  $a_n$'s are
$$\left\{1,\frac{23}{12},\frac{19}{8},\frac{1355}{504},\frac{14761}{5040},\frac{34675}{11088},\frac{20381}{6160},\cdots\right\}$$

Title: Limit of fourier series


问:
If exist a extension of $f$ to a map $D^2\to X$, then $f: S^1\to X$ is homotopic to a constant map

Every map $f: S^1\to X$ is homotopic to a constant map, if and only if, exist a extension of $f$ to a map $D^2\to X$.

Here, $X$ is an arbitrary topological space. I managed to prove the direction $\Rightarrow$, for reference I will also leave some links for similar answers. (1), (2), (3), (4), (5).

I know that there is a list of equivalences, in this case:

The following three conditions are equivalent:

Every map $S^1\to X$ is homotopic to a constant map, with image a point.

Every map $S^1\to X$ extends to a map $D^2\to X$.

$\pi_1(X, x_0) = 0$ for all $x_0\in X$.

As I said, the direction I prove is $1\Rightarrow 2$, and the direction I'm having trouble proving is $2\Rightarrow 1$. From the list of equivalences, it would be enough to prove $2\Rightarrow 3$ and then $3\Rightarrow 1$, but I would like to know the direct demonstration of $2\Rightarrow 1$.

答: Thinking of $D^2$ as consisting of concentric circles with shrinking radii,  we can use a homotopy $H(re^{i \theta}, t) := \tilde{f}(tre^{i \theta})$, where $\tilde{f}$ is the extension of $f$.


问:
Derivation of a recursive binomial coefficient algorithm
I am working through a Haskell book and I see that the binomial coefficients formula

$$\binom{n}{k} = \frac {n!}{k! (n - k)!}$$

is expressible as recurrence relationships

\begin{align}
\binom{n}{k} &= \binom{n - 1}{k - 1} \frac{n}{k} \quad
\text{for} \; n \gt 0 \\
\binom{n}{0} &= 1
\end{align}

and as

\begin{align}
\binom{n}{k} &= \binom{n - 1}{k - 1} + \binom{n - 1}{k} \\
\binom{n}{0} &= 1
\end{align}

My question is how were these derived from the original formula above? Is there a treatment, a proof anywhere of how these were derived that I could see?

Update

Yes, the classic algebra assignment is to take the recurrence relation and simplify it back to the original as a proof that the recursive version is indeed the same as the original

$$\binom{n}{k} = \frac {n!}{k! (n - k)!} = \binom{n - 1}{k - 1} + \binom{n - 1}{k} $$

Instructive, but I would like to know how the recursive version was derived in the first place. I assume that there is was no systematic way, that it was just a guess -- or perhaps some much deeper numerical analysis issue?

Body:
Reducing to the common denominator,
$$\frac{(n-1)!}{k!(n-1-k)!}+\frac{(n-1)!}{(k-1)!(n-k)!}=\frac{(n-k+k)(n-1)!}{k!(n-k)!}=\frac{n!}{k!(n-k)!}.$$