问:
What's wrong in my approach in evaluating $\cos^{-1}(\cos\ 10)$
The Question

$\large\cos^{-1}(\cos 10)$

My Approach

$\large\cos^{-1}(\cos\ (10-2\pi))$ As $2\pi$ is the period of $\cos x.$ Now our angle is-

Now we know that for $\large\cos^{-1}(\cos x) = x$, $\large x\ \epsilon\ [0,\pi]$ but currently our $\large x$ is in the 3rd quadrant, we need to get it into 1st or 2nd quadrant. Therefore let us subtract $\large \pi$ from our current angle i.e $\large 10-2\pi$

$\large\cos^{-1}(\cos\ -((10-2\pi)-\pi))$ Now you must be wondering why did I put this minus sign, w.k.t $\large \cos x$ is -ve in third quadrant, by subtracting $\large \pi$ we brought it into 1st quadrant and $\large \cos x$ is +ve in 1st quad, so to compensate that I have put a minus sign.

Now its within range so we can use $\large\cos^{-1}(\cos x) = x$.

So Final answer $\large -((10-2\pi)-\pi)$ = $\large 3\pi - 10$

But this answer is wrong and the correct answer is $\large 4\pi - 10$.

They have used the property $\large \cos\  (\pi - \theta) = \cos\  (\pi + \theta)$ But why my approach is wrong?

Kindly help.

答:
If $x$ is in the third quadrant, its symmetric w.r.t. the $x$-axis is in the second quadrant, and has the same cosine. This symmetric point has curvilinear abscissa $6\pi - x$, as you've found. To determine the value which is represented by the same point on the trigonometric circle, let's just make some elementary manipulations on inequalities.

If $x=10$ˆ, we have $3\pi < x <\dfrac{7\pi}2$, so
\begin{align}
&&\frac{5\pi}2 &<6\pi -x< 3\pi  \\[1ex]
&\llap{\text{whence}}\qquad& \frac \pi 2 &<(6\pi-x)-2\pi=4\pi -x <\pi&&\qquad
\end{align}
The solution follows.


问:
Cauchy's integral formula when $z_0$ lies outside of contour w/o residue thrm
I understand there are many questions regarding this topic (Complex integral where poles are outside of the given region)

I just wanted to double check with everyone what I gathered so far:

$$f(z_0) = \frac{1}{2\pi i}\int_\Gamma \frac{f(z)}{z-z_0}dz = 0, \\ $$
if $z_0$ lies outside of the given contour and there exists a closed loop around the pole where the loop is analytic everywhere (this is my understanding of holomorphic).

I think this is true due to the path independence but I can't prove it exactly. Can anyone please help?

One specific example will be $$\int_C \frac{ze^z}{2z-3}dz$$ where C be the circle $|z-1.5| = 2$ traversed once in the positive sense.

I calculated that $z_0 = 3/2$ does not lie inside of the curve ($z$ intersects with the real axis at $\pm \sqrt{1.75}$).

Hence, I claim according to the above assumption because the pole does not lie inside of the region of interest ($C$), the integral is $0$.

I stated w/o residue thrm because we didn't learn it. I see many answers use residue thrm in their answer so it will be great if you could explain it w/o residue thrm!

答: The line integral of a holomorphic function $f$ over the boundary of a domain where $f$ is holomorphic equals zero. Your function $\frac{f(z)}{z-z_0}$ is holomorphic inside the domain, so the integral is zero.

ContentLicense: CC BY-SA 4.0


问:
Describing $\mathbb{Z}[x/2]$
In GCD in a subring is GCD in a bigger ring? it is  noted that $\gcd(2, x) = 1$ in $\mathbb{Z}[x]$ (since $\mathbb{Z}[x]$ is a UFD and the prime/irreducible factorizations of $2$ and $x$ share no primes), but $\gcd(2, x) = 2$ in $\mathbb{Z}[x/2]$. I am struggling to understand $\mathbb{Z}[x/2]$.

What is a rigorous definition of $\mathbb{Z}[x/2]$?

Is it isomorphic to a quotient of $\mathbb{Z}[x]$? This won't imply UFD since $\mathbb{Z}[x]/(x^2 + 5) \cong \mathbb{Z}[\sqrt{-5}]$.

Is it a UFD?

What are the units?

What are the primes and irreducibles?

Does all of this generalize to $\mathbb{Z}[ax]$ with $a \in \mathbb{Q}$?

Here are some of my scattered ideas, which may contain incorrect statements (no need to read this). The polynomials in $\mathbb{Z}[x/2]$ ought to look like
$$ \sum_{i = 0}^n a_i \left( \frac x 2 \right)^i = \sum_{i = 0}^n \frac{a_i}{2^i} x^i \in \mathbb{Q}[x]$$
For (1.), my first idea is to conceptualize $\mathbb{Z}[x/2]$ as some sort of monoid ring just like $\mathbb{Z}[x]$ is the monoid ring $\mathbb{N}[\mathbb{Z}]$. But now it looks like $\mathbb{Z}[x/2]$ might be "equal" to $(\mathbb{Z}[1/2])[x]$. Or maybe we can localize $\mathbb{Z}[x]$ at some multiplicative set like $(2\mathbb{Z})[x]$.

On (2.), $\mathbb{Z}[x]/(2x - 1) \cong \mathbb{Z}[1/2]$ ($(2x - 1)$ is the kernel of evaluation at $1/2$: if $f \in \mathbb{Z}[x]$ satisfies $f(1/2) = 0$, then the polynomial $2x - 1$ has coprime coefficients and divides the integer polynomial $f$ in $\mathbb{Q}[x]$ and thereby in $\mathbb{Z}[x]$ by Gauss's lemma), but I have few ideas on how to influence the formal symbol $x$ with a quotient -- I believe Laurent polynomials over a field $F$ come from $F[[x]][y]/(xy - 1)$, which certainly influences $x$ by adding $\frac 1 x$.

For (3.) if we could show Noetherian integral domain then it would suffice to prove that irreducible implies prime. $\mathbb{Z}$ is Noetherian, so $\mathbb{Z}[x_1, \dots, x_n]$ and $\mathbb{Z}[[x_1, \dots, x_n]]$ are Noetherian as well as any quotients of these. Integral domain should be true because the lead coefficients of two nonzero polynomials in $\mathbb{Z}[x/2]$ cannot multiply to zero.

For (4.), there is a lovely classification of the units of a polynomial ring. If the same applies here, the units should be the units of $\mathbb{Z}$, namely $\pm 1$. For (5.), primes and irreducibles will be the same assuming $\mathbb{Z}[x/2]$ is a UFD. The irreducibles in $\mathbb{Z}[x]$ are precisely the integer primes and the polynomials with coprime coefficients that remain irreducible in $\mathbb{Q}[x]$ (and associates of these). So maybe the primes in $\mathbb{Z}[x/2]$ are similar modulo something about $2$.

For (6.), my guess is this all generalizes to $\mathbb{Z}[x/p]$ for primes $p \in \mathbb{Z}$ but not any further than that.

答:
We can define this rigorously as follows: consider the ring homomorphism $\phi: \mathbb Z[x] \longrightarrow \mathbb Q[x]$ defined by $x \mapsto \frac{x}{2}$, i.e. $f \mapsto f\left(\frac{x}{2}\right)$. The image is $\mathbb Z\left[\frac{x}{2}\right]$. This is generally how you adjoin elements from one ring to another - you adjoin a formal variable and evaluate that variable at your chosen element.

Certainly! By the first isomorphism theorem, $\phi$ induces an isomorphism $\mathbb Z[x] / \ker(\phi) \longrightarrow \mathbb Z\left[\frac{x}{2}\right]$. However, you want to say that this is $\mathbb Z[x] / (2x - 1)$, which is not true. I'll prove in a moment that this is not the case, but first of all observe that $2x - 1 \notin \ker(\phi)$. Indeed, $\phi(2x - 1) = x - 1$.

What then is $\ker(\phi)$? Let's take some $f \in \ker(\phi)$ and write $f = \sum f_i x^i$, $f_i \in \mathbb Z$. Then
$$
0=\phi(f) = \sum f_i \left(\frac{x}{2}\right)^i = \sum \frac{f_i}{2^i} x^i.
$$
Remember that the image lives in $\mathbb Q[x]$, where $x$ is still just a formal variable. In particular, the set $\{1, x, x^2, \dots\}$ is linearly independent over $\mathbb Q$. Hence, as $\sum \frac{f_i}{2^i} x^i = 0 \in \mathbb Q[x]$, each coefficient $\frac{f_i}{2^i} = 0$. Then just multiply by $2^i$ and deduce that each $f_i = 0$ as well. Hence, $f = 0$ so $\phi$ is an isomorphism and $\mathbb Z[x] \cong \mathbb Z\left[\frac{x}{2}\right]$. By the way, this means in particular that $2$ is not a unit in $\mathbb Z\left[\frac{x}{2}\right]$ so the quotient $\mathbb Z[x]/(2x-1) \not\cong \mathbb Z\left[\frac{x}{2}\right]$.

This may seem a bit strange. After all, it seems like your intuition was that we are somehow inverting something. And this intuition makes sense, after all we are throwing a $\frac{1}{2}$ in there! However, this formal proof I just gave tells us that no additional relations arise when you replace $x$ with $\frac{x}{2}$. If it's easier to see it this way, let $y = \frac{x}{2}$. Then the inverse map simply sends $y \mapsto 2y$. I think it's easier to see why this map doesn't add any extra units.

3, 4, and 5 all follow from your knowledge of $\mathbb Z[x]$, as we have just shown that $\mathbb Z\left[\frac{x}{2}\right] \cong \mathbb Z[x]$. Let's go through them real quick.

Yes, as $\mathbb Z[x]$ is a UFD.

The isomorphism $Z[x] \longrightarrow \mathbb Z\left[\frac{x}{2}\right]$ restricts to an isomorphism on unit groups. The units of $\mathbb Z[x]$ are $\pm 1$, so the units of $\mathbb Z\left[\frac{x}{2}\right]$ are $\phi(1) = 1$ and $\phi(-1) = -1$.

$\phi$ is a ring isomorphism so these ring theoretic concepts of primality and irreducibility are preserved by it. $f \in \mathbb Z[x]$ is prime iff $\phi(f) = f\left(\frac{x}{2}\right) \in \mathbb Z\left[\frac{x}{2}\right]$ is prime, and prime = irreducible here as we just showed this was a UFD.

Yup, except for $a = 0$ this is an isomorphism for the same reason. Take $\mathbb Z[x] \longrightarrow \mathbb Q[x]$ via $x \mapsto ax$. For $a \neq 0$, the exact same proof works for computing the kernel of this map. This is because $a \neq 0$ is a unit in $\mathbb Q$, therefore not a zero divisor, so if $f_i a^i = 0$ iff $f_i = 0$.

Title: Describing $\mathbb{Z}[x/2]$


问:
How do I prove that all the eigenvalues are positive?

Show that the matrix

\begin{bmatrix}
2 & 0 & 0 & \cdots & 0 & 1\\
0 & 2^2 & 0 & \cdots & 0 & 1\\
0 & 0 & 2^3 & \cdots & 0 & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 2^{n-1} & 1\\
1 & 1 & 1 & \cdots & 1 & a\\
\end{bmatrix}

only has positive eigenvalues if $a\ge 1.$

What should I use to prove this? Should I use the fact that the matrix is symmetric?

答:
Since the matrix is symmetric, it has positive eigenvalues if and only if it is positive definite.  By Sylvester's criterion this is true if and only if all the leading principal minors are positive.  This is clearly true of the first $n-1$ of them, so we must prove that the matrix itself has positive determinant.  Write the matrix as $$M = \begin{bmatrix}D &x\\x^T & a\end{bmatrix},$$ where $$D = \begin{bmatrix}2\\ &\ddots \\&& 2^{n-1}\end{bmatrix}, \quad x = \begin{bmatrix}1 \\ \vdots \\ 1\end{bmatrix}.$$

Let $$U=\begin{bmatrix}I&0\\-x^tD^{-1}&1\end{bmatrix},$$ where $I$ is an $(n-1)\times(n-1)$ identity matrix.  We have $\det(U)=1$, since $U$ is lower triangular, with $1$'s on the diagonal.  Then $$\det(M)=\det(UM)=\det\begin{bmatrix}D&x\\0&a-x^tD^{-1}x\end{bmatrix}=2^{n(n-1)/2}(a-2^{1-n}),$$ so that $a\geq1$ implies $\det(M)>0$ for $n\geq2$.


答:
First, note that $\arccos(\cos(x)) =x$ only if $x \in [0,\pi]$. Note that arccos(x) is inverse of cos(x).

Thus, we need to get $10$ into that range. We can subtract periods to get $10-4\pi$, but this is a negative number $\approx -2$, not in the range, which was your mistake. So, we can use the identity that $\cos(x) = \cos(-x)$, so $\cos(10-4\pi) = \cos(4\pi-10)$. $4\pi - 10$ is in range, so we are good there. Now, we need to calculate $\arccos(\cos(4\pi-10))$. However, since $4\pi - 10 \approx 2.5$, which is in the range $[0,\pi]$, we know that $\arccos(\cos(4\pi-10))$ is just $4\pi-10$.

Title: What's wrong in my approach in evaluating $\cos^{-1}(\cos\ 10)$

Body:
The Question

$\large\cos^{-1}(\cos 10)$

My Approach

$\large\cos^{-1}(\cos\ (10-2\pi))$ As $2\pi$ is the period of $\cos x.$ Now our angle is-

Now we know that for $\large\cos^{-1}(\cos x) = x$, $\large x\ \epsilon\ [0,\pi]$ but currently our $\large x$ is in the 3rd quadrant, we need to get it into 1st or 2nd quadrant. Therefore let us subtract $\large \pi$ from our current angle i.e $\large 10-2\pi$

$\large\cos^{-1}(\cos\ -((10-2\pi)-\pi))$ Now you must be wondering why did I put this minus sign, w.k.t $\large \cos x$ is -ve in third quadrant, by subtracting $\large \pi$ we brought it into 1st quadrant and $\large \cos x$ is +ve in 1st quad, so to compensate that I have put a minus sign.

Now its within range so we can use $\large\cos^{-1}(\cos x) = x$.

So Final answer $\large -((10-2\pi)-\pi)$ = $\large 3\pi - 10$

But this answer is wrong and the correct answer is $\large 4\pi - 10$.

They have used the property $\large \cos\  (\pi - \theta) = \cos\  (\pi + \theta)$ But why my approach is wrong?

Kindly help.


问:
Does every Krull ring have a height 1 prime ideal?
Let $A$ be a Krull ring. According to Theorem 12.3 in Matsumura's Commutative Ring Theory, the family of localizations of $A$ at height 1 prime ideals of $A$ forms a defining family of $A$.

Question: Why such family exists? In other words, why does a Krull ring have at least one height 1 prime ideal?

Remark: By definition a Krull ring is the intersection of DVRs and each DVR has dimension 1, hence its maximal ideal has height 1. However, if we contract this maximal ideal to $A$, it is not necessary that the prime ideal we get will have height 1.

Edit: I realized that the definition of a Krull ring given in Wikipedia is quite different from the one given in Matsumura. In fact, the Wikipedia definition trivially answers my question. Matsumura's definition is: an integral domain is called Krull if it is the intersection of a family of DVRs and every non-zero element in the domain is non-zero in only a finite number of corresponding discrete valuations. How to obtain that such a ring contains a height 1 ideal is not obvious to me.

答:
Any finite-dimensional ring (indeed, any ring with a prime of finite, nonzero height) has prime ideals of height 1, basically by definition of the dimension: if $P$ has finite height $n$, there's a maximal chain $P_0 \subseteq P_1 \subseteq \dotsb \subseteq P_n = P$, and $P_1$ can't contain any more than one prime ideal, since otherwise we could make a longer chain.  Indeed, so does any Noetherian ring, since it satisfies the descending chain condition on prime ideals.

If you're really interested in the non-Noetherian case, things get confusing -- for example, there are domains where every nonzero prime has infinite height.  If $R$ has this property, though, then all localizations of $R$ have this property (the primes of any localization of $R$ corresponding to a downward-closed subset of the primes of $R$).  In particular, no localization of $R$ can be a DVR, since DVRs have finite dimension, so $R$ can't be a Krull ring, even by Matsumura's definition.  This should prove the equivalence of the two definitions, too.

ContentLicense: CC BY-SA 3.0


问:
$L^1$ Approximation of BV functions by piecewise constant function
Let $f \in BV(\mathbb{R}) \cap L^1(\mathbb{R}),$ consider the piecewise constant function  defined by
\begin{eqnarray}
f^{\delta}(x):= \frac{1}{\delta}\int\limits_{k\delta}^{(k+1)\delta}f(y)dy \quad \text{for } x\in [{k\delta},{(k+1)\delta})
\end{eqnarray}
Then clearly $f^{\delta} \in L^{1}(\mathbb{R})$ and $f^{\delta} \rightarrow f$ pointwise a.e.in $\mathbb{R}.$

Now, my doubt is "Do we have $\|f^{\delta}-f\|_{L^1(\mathbb{R})}\leq C \delta$ for some $C$ independent of $\delta?$"
If so how to prove it?

答:
This is true.

We can first split up $\|f^{\delta} - f\|_{L^{1}(\mathbb{R})}$ as follows:
$$
\|f^{\delta} - f\|_{L^{1}(\mathbb{R})}
= \int_{\mathbb{R}} |f^{\delta}(x) - f(x)| \mathrm{d} x
= \sum_{k \in \mathbb{Z}} \int_{k\delta}^{(k + 1)\delta} |f^{\delta}(x) - f(x)| \mathrm{d} x.
$$
Partitioning $\mathbb{R}$ into those intervals enables us to rewrite $f^{\delta}$ using its definition. Thus,
$$
\|f^{\delta} - f\|_{L^{1}(\mathbb{R})}
= \sum_{k \in \mathbb{Z}} \int_{k\delta}^{(k + 1)\delta}
\bigg|\frac{1}{\delta}\int_{k\delta}^{(k + 1)\delta} f(y) \mathrm{d} y - f(x)\bigg|
\mathrm{d} x.
$$
Moving $f(x)$ inside the $y$-integral and using the triangle inequality gives us
$$
\|f^{\delta} - f\|_{L^{1}(\mathbb{R})}
\leq \sum_{k \in \mathbb{Z}} \int_{k\delta}^{(k + 1)\delta}
\frac{1}{\delta}\int_{k\delta}^{(k + 1)\delta} |f(y) - f(x)| \mathrm{d} y
\mathrm{d} x.
$$
The right-hand side only increases if we include all $y$ such that $|x - y| \leq \delta$, so we have
$$
\|f^{\delta} - f\|_{L^{1}(\mathbb{R})}
\leq \sum_{k \in \mathbb{Z}} \int_{k\delta}^{(k + 1)\delta}
\frac{1}{\delta}\int_{x - \delta}^{x + \delta} |f(y) - f(x)| \mathrm{d} y
\mathrm{d} x.
$$
Reassembling the outside integrals yields
$$
\|f^{\delta} - f\|_{L^{1}(\mathbb{R})}
\leq \int_{\mathbb{R}}
\frac{1}{\delta}\int_{x - \delta}^{x + \delta} |f(y) - f(x)| \mathrm{d} y
\mathrm{d} x.
$$
Setting $y = x + h$ and changing the order of integration, we see that
$$
\|f^{\delta} - f\|_{L^{1}(\mathbb{R})}
\leq \frac{1}{\delta}\int_{-\delta}^{\delta}
\int_{\mathbb{R}} |f(x + h) - f(x)| \mathrm{d} x \mathrm{d} h.
$$
Analagously to Proposition 9.3 in Brezis' Functional Analysis, Sobolev Spaces, and Partial Differential Equations, we have that if $f \in BV(\mathbb{R})$, then
$$
\|\tau_{h}f - f\|_{L^{1}(\mathbb{R})} \leq |h|\|\nabla f\|,
$$
where $\tau_{h}f(x) = f(x + h)$ and $\|\nabla f\|$ denotes the total variation of $f$. Using this,
$$
\|f^{\delta} - f\|_{L^{1}(\mathbb{R})}
\leq \frac{1}{\delta}\int_{-\delta}^{\delta} |h| \|\nabla f\| \mathrm{d} h
= \|\nabla f\|\delta.
$$


答:
As mentioned here, if $\mathbf M=\left[\begin{matrix}\bf A & \bf b\\\bf b^\top & \bf d \\\end{matrix}\right]$ then $\det(\mathbf M)=\alpha\det(\mathbf A)$ where $\alpha=\mathbf d-\bf b^\top A^{-1}b$

In our case: $b^\top=[1,1, \cdots 1]$ and $ \alpha = a - \sum_{k=1}^{n-1}2^{-k}$. Then, $a>1 \implies \alpha >0$  and the matrix is positive definite (Sylvester's criterion) , and the eigenvalues are positive.

Title: How do I prove that all the eigenvalues are positive?

Body:
Show that the matrix

\begin{bmatrix}
2 & 0 & 0 & \cdots & 0 & 1\\
0 & 2^2 & 0 & \cdots & 0 & 1\\
0 & 0 & 2^3 & \cdots & 0 & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 2^{n-1} & 1\\
1 & 1 & 1 & \cdots & 1 & a\\
\end{bmatrix}

only has positive eigenvalues if $a\ge 1.$

What should I use to prove this? Should I use the fact that the matrix is symmetric?


问:
Question on the interpretation of topological entropy
The standard intuitive explanation for topological entropy is that it measures the exponential growth rate of the number of distinguishable orbits.  I'm not quite sure why this is the case, any thoughts?  Thanks.

Edit:  The definition I'm using is for a continuous transformation T and an open cover $\alpha$ we define $h(T,\alpha)= \frac{1}{n} lim_{n \to \infty}H(\bigvee_{i=0}^{n-1}T^{-i}\alpha)$.  Where $\vee$ is the adjoin (intersection) of all the covers and $H$ is the minimal covering of the adjoin.  Take the entropy to be the supremum over all the $\alpha$.

答:
There are certainly many definitions of topological entropy that in many situations are equivalent (even forgetting noncompact sets, higher rank actions or even some classes of discontinuous maps).

What happens is that for a continuous map on a compact metrizable space $(X,d)$ the original notion by Adler, Konheim, and McAndrew is equivalent to a notion introduced independently by Bowen and by Dinaburg that answers to your question.

Namely, letting
$$
d_n(x,y)=\max \bigl\{d(f^k (x), f^k(y)) : 0 \le k \le n-1 \bigr\},
$$
the topological entropy of a continuous map $T\colon X\to X$ on a compact metric space is equal to
$$\tag1
h(f)=\lim_{\varepsilon \to 0} \limsup_{n \to \infty} \frac 1 n \log N(n,\varepsilon),
$$
where $N(n,\varepsilon)$ is the largest number of points $p_1,\ldots,p_m \in X$ such that
$$\tag2
d_n(p_i,p_j) \ge \varepsilon \quad \text{for} \ i \ne j.
$$

One usually interprets $(1)$ as the exponential growth rate (that is, the limit on $n$) and $(2)$ as counting the number of distinguishable orbits at least at a distance $\varepsilon$ (as if your vision had only precision $\varepsilon$).


问:
Let $A,B$ be elements of $M_2(\mathbb{R})$. Give an example to show that $A+B$ can be invertible if $A,B$ are both non-invertible
The goal for this problem is to show that even if two matrices $A$ and $B$ are non-invertible, $A+B$ can be invertible. I tried to show this using a proof, but I ended up actually proving that this isn't true. I thought that it might just not be true, but the question before it was to prove the contrapositive, if $A,B$ are both invertible, show, by example, that $A+B$ can be non-invertible. I was able to prove that one, but using the same method I hit the end of the road proving this problem. Here is a basic outline of my steps for the problem. Please show me where I went wrong.

1) $$A = \begin{bmatrix}
e & f\\
g & h
\end{bmatrix}$$ and
  $$B = \begin{bmatrix}
i & j\\
k & l
\end{bmatrix}$$

2) $A+B=C$, so $$C = \begin{bmatrix}
e+i & f+j\\
g+k & h+l
\end{bmatrix}$$

After this, I say that the determinant of $C$ must not be 0 if it is to be invertible. If $A,B$ are non-invertible, then their determinants must be 0. Therefore,

det($A$): $eh-gf = 0$. Let $e=g=h=1$

  $(1)(1) - (1)f = 0$, so $f$ must also be 1.

Therefore, $$A = \begin{bmatrix}
1 & 1\\
1 & 1
\end{bmatrix}$$

I then did the same thing for $B$.

det($B$): $il - kj = 0$. If $l=1$, $i=2$, and $j=1$, then $k$ must be 2 for $B$ to be non-invertible. However, now that
  $$B = \begin{bmatrix}
2 & 1\\
2 & 1
\end{bmatrix}$$

$A+B=C$, where $$C = \begin{bmatrix}
3 & 2\\
3 & 2
\end{bmatrix}$$
  Which is non-invertible as well.

Where did I go wrong?

答:
You tried to construct an example. The example you built didn't work. But there are plenty of others that will. For $A$, top left $1$, everybody else $0$. For $B$, bottom right $1$, all else $0$. </p>

<p>We can start with $A$ all $1$'s like you did. Then for $B$ many things will work, like $1$ down the main diagonal, and $-1$'s at the other corners. Other choices will be fine too, but if we choose very small integers, we may get unlucky.  </p>

Body:
You tried to construct an example. The example you built didn't work. But there are plenty of others that will. For $A$, top left $1$, everybody else $0$. For $B$, bottom right $1$, all else $0$.

We can start with $A$ all $1$'s like you did. Then for $B$ many things will work, like $1$ down the main diagonal, and $-1$'s at the other corners. Other choices will be fine too, but if we choose very small integers, we may get unlucky.


问:
$L^2$ = $L$ equality implications for a formal language
Let's say I have a formal language $L$. I wanted to know which of these three equalities can be derived from the the given fact that $L^2$ = $L$:

$L^+  = L$

$L^* = L$

$\varepsilon \in L$

where $L^+$ and $L^*$ are the Kleene plus and Kleene star. What exactly is correct here? Does the fact that $\varepsilon \in L$ mean that for every $L$ it is equal to $L^2$?

答:
As noted, the empty language $\varnothing$ satisfies $\varnothing^2 = \varnothing$ and $\varnothing = \varnothing^+$, but $\varnothing^* = \{ \varepsilon \} \ne \varnothing$.

Next, if the language $L$ is non-empty, it has to contain $\varepsilon$: Say one of the shortest strings in $L$ is $\alpha \ne \varepsilon$, then there are no strings shorter than $\alpha^2$ in $L^2$, a contradiction.

Furthermore, if a non-empty $L = L^2$, then also $L = L \cdot L = L \cdot L^2 = L^3$, and in general $L = L^n$ for $n \ge 1$. Thus:

$\begin{align*}
  L^*
    &= \{ \varepsilon \} \cup L \cup L^2 \cup \ldots \\
    &= \{ \varepsilon \} \cup L \cup L \cup \ldots \\
    &= \{ \varepsilon \} \cup L \\
    &= L \qquad \text{(since \(\varepsilon \in L\))}
\end{align*}$

The same way:

$\begin{align*}
  L^+
    &= L \cup L^2 \cup L^3 \cup \ldots \\
    &= L \cup L \cup L \ldots \\
    &= L
\end{align*}$

Note that for example if $L = \{ \varepsilon, a \}$ then $L^2 = \{ \varepsilon, a, a^2 \} \ne L^2$. Tracing the above discussion, you see that any language sastisfying $L^2 = L$ must be $\varnothing$, $\{ \varepsilon \}$, or be infinite.


问:
Why is this not a stopping time?
I am going through Shreve's Stochastic Calculus book and he has a random time $\rho$ defined as follows:
$$\rho(HH) = 0, \quad \rho (HT) = 0, \quad \rho(TH) = 1, \quad \rho(TT) = 2$$

which he claims is not a stopping time. I was under the impression that this is, as $\rho(HX) = 0$, regardless of time outcome of $X$. Would anyone be able to clear up where I have misunderstood? Thanks.

答: It is not a stopping time because $\{\omega : \rho(\omega) \le 1 \} = \{HH,HT,TH\}$ is not in $\mathcal F_1 = \{\emptyset,\Omega,\{HH,HT\},\{TH,TT\}\}.$

Title: Why is this not a stopping time?


问:
The graph of $x^{n}+y^{n}=r^{n}$ for sufficiently large $n$
The graph of the function $x^{n}+y^{n}=r^{n}$ for certain large values of $n$ looks suspiciously like a square.

See this page from wolframalpha. Have any results been proven regarding this observation? What do we call this figure anyway?

答:
You have just rediscovered the max-norm.

More precisely, you have noted that as $p$ becomes large, the unit circle in the $l_p$ norm looks similar and similar to the one of the $l_\infty$ norm.


答: Let $A=\left(\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right)$ and  $B=\left(\begin{smallmatrix}0&0\\0&1\end{smallmatrix}\right)$.  Each is non-invertible (with determinant $0$), but their sum is the identity and is invertible.</p>

Title: Let $A,B$ be elements of $M_2(\mathbb{R})$. Give an example to show that $A+B$ can be invertible if $A,B$ are both non-invertible

Body:
The goal for this problem is to show that even if two matrices $A$ and $B$ are non-invertible, $A+B$ can be invertible. I tried to show this using a proof, but I ended up actually proving that this isn't true. I thought that it might just not be true, but the question before it was to prove the contrapositive, if $A,B$ are both invertible, show, by example, that $A+B$ can be non-invertible. I was able to prove that one, but using the same method I hit the end of the road proving this problem. Here is a basic outline of my steps for the problem. Please show me where I went wrong.

1) $$A = \begin{bmatrix}
e & f\\
g & h
\end{bmatrix}$$ and
  $$B = \begin{bmatrix}
i & j\\
k & l
\end{bmatrix}$$

2) $A+B=C$, so $$C = \begin{bmatrix}
e+i & f+j\\
g+k & h+l
\end{bmatrix}$$

After this, I say that the determinant of $C$ must not be 0 if it is to be invertible. If $A,B$ are non-invertible, then their determinants must be 0. Therefore,

det($A$): $eh-gf = 0$. Let $e=g=h=1$

  $(1)(1) - (1)f = 0$, so $f$ must also be 1.

Therefore, $$A = \begin{bmatrix}
1 & 1\\
1 & 1
\end{bmatrix}$$

I then did the same thing for $B$.

det($B$): $il - kj = 0$. If $l=1$, $i=2$, and $j=1$, then $k$ must be 2 for $B$ to be non-invertible. However, now that
  $$B = \begin{bmatrix}
2 & 1\\
2 & 1
\end{bmatrix}$$

$A+B=C$, where $$C = \begin{bmatrix}
3 & 2\\
3 & 2
\end{bmatrix}$$
  Which is non-invertible as well.

Where did I go wrong?


问:
What are the unknown angles?

I found this problem on facebook. I have tried for hours to solve the problem but could not solve it. My approach is finding the first equation $x+y=30$. But can not find second equation. I am new here. Please help me.

答:
Apply the sine rule to the three triangles making up the large one to establish

$$\frac{\sin 2x}{\sin 30^\circ}\cdot \frac{\sin x}{\sin(30^\circ-x)}\cdot \frac{\sin(90^\circ-2x)}{\sin30^\circ}=1
$$

which has the solution $x= 12^\circ$, and in turn $y=18^\circ$.

Title: What are the unknown angles?


问:
How to define a sequence $(a_n)_{n\geq 0}$ in $\mathbb{R}$ such that each convergent sub-sequence has limit 1, but $(a_n)_{n\geq 0}$ doesn't converge?
Part of a weekly submission assignment is about:

How to define a sequence $(a_n)_{n\geq 0}$ in $\mathbb{R}$ such that each convergent sub-sequence of $(a_n)_{n\geq 0}$ has limit 1, but $(a_n)_{n\geq 0}$ does not converge?

Given that $(V, d_v)$ is a metric space, $(a_n)_{n\geq 0}$ is a sequence in $V$ with $ a \in V$

However, me and my fellow students have been trying to figure it out for some time now, but we can't...


问:
Doubts regarding Dedekind construction of R starting from Q (Rudin PMA)
I followed the 9 steps  given by Rudin but in the end I didn't fully understand. My main questions are the following (in my understanding 2. is the follow up of 1.):

since the members of R have been defined as cuts of Q (thus containing rational numbers), how does this definition of R let appear the irrational numbers in R in the end?

In step 9 (the end indeed!) Rudin writes:

It is this identification of Q with Q' which allows us to regard Q as a subfield of R

What's the link between Q->Q'->R? Is Q' actually R? I understand that Q' contains cuts and it is isomorphic to Q. R also contains cuts by definition but how is it that Q is a subfield of R?

I hope my doubts are clear and I thank you in advance!

p.s.

Q' is what Rudin writes for Q* but in the post the star symbol gave me trouble forcing italics.

答:
Let $C=\{q\in\Bbb Q:q<0\}\cup\{q\in\Bbb Q:q^2<2\}$; then the cut $C$ is the irrational number $\sqrt2$: it appears in the right place in the linear order on $\Bbb R$, and its square, using the definition of multiplication in $\Bbb R$, is $\{q\in\Bbb Q:q<2\}$, i.e., the member of $\Bbb Q'$ corresponding to $2\in\Bbb Q$. The other irrational numbers also appear as cuts: there is one for each cut $C$ with the property that $\Bbb Q\setminus C$ does not have a smallest element.

For your second question, the assertion is not that $\Bbb Q$ is literally a subfield of $\Bbb R$; after all, it is not even a subset of $\Bbb R$, so it can hardly be a subfield. However, $\Bbb Q'$ is a subfield of $\Bbb R$, and $\Bbb Q$ is isomorphic to $\Bbb Q'$, so from a purely algebraic and order-theoretic point of view $\Bbb Q'$ is indistinguishable from $\Bbb Q$. Thus, for all practical purposes we can think of $\Bbb Q'$ as actually being $\Bbb Q$, as if $\Bbb Q$ itself were a subset (and subfield) of $\Bbb R$, even though in this construction of $\Bbb R$ that is not literally the case.


答:
Maybe this sequence can help you.
\begin{equation}
a_n=\left\{
\begin{array}{ccc}
1 & \text{if} & n \ \text{is even}\\
n & \text{if} & n \ \text{is odd}
\end{array}
\right.
\end{equation}
Observe that $a_n$ doesn't converges because is an unbounded sequence. Now, if $(a_{n_k})_{k\in\mathbb{N}}$ is a convergent subsequence to $x_0\in\mathbb{R}$ then we have the next cases

If $x_0<1$ take $\varepsilon>0$ such that $(x_0-\varepsilon,x_0+\varepsilon)\subseteq (-\infty,1)$. All the values of the sequence are in $[1,\infty)$. Therefore is impossible that the subsequence converges to $x_0$.

If $x_0>1$ take $\varepsilon>0$ such that $(x_0-\varepsilon, x_0+\varepsilon)\subseteq (1,\infty)$. Again, is impossible that the subsequence converges to $x_0$ because we can found an infinitely many terms in the subsequence that are greather than $x_0+\varepsilon$ or less than $x_0-\varepsilon$.

This proves that if $(a_{n_k})$ is a convergent subsequence, the only posibility is that $a_{n_{k}}$ converges to $x=1$.

Title: How to define a sequence $(a_n)_{n\geq 0}$ in $\mathbb{R}$ such that each convergent sub-sequence has limit 1, but $(a_n)_{n\geq 0}$ doesn't converge?

Body:
Part of a weekly submission assignment is about:

How to define a sequence $(a_n)_{n\geq 0}$ in $\mathbb{R}$ such that each convergent sub-sequence of $(a_n)_{n\geq 0}$ has limit 1, but $(a_n)_{n\geq 0}$ does not converge?

Given that $(V, d_v)$ is a metric space, $(a_n)_{n\geq 0}$ is a sequence in $V$ with $ a \in V$

However, me and my fellow students have been trying to figure it out for some time now, but we can't...


问:
Comparing $2^n$ to $n!$
Comparing $\;2^n\;$ and $\;n!\;$

I need to rewrite one or the other so they can be more easily compared (i.e. have similar form). Because of the factorial, I'm a little lost as to how to compare the two functions. Normally, I would take the logarithm when trying to re-express a power like $2^n$, but I don't think this will work, since I don't know how to take the logarithm of a factorial, if at all.

Might the series expansions of either functions be the right approach? If not could, someone point me in the right direction?

答:
Method 1: You can take a graphical approach to this problem:

It can be seen that the graphs meet at (0, 1), $2^x$ is greater until they intersect when $x \approx 3.459$, and then the factorial becomes much greater. Alternatively, plot $x! - 2^x$ to see a demonstration of the difference.

Method 2: use Stirling's approximation,

$$x! \sim \left(\frac{x}{e}\right)^x\sqrt{2\pi{x}}\left(1 + \frac{1}{12x}\right)$$

Now you can take logarithms of both expressions. Comparing the graphs of $y = x$ and the base 2 logarithm of the above expression, it can be seen that the latter quickly rises above the former.

Title: Comparing $2^n$ to $n!$


问:
Proving validity when replacing every occurrence of a variable
I have a problem finding a correct way to solve the next problem:
Let $φ$ be an argument with $n$-variables, $φ'$ is obtained from $φ$ by replacing every occurrence of a variable Vi by $¬$Vi, there is a need to prove: $φ$ is valid iff $φ'$ is valid.

I tried setting $φ$ as valid and using the rules ,mods($φ$)=World and mods($¬φ$)={} but I do not understand how will $φ'$ still be valid when we might change the premises themself by altering Vi to ¬Vi.

答:
Note that the formula $¬\varphi$ is not the same as the formula $\varphi'$ obtained from $\varphi$ by replacing each variable by its negation. E.g., we have

$$¬(x \wedge y) \sim ¬x \vee ¬y.$$

Try to show that whenever $\varphi$ is satisfied under some valuation $v$, then $\varphi'$ is satisfied under the valuation $v'$ mapping each variable to true if $v$ maps it to false, and mapping to false if $v$ maps it to true. Then show that every valuation is of the form $v'$ for some $v$.


问:
Why doesn't "Proof by contradiction" lead to "Principle of Explosion"?
I once asked here a question about how I can get from a false premise to a true premise, and people noted that it is called "Principle of Explosion" - meaning that from a false premise you can get to whatever you want. Link is can be found here: Using deductive reasoning to prove a true statement with false premises?.

But, now came to my mind the proof that $\sqrt{2}$ isn't rational, which is a proof by contradiction (the famous one) and I thought to myself, assuming $\sqrt{2}$ is rational, is as we now know a false premise as it is not rational, meaning that we have again the principle of explosion! (Or in Latin "Ex falso quodlibet").

What is going on here? Why can we assume $\sqrt{2}$ is rational and say that we come to a contradiction when in the first place we assumed a false statement!

For me, it is like saying "Assuming $1=2$, now I can prove that $1=3$" (Hey that is a contradiction no? we said $1=2$ ) thus $1 \ne 3$... but as we know $1=2$ isn't true also!.

Maybe the example of $\sqrt{2}$ is pretty obvious that there is a dichotomy between those relations, either it is rational or not. But there are proofs which do no deal with a dichotomic "property" of something... or proof by contradiction is always for dichotomic properties? (like, being even or odd, rational and irrational, etc...)

Thank you!

答:
A proof by contradiction has an important step that you omitted: you start from a false premise, reason to a contradiction, and then discard the false premise as incorrect.  It is in fact exactly the avoidance of "proof by explosion."

In other words, we assume various proven facts, call them $A,B,C...$, as well as facing a claim that $P \implies Q$.  If you start from an assumption of $\neg (P \implies Q)$, and use it to derive $\neg C$, you have shown that $\neg (P \implies Q) \implies \neg C$, and the contrapositive of that is $C \implies (P \implies Q)$. Since you know $C$ is true, you can conclude that $P \implies Q$.


问:
How to multiply $2 \sin(x) \cdot \sin^{2} (x)$
I am just confused on the way you would multiply this out, if you could help that would be fantastic. The pre calc method for doing these is the solution I am looking for</p>

答: First focus on your sin(x). When multiplying, $sin^n(x) * sin^m(x)$ = $sin^{n+m}(x)$. Therefore, $sin^1(x) * sin^2(x)$ = $sin^3(x)$. Then multiply by 2 to get $2sin^3(x)$.


问:
Is a Knapsack solution considered infeasible if one of its values is a fraction?
In the realm of Linear Programming - more specifically, looking into Branch and Bound Algorithms - if finding a solution to the Knapcack problem involes using a fraction of an item to fit inside the sack, does that mean its solution is infeasible? E.g. it cannot be used.

For example, I am working on the following question:
$$
maximize \quad 8x_1\ +\ 7x_2\ +\ 10x_4\ +\ 3x_5\\
subject\ to \quad 2x_1\ +\ 2x_2\ +\ 4x_3\ +\ 5x_4\ +\ 3x_5\ \leq8\\ x_1,x_2,x_3,x_4,x_5\ =\ 0\ or\ 1
$$

In terms of branch-and-bounding the problem, I have found a maximised solution for the above question, and then a solution for each of its two children.

The first solution has the most maximized value of 23, but $x_4$=0.8, so I had to branch.

For its children's solutions: one has a value of 21.5 (less than the original and it also involves a fraction); the other has a value of 22, but none of its variables are fractions. Does this mean the branch with the value of 22 is the solution to the problem?

Note: the reason I ask, even though I know that the Knapsack problem comes under Integer Programming (i.e. values must be integers) is because my course has not been explicit enough with its teachings, I feel. I just want to be sure.

答: The knapsack problem states that there are discrete items numeberd by $i$, with weight $w_i$ and value $v_i$, and a knapsack of capacity $C$. What you want is to find the maximum of $\sum_i x_i v_i$ subject to $\sum_i x_i w_i \le C$ and $x_i \in \{ 0, 1\}$. It is an integer linear problem. So yes, fractional $x_i$ aren't solutions.


问:
Distribution of number of trials before $k$ distinct successes
I am familiar with the Negative Binomial distribution $NB(p, k)$, which gives the number of failures before $k$ successes occur in a Bernoulli process with parameter $p$. I am wondering, however, if there the distribution of the number of failures before getting $k$ distinct successes is a well-known distribution.

More precisely, suppose we have some set of $N$ elements; without loss of generality, we may assume this set is $[N] = \{1, \cdots, N\}$. We have a subset $S \subseteq [N]$ which contains the successes, while every element in $[N] \setminus S$ is a failure. Now in our process, we select a random element uniformly at random from $N$, and keep on doing this until we have attained some $k \leq |S|$ successes. Let $X$ be the total number of trials performed before we stop the process. What is the distribution of $X$, in terms of $k$, N, and p = $|S| / N$? If we can't concisely describe this distribution, can we at least come up with some (relatively tight) tail bounds? And how valid is the approximation $X \approx NB(p, k)$?

I know this problem can simplify to the coupon collector's problem in the case that $S = [N]$ and $k = N$, but I'm more interested in the case where $0 < p < 1$ and $k \ll |S| = pN$.

答:
I will try and find $P(X = m)$. If $m < k$, then this probability is clearly $0$. Otherwise, the probability is the sum over all distinct $k-$tuples from $S$ of the probability of that exact tuple being the "winning" numbers. Since this is equal for all the different tuples, the probability is $\binom{s}{k}$ times the probability of $\{1, \cdots, k\}$ being the winning numbers and none of $\{k+1, \cdots, s\}$ showing up, where $s = |S|$. Then because the last one has to be a number in $\{1, \cdots, k\}$ that hasn't been seen before, this is equal to

$$\binom{s}{k}\frac{k}{n}\cdot P(\text{all of $\{1, 2, ..., k-1\}$ getting chosen in $m-1$ trials and none of $\{ k, k+1, ..., s \}$ being chosen})$$

This inner probability is equal to
$$\frac{|\text{set of $(m-1)$-tuples such that all of $\{1, 2, ..., k-1\}$ is there}|}{n^{m-1}}$$

but instead of having $n$ choices for each element in the tuple, there are $n-s+k-1$ choices, which accounts for $\{k, \cdots, s\}$ not being allowed.

Then, after taking the complement of this and using inclusion-exclusion, this is $$\frac{\sum_{i=0}^{k-1} (-1)^{i}\binom{k-1}{i}(n-s+k-1-i)^{m-1}}{n^{m-1}} = \frac{\sum_{i=0}^{k-1} (-1)^{k-1-i}\binom{k-1}{i}(n-s+i)^{m-1}}{n^{m-1}}$$

So the final probability is

$$P(X=m)=\frac{k\binom{s}{k}}{n^m} \sum_{i=0}^{k-1} (-1)^{k+i+1}\binom{k-1}{i}(n-s+i)^{m-1}$$

I'm not sure how valid the negative binomial distribution is for approximating this though.


问:
Representation theory of truncated polynomials in two variables
Let $k$ be an algebraically closed field. I was wondering what is known for the representation theory of the algebra $k[x,y]/(x^2,y^2)$. I think it is clear that there are infinitely many non-isomorphic indecomposable finite dimensional representations, but is there a complete classification?

答:
Yes, there is a complete classification.

If $k$ has characteristic $2$, then $k[x,y]/(x^2,y^2)$ is isomorphic to the group algebra $k[C_2\times C_2]$, and one place that has a fairly detailed derivation of the classification is Section 4.3 of

Benson, D. J., Representations and cohomology. I: Basic representation theory of finite groups and associative algebras, Cambridge Studies in Advanced Mathematics, 30. Cambridge University Press. (1991).

The classification there also works for $k[x,y]/(x^2,y^2)$ for a field $k$ of arbitrary characteristic.


问:
Infinite monkey theorem and numbers
I had a discussion with a friend about the monkey infinite theorem, the theorem says that a monkey typing randomly on a keyboard will almost surely produce any given books (here let's say the bible).

I believe this experience can be reduced to choosing a random sequence among real numbers:, strictly speaking the monkey could type a sequence looking like 1/3. (Characters replaced by numbers), my friend told that it is not possible, but for me this is what pure randomness is.

I also believe that not all infinite sequences contain the bible, so it doesn't make any sense to say that the monkey will almost surely write it.

My friend is sure that the monkey will ALWAYS write everything possible, while for me this implies to always choose a "disjunctive sequence".

I understand that from a probabilistic point of view, the chance that the monkey doesn't write a particular sequence drop to 0 as the number of character increase. But probability is also the ratio of specific events on all the events, since all the sequences are not disjunctive sequence, this probability cannot be equal to 1.

Who is right here?

答:
The 2 examples you made (the bible and the sequence of the number $\frac{1}{3}$) are similar, but deeply different. In fact, the bible is a finite sequence of characters, whereas the decimal digits of $\frac{1}{3}$ are infinitely many.

So, the probability that the monkey writes exactly the bible during its typing is 1, while the probability that the monkey will write exactly all the digits of $\frac{1}{3}$ is 0.

Digits of $\frac{1}{3}$

In order to write exactly the digits of $\frac{1}{3}$, from a certain digits the monkey must push the key 0 and then the key 3 infinitely many times... up to infinity! This has probability (not rigorous) $\frac{1}{10} * \frac{1}{10} * \dots * \frac{1}{10} = (\frac{1}{10})^\infty = 0$.

Sequence of letters of the bible

In order to write exactly the sequence of letters of the bible, instead, the monkey should guess a finite number of characters.

The first thing we can notice is that the probability that this happens is larger than 0. In fact, assuming that the bible is 1.000.000 letters long, the probability that the bible starts from the first letter of the first page that the monkey writes is something like $p = \frac{1}{26} * \frac{1}{26} * \dots * \frac{1}{26} = (\frac{1}{26})^{1.000.000}$, which is a very small number, but it is larger than 0.

Then, think that the bible could start from the second letter of the first page, so we have another $p$ probability that the monkey will write exactly the bible. And consider that the monkeys has infinitely many letters from which it can start to write the bible, so it has infinitely many chances to succeed in the challenge, each of them having $p$ probability to work.

Note that these infinitely many events (each corresponding to which letter the monkey is starting to write the bible from) are not independent, so the total probability will not be $\infty * p = \infty$, but they are enough to guarantee that the monkey will achieve it anyway, with probability 1.

The rigorous proof that this probability is 1 is not even difficult, but I will not write it here. You can find it on Wikipedia.

The idea of the proof is to estimate the probability that the monkey will not write the bible and eventually you can proof that that probability is 0, meaning that it is almost impossible (but still not impossible) that the monkey doesn't write the bible.

Title: Infinite monkey theorem and numbers


问:
Counting pairs of animals

Old MacDonald has $5$ chickens, $4$ donkeys, and $7$ emus. How many ways can he pair up the animals so that every pair consists of animals of different species? (The order of the animals within each pair does not matter, and the order among the pairs does not matter. Assume that all animals are distinguishable.)

The answer I have currently is $83$ since there are $5*4=20$ possible pairs for (Chicken, Donkey), $5*7=35$ possible pairs for (Chicken, Emu), $4*7=28$ possible pairs for (Donkey, Emu). Then I added them up: $20+35+28=83$. Is this correct?

答:
The answers given so far are wrong: they count the number of possible pairs but not the number of ways to pair all the animals up at once. (We can think of these as matchings.)

Notice that the 7 emus must be paired with 4 chickens and 3 donkeys, leaving 1 chicken and 1 donkey to form a pair. Any other option (e.g. pairing the emus with 5 chickens and 2 donkeys) leaves unequal numbers of chickens and donkeys, who then can't be paired up.

Each valid matching is specified in exactly one way by a sequence of choices as follows:

Divide the emus into a set of 3 and a set of 4. There are $\binom73=\binom74=35$ ways to do this.

Pair the emus from the set of 3 with donkeys one at a time: $4\times3\times2=24$ ways.

Pair the emus from the set of 4 with chickens: $5\times4\times3\times2=120$ ways.

Pair the remaining donkey and chicken together (1 way).

So the total number of ways to pair the animals up is $35\times24\times120=100800$.

Title: Counting pairs of animals


问:
How can I calculate the average outcome / expected value from a list of probabilities?
I have a question regarding calculating the expected outcome from the list or probabilities.

Suppose each can play a max of 6 games. If you win the game you move to the next round. Below I have a sample of a team the probability of reaching the certain round, how can I calculate to get the average expected number of games won?

```<code>Number of Games    1      2      3        4       5       6
% Chance to win  0.97   0.92    0.83    0.679   0.545   0.384
</code>```

答:
As the answer above mentions you could manually calculate each of the probabilities for the total number of wins.

However you can also be lazy. The "tails sum" formula states that if a random variable, $X$, is Natural valued, that is the non-negative integers, then the expected value of $X$ is $\sum_{n\in\mathbb{N}} \mathbb{P}[X \geq n]$

So in your case you could have just done $0.97+0.92+0.83+0.679+0.545+0.384 = 4.328 $


问:
$a=\sum_{i=0}^n a_i 10^i$. Prove, that $3|a$ only when $\sum_{i=0}^n a_i \equiv_3 0$
Let$$a=a_n 10^n + a_{n-1} 10^{n-1} + ... + a_1 10 + a_0 $$
And $$ a_i \in \{0,...,9\} \text{ for } i=0,1,...,n $$

Prove, that:
$$3|a \Leftrightarrow a_n+a_{n-1} + ... + a_1 + a_0 \equiv_3 0$$

Prove, that:
$$11|a \Leftrightarrow (a_1+a_3+...) - (a_0+a_2+...) \equiv_{11} 0$$

I don't know how to do either 1 or 2. A push in the right direction would be greatly appreciated. Thank you.

Body:
hint

$$10\equiv 1 \mod 3$$
$$10^k\equiv 1 \mod 3$$

$$10\equiv -1\mod 11$$
$$10^k \equiv (-1)^k \mod 11$$


问:
Composition of uniform function and continuous function
The following is given:
$$g: \mathbb{R}\to\mathbb{R}\colon\quad x \mapsto f\left(\frac{1}{1+x^2}\right),$$
with $f:\mathbb{R}^+\to\mathbb{R}$ continuous.

Is $g$ uniform continous?

I answered there yes and I thought I could prove it by dividing the problem by looking in an interval and by looking outside that interval where the interval is big enough so $g(x)$ goes to $f(0)$. Then using the triangular inequality I proved the problem. However, the next question is: is $h\colon\mathbb{R}\to\mathbb{R}: x \mapsto g(x)^2$ also uniform continuous?

My gut feeling says no but also yes because I can use the same method, but I can not proof this or give a counter example.

答:
We need $f$ defined and continuous at $x = 0$ for $g$ to be uniformly continuous.  For example, if $f:\mathbb{R}^+ = (0,\infty) \to \mathbb{R}$ with $f(x) = 1/x$, then $f$ is continuous on its domain, but $g(x) = 1 + x^2$ is not uniformly continuous on $\mathbb{R}$.

The same applies to $h = g^2$.

Otherwise, if $f$ is continuous on $[0,\infty)$ then both $g$ and $h=g^2$ are uniformly continuous on $\mathbb{R}$. The proofs are similar and one for $h$ follows.

Since $\lim_{x \to \pm \infty}h(x) = f^2(0)$, there exists $X>0$ such that $|h(x) - f^2(0)| < \epsilon/3$ for all $x \in [X, \infty]$ and $x \in (-\infty,-X]$.

Since $h$ is continuous  on the compact set $[-X,X|$, it is uniformly continuous there and there exists $\delta > 0$ such that $|h(x) - h(y)| < \epsilon/3$ for $x,y \in [-X,X]$ when $|x-y| < \delta$.

Now we can show that for all $x, y \in \mathbb{R}$ such that $x - y| < \delta$, we have $|h(x) - h(y)| < \epsilon$.   There are a few cases to consider. For example if $x \in [-X,X]$, $y \in [X,\infty)$ and $|x-y| < \delta$, then

$$|h(x) - h(y)| \leqslant |h(x) - h(X)| + |h(X) - f^2(0)| + |f^2(0) - h(y)| < \frac{\epsilon}{3} + \frac{\epsilon}{3}+ \frac{\epsilon}{3} = \epsilon$$


问:
What is the sum of number of digits of the numbers $2^{2001}$ and $5^{2001}$
What is the sum of number of digits of the numbers $2^{2001}$ and $5^{2001}$? (Singapore 1970)

I attempted to solve this question by working out what each digit must be, and maybe find some pattern, but I couldn't find any, apart from the fact that $2^{2001}\mod{10}\equiv 4$ and $2^{2001}\pmod{10}\equiv 5$. Could you please explain to me how to solve this question? This question is multiple choice with options $1999, 2003, 4002, 6003, 2002$

答:
It looks like you don't need logarithms or any calculator to solve this problem. Let's start.

First, observe that the following inequalities hold:

$$10^m<\underbrace {2^{2001}}_{m+1 ~ \text{digits}}<10^{m+1}$$

$$10^n<\underbrace{5^{2001}}_{n+1 ~ \text{digits}}<10^{n+1}$$

You get,

$$10^{m+n}<10^{2001}<10^{m+n+2}$$

$$2001=m+n+1$$

$$m+n=2000$$

Finally,  the sum of digits of $2^{2001}$ and $5^{2001}$ is equal :

$$\begin{align}\color {gold}{\boxed {\color{black}{m+1+n+1=m+n+2\\ \qquad \qquad \qquad\thinspace=2000+2 \\\qquad \qquad \qquad \thinspace=2002.}}}\end{align}$$


问:
Can we construct groups of several discrete "loops"?
I know that we have cyclic groups.

Maybe the groups which in the most easy-to-access way can explain the concept of a generator.

But do there also exist groups of two or more loops?

For example $\cases{{g_1}^{n_1} = e\\{g_2}^{n_2} = e}$

With $$\{g_k,{g_k}^{1},\cdots,{g_k}^{n_k}\} \text{  distinct}$$

and $${g_1}^{m}{g_2}^n={g_2}^n, \forall n\neq kn_2$$

In other words the only way to enter a loop is through $e$ and trying to loop in the other direction if we already are in another loop will do nothing.

Kind of like a state machine that starts counting one direction and then cannot count in the other direction until we have looped a whole circle around to $e$.

Is this possible or will this violate some axiom for groups?

Is there some other algebraical concept which it would be possible for?

答:
Let's examine the structure you propose. This is how I interpret it: The set is $M=\{e, g_1, g_1^2, \ldots, g_1^{n_1-1}, g_2, g_2^2, \ldots, g_2^{n_2-1}\}$. Composition is non-commutative and defined as:
$$
ex = xe = x
$$
$$
g_i^a g_i^b = g_i^{a+b\bmod n_i}, \quad i\in\{1,2\}
$$
$$
g_i^a g_j^b = g_j^b, \quad i\ne j\in \{1,2\}
$$
The last one makes our object lose a lot of algebraic structure. It is not associative since $g_1(g_2g_1)=g_1^2$ but $(g_1g_2)g_1=g_1$. It is not cancellative since $eg_2 = g_1g_2$ but $e\ne g_1$. It doesn't have right division since $x g_1 = g_2$ has no solutions in $x$.

What it does have is an identity element, and all elements have a unique two-sided inverse. It also has unique left division for whatever that's worth: E.g. $g_1^a x = g_2^b$ is satisfied exactly by $x = g_2^b$.

I don't think there is a widely used term for this sort of structure, other than just discribing it when you need to. This other answer cites a paper that simply calls it a magma with inverses, but note that this would perhaps more often be called a unital magma with inverses. If I needed an ad hoc name for your specific construction, I might call it something like "the disjoint product of $\langle g_1 \rangle$ and $\langle g_2 \rangle$" (at least if that name isn't taken for something else).

[EDIT/erratum: I erroneously called the structure a loop at first. That's wrong, because a loop is a quasigroup with identity, which means it satisfies the "Latin square property"; in particular, it is cancellative].

But indeed, you should consider Steven Stednicki's comment ("What are you trying to 'do' with the object"?). This algebraic structure might not be what you're actually after, even though it matches your formal discription.


问:
Prove that there exists N ∈ (N) (naturals) such that $a_n$ > 0 for all n ≥ N.
Let $(a_n)$ be a convergent sequence. Suppose that $\lim_{n\to\infty} a_n > 0$. Use the definition of a limit to prove that there exists N ∈ (N) (naturals) such that $a_n$ > 0 for all n ≥ N.

Use definition of a limit ($|a_n−L|<ϵ$).

Please check my proof!

My Proof:

Let $\epsilon$ be an arbitrary positive number. Let L = $\lim_{n\to\infty} a_n > 0$. Choose N ∈ (N) (naturals) such that n ≥ N implies that $|a_N−L|<\frac{\epsilon}{2}$ by definition of a limit. Assume n ≥ N. Therefore, $|2a_N−2L|=|2(a_N−L)|=|2||a_N-L|<\frac{\epsilon}{2}*2=\epsilon$. Since L > 0, $a_n$ > 0.

答: I don't think your proof is quite right. Instead, try expanding the definition of a limit a little bit. The key will be to pick some $\varepsilon >0$ that is strictly less than the limit $L$, say $\varepsilon = \frac{L}{2}$ (we can do that because $L$ is positive). Now note that there exists $N\in \mathbb{N}$ such that for all $n\geq N$ we have  $$\lvert a_n - L\rvert < \varepsilon \iff - \varepsilon < a_n - L < \varepsilon.$$ Adding $L$ results in $$L-\varepsilon < a_n < L + \varepsilon.$$ Since we chose our $\varepsilon = \frac{L}{2}$, we have $$L-\varepsilon = L - \frac{L}{2} = \frac{L}{2} > 0,$$ because $L>0$. Hence $0< a_n$ for all $n\geq N$, as required.


问:
Can the real projective plane extended to a 3 space by R partition R4?
I know $\mathbb{R}P^2$ immersed in $\mathbb{R}^3$ self-intersects: Roman surface, Boy surface. Is there an embedding of $\mathbb{R}P^2\otimes\mathbb{R}$ in $\mathbb{R}^4$ that partitions it, analogous to a sphere partitioning $\mathbb{R}^3$ into an inside and an outside? I don’t see how you could do that with a non-orientable surface, but I don’t know what the embeddings look like.

答: You can’t do that with any sort of $2$-dimensional manifold, orientable or no. Magic words are: Alexander duality.


答:
I think you nearly have the idea - the only problem is that you never define $\varepsilon$ and then later assume things about it; in particular, you start with
$$|a_N - L | <\varepsilon /2$$
and then later conclude that
$$a_N > 0$$
from this - but this does not follow from what you have written, which implies that $\varepsilon$ could be chosen to be any positive number - and chosen before even inspecting the limit $L$ or the sequence $a_n$. Perhaps you're used to starting analysis proofs this way - as most proofs that a sequence converges do begin with letting $\varepsilon$ be an arbitrary positive number - but it's not appropriate here.

Rather, I would just eliminate $\varepsilon$ and replace its occurrences with, say, $L$. As an example:

Let $L=\lim_{n\rightarrow\infty}a_n$. Using the definition of a limit and the fact that $L>0$, we may choose $N\in\mathbb N$ such that if $n\geq N$ then $|a_n - L| < L$. This implies that $L-a_n < L$ and thus that $a_n>0$.

Where I've simplified some of the algebra as well - though the algebra is not the point. Note that here we are referencing the definition of a limit:

If $L=\lim_{n\rightarrow\infty}a_n$, then for every $\varepsilon>0$ there exists some $N\in\mathbb N$ such that if $n\geq N$ then $|L-a_n|<\varepsilon$.

We then supply some particular value of $\varepsilon$ to use - in this case, we use $L$. Essentially, if you want to prove that a limit exists, you need to write a proof that works for all $\varepsilon > 0$. If you already know that a limit exists, you can use this by choosing some particular $\varepsilon$ that works for your case - and choosing $\varepsilon = L$ gives the existence of some $N$ that works for your proof.

It's possible that your idea was to imagine, in your proof, that $\varepsilon$ will be really small (since we are free to choose it as whatever we want) - and the fact that $a_n$ stays arbitrarily close to $L$ after some time would imply that $a_n$ is positive. However, to capture this intuition formally, it is necessary to choose some particular radius around $L$ where all the values are positive.

Title: Prove that there exists N ∈ (N) (naturals) such that $a_n$ > 0 for all n ≥ N.

Body:
Let $(a_n)$ be a convergent sequence. Suppose that $\lim_{n\to\infty} a_n > 0$. Use the definition of a limit to prove that there exists N ∈ (N) (naturals) such that $a_n$ > 0 for all n ≥ N.

Use definition of a limit ($|a_n−L|<ϵ$).

Please check my proof!

My Proof:

Let $\epsilon$ be an arbitrary positive number. Let L = $\lim_{n\to\infty} a_n > 0$. Choose N ∈ (N) (naturals) such that n ≥ N implies that $|a_N−L|<\frac{\epsilon}{2}$ by definition of a limit. Assume n ≥ N. Therefore, $|2a_N−2L|=|2(a_N−L)|=|2||a_N-L|<\frac{\epsilon}{2}*2=\epsilon$. Since L > 0, $a_n$ > 0.


问:
Given that  and  are positive integers with $\mid$, how would one prove that $\gcd(, ) = $?
I'm having trouble showing that if  and  are positive integers with |, then (, ) = a.

I'm aware that the greatest common divisor of two numbers is the largest number that is a divisor of both numbers. Ex) $\gcd(4,2) = 2 $. But in this case $a=4$ and $b=2$ so the expression above would not work. I suppose what I'm asking is if variable $a$ is always the smallest of the two numbers given.

Note

I now realize that I should have given the problem some more thought before posting on here. I had been overlooking the | portion of the expression. Thank you for the help!

答:
Recall the definition of $\gcd$. Let $x,y \in \mathbb{Z}$. If $d = \gcd(x,y)$, then there exists integers $e,f$ such that $x = ed$ and $y = df$ where $d$ is the largest possible integer with that property. So turning to your question:

Let $a,b \in \mathbb{Z}$ such that $a|b$. Then there exists some $c \in \mathbb{Z}$, where $b = ac$. From the definition of $\gcd$ then, we have that $a = a \cdot 1$ and $b = ac$, so then $a = \gcd(a,b)$.

Notice, this is also due to the fact there cannot be any divisor of $a$ larger than $a$ itself. So we're showing that $a$ is indeed a divisor of both integers and in fact the greatest possible divisor of $a$ and hence the greatest possible divisor of both $a$ and $b$.


问:
You have triangle $\triangle ABC$ which is orthogonal $\angle C=90^o$ with circle with center O inscribed in it. Find angle $\angle MCN$
You have triangle $\triangle ABC$ which is orthogonal $\angle C=90^o$ with circle with center O inscribed in it. If $KL$ is the diameter, $KL\parallel AB$, $KM\perp AB$ and $LN\perp AB$ Find angle $\angle MCN$.

Measuring this angle it works out to be $45^o$, but I don't know how to work it out. I attempted to utilize $KLNM$ being a rectangle (something which is proved since $KM\perp AB$ and $LN\perp AB$ which means $KM\parallel LN$ and also $KL\parallel MN$ hence $KLMN \#$ and since $KM\perp MN$ then $KMNL$ is a rectangle), but this didn't work out for me. Could you please explain to me how to solve the question?

答: Produce line $MK$ to intersect $AC$ at $P$, and produce $NL$ to intersect $BC$ at $Q$. Triangles $MPC$ and $NQC$ are isosceles. Drop altitude $CH$ to base $AB$ to conclude the proof.


问:
Five digit password whose digits are all among $\{0,2,3,5,7\}$
A password of computer used five digits where they are from 0 and 9. What is the probability that the digits of the password  are all among $\{0,2,3,5,7\}$?

Edit:

I tried this way but not sure is it correct or not.

Number available 0 to 9 .

total = 10 numbers .

The code is of 5 numbers .

first number can be selected in 10c1 ways .

Since the numbers are not unique and repetition is allowed we can select the second number in 10c1 ways and so on .

So total possible ways in which this code can be made is (10c1)(10c1)(10c1)(10c1)(10c1).

Now the possibility of finding the code with only prime numbers and zero
.

So total numbers to select from 0,2,3,5,7 .

Total numbers 5 .

The first number can be selected in 5c1 ways again repetition allowed so second number can be selected in 5c1 ways and so on .

So numbers of ways in which the code with odd numbers is selected is (5c1)(5c1)(5c1)(5c1)(5c1) .

Probability = (number of required outcomes )/ (total possible outcomes)

= (5c1)(5c1)(5c1)(5c1)(5c1) / (10c1)(10c1)(10c1)(10c1)(10c1)

= 1/32

答:
Hint: Since you want each digit to be either $0$ or a prime number, then how many desired options are there for the first digit? What about the second? The third? Fourth? Fifth? Then how many passwords are of the desired form?

Now, how many total options are there for the first digit? What about the second? The third? Fourth? Fifth? Then how many possible passwords are there?

What then is the probability that a given password is of the desired form?

Edit: Perfect (except you say "odd numbers" when you mean "prime or zero digits", though it doesn't alter the numerical result)! Well done.


问:
Interval optimization with length inequalities
What is the largest constant $c$ such that for any integrable function $f:[0,1]\rightarrow\mathbb{R}_{\ge 0}$ with $\int_0^1 fdx = 1$, it is possible to divide $[0,1]$ into five intervals $I_1,\dots,I_5$ (from left to right) such that

$|I_j|+|I_{j+2}| \ge |I_{j+1}|$ for $j=1,2,3$ ($|\cdot|$ denotes the length of the interval).

$\int_{I_j}f dx \ge c$ for $j=1,\dots,5$?

Without the first condition, the answer is clearly $c = 1/5$. With the first condition, by starting with the $c = 1/5$ solution and modifying it to fit the condition, it looks like getting $c = 1/10$ is possible, although this does not seem to be the right answer.

答:
Some thoughts / partial answer: the function $f$ below sets a limit of $c \le 1/6$.

Define $Z = (0.1, 0.9)$.  I chose $Z$ to stand for Zero.

$f(x) = 5$ for $x \notin Z$.

$f(x) = 0$ for $x \in Z$.

So this function has two narrow parts ("end zones") with area $= 1/2$ each, one on each end.  Consider the four boundary points (besides $0,1$).

If two of them are in $Z$ then the min area $= \min_i \int_{I_i} f\ dx = 0$.

If none of them is in $Z$ then we cannot meet the first constraint unless all four of them are in the same end zone.  This limits the min area to $1/8$.

If one of them is in $Z$ then the best we can do is to have two in the one end zone and one in the other.  By having $3$ intervals in one of the end zones, this limits the min area to $1/6$.


问:
Is there alternative factoring of a quintic equation?
In a paper here
the author appears to be able to factor a Bring-Jerrard quintic making

$$P=2mn(m^2-n^2)(m^2+n^2)=2m^5n-2mn^5\\ \implies n^5-m^4n+\frac{P}{2m}=0 \rightarrow x^5+px+q=0$$
become
$$(x^3+bx^2+cx+d)(x^2+ex+f)=0$$
but I haven't been able to follow how he got there. If I could, I would have what I need to find the one or more valid values of $n$ in the equation:

$$n^5-m^4n+\frac{P}{2m}=0$$

given that I will know the values of $P$ and $m$.

Can anyone help me figure out how the 'factored' equation would look in terms of $p,q$?

答:
Yes the paper has an error and was not properly checked. There is an alternative approach here:

http:/?utm_medium=email&utm_source=researchgate&utm_campaign=re322&utm_term=re322_x&utm_content=re322_x_p2&cp=re322_x_p2&uid=cx4ZWZ6hcdiX0hKERkrM99zJiDWztOEoeI4v&ch=reg


问:
What's wrong with this Buffon's needle geometric "solution"?
Suppose that one end of a needle is red and the other is blue. Let's assume the blue end is the right one when the needle is dropped on the floor. Now the blue end is on one of the strips (at some height) and the needle can lie only across the line that is the left edge of that strip. This image shows possible positions for both ends and this image shows area of positions of the red end if the needle lies across the left line of the strip (darker area). If $l$ is the length of the needle and $t$ is the width of the strip, the probability in question is equal to:
$$
\frac{\frac{1}{2}l^2\pi}{2lt}=\frac{l\pi}{4t}
$$
(In the images the radius of the semicircles is $l$ and the length of "the figure" is $t$).

What's wrong with this approach?

答: This argument incorrectly assumes that the red end of the needle is uniformly distributed in the given region. If the angle of the needle is uniformly distributed, then the vertical distance between its endpoints is not. The red endpoint is more likely to fall within a small horizontal strip near the top or bottom of the region than in a strip of the same height near the middle.


问:
Arithmetic in Lambert-W number system
$\DeclareMathOperator{\sign}{sign}$
The principal branch of Lambert-W function is defined like so:
$$W(x) = f^{-1}(x)\\
  \text{where} f(x)=xe^x$$
So $W(e) = 1$, $W(2e^2)$ is 2, $W(3e^3)$ is 3 etc.

We define Lambert-W number system to represent a real number $X$:
$$X \rightarrow \{s=\sign(x),x=W(|x|)\}$$
For instance, $-2e^{2}$ is represented $\{-1,2\}$, and $1$ is represented as $\{1,W(1)\}$.

Additionally, we define double Lambert-W number system as this:
$$X \rightarrow \{s=\sign(x),x=W(W(|x|))\}$$
For instance, $(2e^{2})\cdot{e^{(2e^{2})}}$ is represented as $\{1,2\}$.

My question: How to add, subtract, multiply and divide numbers represented in these two systems?

答:
$\DeclareMathOperator{\sign}{sign}$
Let us consider $x \mapsto (\sign(x),W(|x|))$ and $y \mapsto (\sign(y),W(|y|))$.
This means that
$$ |x| = f(f^{-1}(|x|)) = f(W(|x|)) , $$
so
$$ x = \sign(x) |x| = \sign(x) f(W(|x|)) = \sign(x) W(|x|) e^{W(|x|)} . $$

At the same way,
$$ y = \sign(y) W(|y|) e^{W(|y|)} . $$

So, knowing the representation of $x$ and $y$ in the Lambert-W number system, we can reconstruct the value of $x$ and $y$.

Now assume that $\sign(x)=\sign(y)$.

$$ x+y = \sign(x) \left( W(|x|) e^{W(|x|)} + W(|y|) e^{W(|y|)} \right) , $$
which means that
$$ |x+y| = W(|x|) e^{W(|x|)} + W(|y|) e^{W(|y|)} . $$

The representation of $x+y$ in the Lambert-W number system will be then
$$ x+y \mapsto \left(\sign(x+y), W(|x+y|)\right) = \left(\sign(x+y), W\left( W(|x|) e^{W(|x|)} + W(|y|) e^{W(|y|)} \right) \right) . $$

We just proved that, if $(s_1,x_1)$ and $(s_2,x_2)$ are two numbers in the Lambert-W number system, if $s_1 = s_2$, we can write
$$ (s_1,x_1)+(s_2,x_2) = \left(s_1,W\left(x_1 e^{x_1} + x_2 e^{x_2} \right) \right) . $$

When $s_1 \neq s_2$, that is when $\sign(x) \neq \sign(y)$, you can do a similar calculation using the fact that the function $f$ is monotone, so if $x > y$, then $x e^x > y e^y$.

You can derive the other operations with similar arguments.

Title: Arithmetic in Lambert-W number system


问:
$a_{n+1}$ = $\frac{1}{2} (a_n + \frac{2}{a_n})$ for n ≥1
In this problem, we give an algorithm for computing $\sqrt{2}$. Let $a_1$ = 2, and define $a_{n+1}$ = $\frac{1}{2} (a_n + \frac{2}{a_n})$ for n ≥1.

(a) Prove that $a_n^2$ ≥ 2 for all n. (Use proof by induction.)

Here I have:

Let $a_{n+1} = \frac{1}{2}(a_{n}+\frac{2}{a_{n}})$ for all n ≥1.

Therefore, $a_{n+1-1} = \frac{1}{2}(a_{n-1}+\frac{2}{a_{n-1}}) = a_n.$

Therefore, $a_n^2 = (\frac{1}{2}(a_{n-1}+\frac{2}{a_{n-1}}))^2 = (\frac{1}{2}a_{n-1}+\frac{1}{a_{n-1}})^2 = \frac{1}{4}a_{n-1}^2 + \frac{a_{n-1}}{2a_{n-1}} + \frac{a_{n-1}}{2a_{n-1}} + \frac{1}{a_{n-1}^2} = \frac{a_{n-1}^4+4}{4a_{n-1}^2}+1$.

Additional Thoughts: if n≥1 for $a_{n+1}$ then n-1≥0 for $a_n$. Since $a_1$=2, then all $(n-1)^2$≥0, so $n^2-2n+1$≥0 = $n^2-2n$≥1 = $n(n-2)$≥1,  such that all n≥3 or n≥1.

I have no clue if this is how I should be approaching it or not, I just wanted to try something. Please let me know what I need to do for this and the following parts, such as hints! Thanks.

(b) Use part (a) and equation (1) to prove that $a_n − a_{n+1}$ ≥ 0 for all n.

(c) Conclude that the sequence $(a_n)$ converges.

(d) Prove that $\lim_{n\to\infty} a_n = \sqrt{2}$.

(e) Modify the sequence $(a_n)$ so that it converges to $\sqrt{c}$. No formal proof is required
for this part, but you should give a brief justification.

答:
This kind of problem has been solved quite a few times on this site, but I want to try to show you how you could simply have followed the road map laid out in the statement of the problem.

In (a) you should at least try to follow the hint. Clearly $a_1^2=4\ge 2$, so you can immediately go on to the induction step: assume that $a_n^2\ge 2$, and try to use that to show that $a_{n+1}^2\ge 2$. There is only one reasonable way to approach this: start by expressing $a_{n+1}^2$ in terms of $a_n$.

$$\begin{align*}
a_{n+1}^2&=\frac14\left(a_n+\frac2{a_n}\right)^2\\
&=\frac14\left(a_n^2+4+\frac4{a_n^2}\right)\\
&=\frac{a_n^2}4+1+\frac1{a_n^2}\,.
\end{align*}$$

At this point you could try applying the induction hypothesis in a very straightforward way:

$$\begin{align*}
\frac{a_n^2}4+1+\frac1{a_n^2}&\ge\frac24+1+\frac1{a_n^2}\\
&=\frac32+\frac1{a_n^2}
\end{align*}$$

Unfortunately, $\frac1{a_n^2}<\frac12$, so that doesn’t yield the desired result. We need to dig a little deeper. We’re assuming that $a_n^2\ge 0$, and we want to know that $a_{n+1}^2-2\ge 0$, so perhaps we should focus more narrowly on $a_n^2-2$ and $a_{n+1}^2-2$. Let $d=a_n^2-2$; then

$$\begin{align*}
a_{n+1}^2-2&=\frac{a_n^2}4+1+\frac1{a_n^2}-2\\
&=\frac{2+d}4-1+\frac1{2+d}\\
&=\frac{(2+d)^2-4(2+d)+4}{4(2+d)}\\
&=\frac{d^2}{4(2+d)}\\
&\ge 0\,,
\end{align*}$$

as desired.

In (b) there is also a very reasonable place to begin, even before you have any real idea of how to proceed. Clearly we want to look at $a_n-a_{n+1}$, and the equation defining the relationship between $a_{n+1}$ and $a_n$ is the obvious place to start: we’re given that $a_{n+1}=\frac12\left(a_n+\frac2{a_n}\right)$, we have $2a_{n+1}=a_n+\frac1{a_n}$, and therefore

$$a_n-a_{n+1}=a_{n+1}-\frac1{a_n}\,.$$

The hint indicates that (a) is useful, so what does it say about $a_n$ and $a_{n+1}$? It says that $a_n^2\ge 2$ and $a_{n+1}^2\ge 2$, and since it’s clear (and easily proved by induction on $k$) that all of the numbers $a_k$ are positive, this means that $a_{n+1}\ge\sqrt2$ and $a_n\ge\sqrt2$. Finally, the fact that $a_n\ge\sqrt2$ implies that

$$\frac1{a_n}\le\frac1{\sqrt2}<\frac2{\sqrt2}=\sqrt2\,,$$

and we conclude that $a_n-a_{n+1}>0$ for all $n$.

Now (c) should be immediate: the sequence $\langle a_n:n\ge 1\rangle$ is monotone decreasing and bounded below, so … ?

(d) uses a standard trick that you can employ whenever you know that a sequence actually has a limit: if $\lim\limits_{n\to\infty}a_n=L$, then $\lim\limits_{n\to\infty}a_{n+1}=L$ as well. Thus,

$$\begin{align*}
L&=\lim_{n\to\infty}a_{n+1}\\
&=\lim_{n\to\infty}\frac12\left(a_n+\frac2{a_n}\right)\\
&=\frac12\left(\lim_{n\to\infty}a_n+\frac2{\lim\limits_{n\to\infty}a_n}\right)\\
&=\frac12\left(L+\frac2L\right)\,,
\end{align*}$$

and you can now solve for $L$.

Finally, (e) just asks you to go back through the work done in (a)-(d) and see what has to be modified to make the sequence converge to $\sqrt{c}$ instead.


问:
How to force wxMaxima to calculate subfunctions?
I have the following code:

```<code>B1(a,b,c,d) := (1-c) + (1-b) $
B2(a,b,c,d) := -(1-c) * (1-b) $
B3(a,b,c,d) := c*d + b*a $
B4(a,b,c,d) := -c*d*(1-b) - b*a*(2-c) $
B5(a,b,c,d) := b*a*(1-c) $
_estP(x1,x2,x3,x4,x5) := x1*PTL1 + x2*PTL2 + x3*DTL1 + x4*DTL2 + x5*DTL3 $
estP(a,b,c,d) := _estP( B1(a,b,c,d), B2(a,b,c,d), B3(a,b,c,d), B4(a,b,c,d), B5(a,b,c,d));
quadrat(a,b,c,d) := ((PTL0 - estP(a,b,c,d))/(b*c))^2;
</code>```

now I want to see the result of my calculation, i.e. a function of a, b, c and d.

but all I see is precisely what I typed (but formatted) - "quadrat(a,b,c,d):=((PTL0-estP(a,b,c,d))/(b*c))^2", meaning wxmaxima still uses my subfunctions like estP instead of performing full calculation.

I'd be thankful for any tip showing me how to get the result in a neat form (or at all).

答:
I copied your definitions into wxMaxima and at the end I typed quadrat(a,b,c,d); and I got:

(-(1-b)*(c-1)*PTL2-(-c-b+2)*PTL1+PTL0-a*b*(1-c)*DTL3-(-(1-b)*c*d-a*b*(2-c))*DTL2-(c*d+a*b)*DTL1)^2/(b^2*c^2)

which is probably what you are looking for :)


答:
If $r$ is the inradius, please note that $CS = KM = r, CT = LN = r$

Also note that $PS = PK$ and $QT = QL$

So we have $PC = PM, QC = QN$

That makes $\triangle PCM$ and $\triangle QCN$ isosceles triangles.

As $\angle BQN = \angle A, \angle BCN = \frac{\angle A}{2}$

And as $\angle APM = \angle B, \angle ACM = \frac{\angle B}{2}$

$\therefore \angle MCN = 45^0$.

Title: You have triangle $\triangle ABC$ which is orthogonal $\angle C=90^o$ with circle with center O inscribed in it. Find angle $\angle MCN$

Body:
You have triangle $\triangle ABC$ which is orthogonal $\angle C=90^o$ with circle with center O inscribed in it. If $KL$ is the diameter, $KL\parallel AB$, $KM\perp AB$ and $LN\perp AB$ Find angle $\angle MCN$.

Measuring this angle it works out to be $45^o$, but I don't know how to work it out. I attempted to utilize $KLNM$ being a rectangle (something which is proved since $KM\perp AB$ and $LN\perp AB$ which means $KM\parallel LN$ and also $KL\parallel MN$ hence $KLMN \#$ and since $KM\perp MN$ then $KMNL$ is a rectangle), but this didn't work out for me. Could you please explain to me how to solve the question?


答:
Yes there were some technical errors in that paper. I wrote some other papers possible factorization methods but did not actually succeed in getting a general solution. There was need for change in approach. When the general quintic equation is changed to identity for one actually succeeds and obtains an algebraic solution. Try this link:

http:/?utm_medium=email&utm_source=researchgate&utm_campaign=re322&utm_term=re322_x&utm_content=re322_x_p2&cp=re322_x_p2&uid=cx4ZWZ6hcdiX0hKERkrM99zJiDWztOEoeI4v&ch=reg

Title: Is there alternative factoring of a quintic equation?

Body:
In a paper here
the author appears to be able to factor a Bring-Jerrard quintic making

$$P=2mn(m^2-n^2)(m^2+n^2)=2m^5n-2mn^5\\ \implies n^5-m^4n+\frac{P}{2m}=0 \rightarrow x^5+px+q=0$$
become
$$(x^3+bx^2+cx+d)(x^2+ex+f)=0$$
but I haven't been able to follow how he got there. If I could, I would have what I need to find the one or more valid values of $n$ in the equation:

$$n^5-m^4n+\frac{P}{2m}=0$$

given that I will know the values of $P$ and $m$.

Can anyone help me figure out how the 'factored' equation would look in terms of $p,q$?


问:
Estimating $\frac{\sum_{r = 1}^{N} r^n}{N^n}$, getting a good numerical approximaion for large $N$
I am trying to find an accurate estimate for

$$f(n,N) = \frac{\sum^{N}_{r = 1} r^n}{N^n}$$

for large values of $N$.

I know the result that as ${N\to \infty}$, the $\frac{f(n,N)}{N}$ assumes value of $\frac{1}{n+1}$.

So I try to approximate $f(n,N)$ as $\frac{N}{n+1}$. Now what I have found running some python program that a better estimation for the sum is $\frac{N}{n+1} + \frac{1}{2}$ which fits much more exactly with correct value.

```<code>from math import *

n = 40
N = 1000000

s = 0
for i in range(N+1):
    s += i**n

s /= N**n
print("s:  ", s)

s1 = N/(n+1)
print("s1: ", s1)

s2 = (N)/(n+1) +1
print("s2: ", s2)

av = (s1+s2)/2
print("av: ", av)

perr_s_av = fabs(av-s)/s * 100
print("error (av): ", perr_s_av)

perr_s_s1 = fabs(s1-s)/s * 100
print("error (s1): ", perr_s_s1)
</code>```

The output is this

```<code>s:   24390.743905772357
s1:  24390.243902439026
s2:  24391.243902439026
av:  24390.743902439026
error (av):  1.3666376215028742e-08
error (s1):  0.0020499716419575546
</code>```

It is clear that the value $\frac{N}{n+1} + \frac{1}{2}$ is numerically much better approximation, why is this so, am I missing something

EDIT:

My assumption is that the next term of the polynomial $f(n, N)$ must be $\frac{1}{2}$ which is leading to this result, ie the polynomial form of $f(n, N) = \frac{N}{n+1} + \frac{1}{2} + \text{lower powers of N}$

答:
The way to compute this sum, is to find a polynomial of degree $N+1$ such that
$$ P(X+1)-P(X)=X^n.$$
Assuming that such a polynomial has been found, note that since we only ask for the difference of $P(X+1)-P(X)$ do be something, we can assume that $P(0)=0$.
Then You see that
$$
\sum_{r=1}^N r^n = \sum_{r=1}^N P(r+1) -P(r) = P(N+1)-P(0)=P(N+1).
$$
Suppose $P(X)= a_{n+1} X^{n+1} +a_{n}X^n + \ldots +a_1 X.$
Then, expanding the leading terms by the binomial rule,
$$
P(N+1)= a_{n+1} N^{n+1} + \left((n+1) a_{n+1} +a_{n}\right) N^{n} + \ldots
$$
and therefore

$$
\frac{P(N+1)}{N^n} = a_{n+1} N + \left((n+1) a_{n+1} +a_{n}\right) + O(N^{-1}). \quad\quad (1)
$$

So what you are asking is: what are the first two terms of $P$?
To make our life simpler, let us differentiate the relationship $n$ times. This gives
$$
P^{(n)}(X+1)-P^{(n)}(X) = (n+1)!a_{n+1} (X + 1)+n!a_n - \left((n+1)!a_{n+1} X+n!a_n\right) = n!
$$
so $$a_{n+1} = \frac{1}{n+1}.$$
Now differentiate only $(n-1)$ times :
$$
P^{(n-1)}(X+1)-P^{(n-1)}(X)= \frac{(n+1)!}{2} a_{n+1} (X + 1)^2 + n!a_n (X + 1)
- \left(\frac{(n+1)!}{2} a_{n+1} X^2 + n!a_n X  \right) = n! X
$$
So using what we already know,
$$
 \frac{1}{2} (X + 1)^2 + a_n (X + 1)
- \left(\frac{1}{2}X^2 + a_n X  \right) = X
$$
in other words
$$
X+\frac{1}{2} +a_n = X,
$$
or $a_n=-\frac{1}{2}$. Plugin this back into (1), you find

$$
\frac{P(N+1)}{N^n} = \frac{1}{n+1} N + \frac{1}{2} + O(N^{-1}). \quad\quad (1)
$$


问:
Understanding pdf of ratio of two random variables
I am not getting where i am doing mistake in deriving the PDF of ratio of two normalrandom variables. Here is what i had did:

Say we have two random variables $X$ and $H$ with PDF as $f_{X}(x) = \frac{1}{\pi \sigma_x}K_0\left(\frac{|x|}{\sigma_x}\right)$ and $f_H(h)=\frac{1}{\sqrt{2\pi}\sigma_h}e^{-\frac{h^2}{2\sigma_h^2}}$. Then using formula for PDF of ratio of two random variables from M.D.Springer book(P.No.92), the PDF of $Z=\frac{\alpha X}{H}$ is obtained as

$f_Z(z)=\int_{-\infty}^{+\infty}h \frac{1}{\pi \sigma_x}K_0\left(\frac{|hz|}{\alpha\sigma_x}\right)\frac{1}{\sqrt{2\pi}\sigma_h}e^{-\frac{h^2}{2\sigma_h^2}}dh$-------(1)

However in the research paper (which i am refereeing for the derivation), the PDFof $Z$ is given as
$f_Z(z)=\int_{-\infty}^{+\infty}|h| \frac{1}{\alpha\pi \sigma_x}K_0\left(\frac{|hz|}{\alpha\sigma_x}\right)\frac{1}{\sqrt{2\pi}\sigma_h}e^{-\frac{h^2}{2\sigma_h^2}}dh$------(2)

I am not getting how the equation (2) is coming. Any help in this regard is highly appreciated.

答:
Let $Z=\alpha X/H$ with $\alpha>0$. Then
$$
\begin{aligned}
\mathsf P(Z\leq z)%
&=\mathsf P(\alpha X/H\leq z)\\
&=\mathsf P(\alpha X/H\leq z,H>0)+\mathsf P(\alpha X/H\leq z,H<0)\\
&=\mathsf P(X\leq zH/\alpha,H>0)+\mathsf P(X\geq zH/\alpha,H<0).
\end{aligned}
$$
Assuming $X$ is independent of $H$ we then have
$$
\begin{aligned}
\mathsf P(Z\leq z)%
&=\int_0^\infty\int_{-\infty}^{zh/\alpha}f_X(x)f_H(h)\,\mathrm dx\mathrm dh+\int_{-\infty}^0\int_{zh/\alpha}^\infty f_X(x)f_H(h)\,\mathrm dx\mathrm dh.
\end{aligned}
$$
Differentiating we have by the Leibniz integral rule:
$$
\begin{aligned}
f_Z(z)=\partial_z\mathsf P(Z\leq z)%
&=\int_0^\infty \frac{h}{\alpha}f_X(zh/\alpha)f_H(h)\,\mathrm dh-\int_{-\infty}^0 \frac{h}{\alpha} f_X(zh/\alpha)f_H(h)\,\mathrm dh\\
&=\int_{-\infty}^\infty \frac{|h|}{\alpha}f_X(zh/\alpha)f_H(h)\,\mathrm dh.
\end{aligned}
$$
Substituting in the appropriate quantities for $f_X$ and $f_H$ then gives $(2)$.


问:
How thick should a cylindrical coin be for it to act as a fair three-sided die?
When flipping a coin of radius $r>0$ and thickness $t>0$ in the real world, there is some non-zero probability of getting neither heads nor tails, but instead landing on the thin lateral side. My question is, how thick does this lateral face need to be relative to $r$ so that the coin has equal chance to land on each of its 3 faces? That is, how do we choose $t$ so that the chance of landing on the lateral is $\frac{1}{3}$?

Clearly this problem requires some simplification, so we may assume the coin does not bounce and lands on a flat surface, and, well, any other assumptions implicit in the next paragraph:

I have heard from a friend that there is a method of solving this which entails centering the coin in a sphere, and adjusting $t$ so that a randomly chosen line segment between the center and a point on the sphere has an equal probability of intersecting each face of the coin. Is this a common modeling technique? How should we choose the radius of this sphere? Does it not matter given that the coin is fully contained?

And secondly, how can we proceed to find $t$ as a function of $r$ once this model has been established?

答:
My solution is similar, and I found many other sources of the same solution later in the literature. I started with a rectangle, as you'd see the coin from the edge side. It's circular, and therefore symmetrical by any line that goes through the centre, and symmetrial around the point in the centre. Therefore I figure a rectangular depiction of the coin is sufficient to create a model.

To make it equally equal that the coin fall on its edge, or on either side, it is sufficient that a line through the edge of the rectangle and the centre of the rectangle, on two edges of the shorter side, describe a 60 degree angle at the centre. This gives 120 degrees of probability of either sides of the rectangle to fall on, and 60 degrees probability on the edge, but also on the other side of the rectangle, so the two edges add up to 120 degrees as well.

Geometry gives you r^2 = D^2 + s^2, where r is half the diameter of the rectangle, D is half the diameter of the coin, and t is half the height of the cylinder. If you take r as given, for instance as 1, then s=1/2, since sin (30) = 1/2, and and D is sqrt(3)/2.

There is an other solution: the equal area solution. This presupposes that equal area surfaces will attract, so to speak, equal area of landings.

The area of one circular side is PiDD=2PiDt, which is the area of the planer edge of a cylinder. Divide both sides by PiD, and you get D = 2t, which means the diameter of the coin is twice the distance between the two faces of the coin, that is, it is the height of the cylinder.

Experimental results do not justify any of the three solutions. This is so because of the third law of thermodynamics: entropy, according to which every object will seek, so to speak, its lowest possible energy state. In translation this says, that if the coin falls on its edge, or on its side, in an elastic collision of the surface it lands on, and therefore it bounces, then chances are bigger that it will come to rest on a circular side than on its edge surface.

I worked out the mathematical models myself; my friend Geza Gyuru has come up with why the math models do not yield an experimental result due to the third law of thermodynamics.


问:
arithmetic operations on alephs in WolframAlpha
Can someone please enlighten me if or how both of these facts can be true? I entered $\aleph_0-\aleph_0$ into WolframAlpha, which gave the result as undefined. But it also plotted the result on a number line at zero. Similarly, $\frac{\aleph_0}{\aleph_0}$ is 1 on the number line, but undefined. Is this just a bug? </p>

答:
You ought to consider what the aleph numbers really mean: $\aleph_0$ is the cardinality of the naturals, and is, loosely speaking, a "type of infinity". Normal arithmetic operations are not defined for these numbers because it doesn't make any sense to use them; WA is probably just using a rule (x-x=0, x/x=1) that doesn't apply in this case.

To see what I mean, consider the set of naturals $\mathbb{N}$. It has cardinality $\aleph_0$. Now consider the set of even numbers $\mathbb{2N}$: it also has cardinality $\aleph_0$. If you take $\mathbb{N}$ and exclude (subtract) all even numbers, you get another set, the set of odds, which also has cardinality $\aleph_0$. So $\aleph_0-\aleph_0=\aleph_0$! Obviously this is silly, but you could construct a case where this subtraction gave you any number.


答:
Part (a) is actually very easy: the inequality $a_{n+1}^2\ge2$ is equivalent to
$$
\frac{1}{4}\Bigl(a_n^2+4+\frac{4}{a_n^2}\Bigr)\ge2
$$
that becomes $a_n^4+4a_n^2+4\ge8a_n^2$, hence
$$
a_n^4-4a_n^2+4\ge0
$$
which is true for every $a_n$. Since the base step is true, you're done.

Now you have to prove that $a_n>0$ for every $n$, which is also easy.

Next,
$$
a_n-a_{n+1}=a_n-\frac{1}{2}a_n-\frac{2}{a_n}=\frac{a_n^2-2}{2a_n}
$$
which is $\ge0$ because of part (a).

The rest is standard.

Body:
In this problem, we give an algorithm for computing $\sqrt{2}$. Let $a_1$ = 2, and define $a_{n+1}$ = $\frac{1}{2} (a_n + \frac{2}{a_n})$ for n ≥1.

(a) Prove that $a_n^2$ ≥ 2 for all n. (Use proof by induction.)

Here I have:

Let $a_{n+1} = \frac{1}{2}(a_{n}+\frac{2}{a_{n}})$ for all n ≥1.

Therefore, $a_{n+1-1} = \frac{1}{2}(a_{n-1}+\frac{2}{a_{n-1}}) = a_n.$

Therefore, $a_n^2 = (\frac{1}{2}(a_{n-1}+\frac{2}{a_{n-1}}))^2 = (\frac{1}{2}a_{n-1}+\frac{1}{a_{n-1}})^2 = \frac{1}{4}a_{n-1}^2 + \frac{a_{n-1}}{2a_{n-1}} + \frac{a_{n-1}}{2a_{n-1}} + \frac{1}{a_{n-1}^2} = \frac{a_{n-1}^4+4}{4a_{n-1}^2}+1$.

Additional Thoughts: if n≥1 for $a_{n+1}$ then n-1≥0 for $a_n$. Since $a_1$=2, then all $(n-1)^2$≥0, so $n^2-2n+1$≥0 = $n^2-2n$≥1 = $n(n-2)$≥1,  such that all n≥3 or n≥1.

I have no clue if this is how I should be approaching it or not, I just wanted to try something. Please let me know what I need to do for this and the following parts, such as hints! Thanks.

(b) Use part (a) and equation (1) to prove that $a_n − a_{n+1}$ ≥ 0 for all n.

(c) Conclude that the sequence $(a_n)$ converges.

(d) Prove that $\lim_{n\to\infty} a_n = \sqrt{2}$.

(e) Modify the sequence $(a_n)$ so that it converges to $\sqrt{c}$. No formal proof is required
for this part, but you should give a brief justification.


问:
Proving that $f$ verifies $f(1−x) + f(x) = 1$
I have a function which is defined for $x \in (0,1)$ and for $p>1$ with the expression

\begin{align*}
f(x) = \sum_{k=0}^{p-1} \frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}\left(\prod_{i=k-p+1, i\neq0}^{k+p} (x+k-i)\right)
\end{align*}

I would like to show that $f(1-x) + f(x) = 1$.

What I have done so far

With a little bit of algebra I have

\begin{align*}
f(x)
&= \frac{1}{(2p-1)!}\sum_{k=0}^{p-1} (-1)^{p+k} \binom{2p-1}{p+k} \left(\prod_{i=k-p+1, i\neq0}^{k+p} (x+k-i)\right)
&& \text{using }\binom{n}{m} = \frac{n!}{m!(n-m)!}
\\
&= \frac{1}{(2p-1)!}\sum_{k=0}^{p-1} (-1)^{p+k} \binom{2p-1}{p+k} \left(\prod_{j=1-p, j\neq-k}^{p} (x-j)\right)
&& \text{using the change of indices j=i-k}
\\
&= \frac{1}{(2p-1)!}\sum_{k=0}^{p-1} (-1)^{p+k} \binom{2p-1}{p+k} \left(\prod_{j=1-p}^{p} (x-j)\right) \frac{1}{x+k}
&& \text{completing the product}
\\
&= \frac{1}{(2p-1)!} \left(\prod_{j=1-p}^{p} (x-j)\right) \left(\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{x+k}\binom{2p-1}{p+k}\right)
&& \text{factoring the product from the sum}
\end{align*}

Then I introduce the functions
\begin{align*}
F(x) &= (2p-1)! f(x)  = G(x) H(x)\\
G(x) &= \prod_{j=1-p}^{p} (x-j) \\
H(x) &= \sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{x+k}\binom{2p-1}{p+k}
\end{align*}

So the initial question reduces to proving that $F(1-x) + F(x) = (2p-1)!$

Next, playing with the indices of the product I have been able to show that $G(1-x) = G(x)$.

So that now I only need to show $G(x) \left( H(1-x) + H(x)\right) = (2p-1)!$

EDIT:

I have just been able to simplify $H(1-x) + H(x)$ as follows :

\begin{align*}
H(x)
&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{x+k}\binom{2p-1}{p+k}
\\
&=\sum_{s=p}^{2p-1}\frac{(-1)^{s}}{x-p+s}\binom{2p-1}{s}
&& \text{using a change of index } s=p+k
\end{align*}

And

\begin{align*}
H(1-x)
&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{1-x+k}\binom{2p-1}{p+k}
\\
&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{1-x+k}\binom{2p-1}{p-1-k}
&& \text{using } \binom{2p-1}{p+k} = \binom{2p-1}{p-1-k}
\\
&=\sum_{s=0}^{p-1}\frac{(-1)^{s}}{x-p+s}\binom{2p-1}{s}
&& \text{using a change of index } s=p-1-k
\end{align*}

So finally

\begin{align*}
H(1-x) + H(x)
&=\sum_{s=0}^{2p-1}\frac{(-1)^{s}}{x-p+s}\binom{2p-1}{s}
\end{align*}

EDIT2:

I have just made the same work on $G(x)$ as follows

\begin{align*}
G(x)
&=\prod_{j=1-p}^{p} (x-j) \\
&=\prod_{s=0}^{2p-1} (x-p+s)
&& \text{using a change of index } s=p-i
\end{align*}

Now I think the constant $(2p-1)!$ in the relation $F(1-x)+F(x)=(2p-1)!$ comes from the binomial factor in $H(1-x) + H(x)$ but I don't see how to prove that...

Any lead ?

答:
Here we have a partial fraction decomposition in disguise.

We obtain
\begin{align*}
\color{blue}{f(x)}&=\sum_{k=0}^{p-1}\frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}
\prod_{\substack{j=k-p+1\\j\neq 0}}^{k+p}(x+k-j)\\
&=\prod_{j=-p+1}^{p}(x-j)\sum_{k=0}^{p-1}
\frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}\,\frac{1}{x+k}\tag{1}\\
&\color{blue}{=\prod_{j=-p}^{p-1}(x+j)\sum_{k=0}^{p-1}
\frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}\,\frac{1}{x+k}}\tag{2}\\
\end{align*}

Comment:

In (1) we expand the summands with $(x+k)/(x+k)$ and shift the product by $k$ to start with $j=-p+1$. This way the product can be factored out from the sum.

In (2) we substitute $j$ with $-j$ in the product.

We start with (2) and consider $f(1-x)$. We obtain
\begin{align*}
\color{blue}{f(1-x)}&=\prod_{j=-p}^{p-1}(1-x+j)\sum_{k=0}^{p-1}
\frac{(-1)^{p+k}}{(p-1-k)!(p+k)!}\,\frac{1}{1-x+k}\\
&=\prod_{j=-p}^{p-1}(x-j-1)\sum_{k=0}^{p-1}
\frac{(-1)^{k-1}}{k!(2p-1-k)!}\,\frac{1}{-x+p-k}\tag{3}\\
&=\prod_{j=-p+1}^{p}(x+j-1)\sum_{k=0}^{p-1}
\frac{(-1)^{k}}{k!(2p-1-k)!}\,\frac{1}{x+k-p}\tag{4}\\
&\,\,\color{blue}{=\prod_{j=-p}^{p-1}(x+j)\sum_{k=-p}^{-1}
\frac{(-1)^{p+k}}{(p+k)!(p-1-k)!}\,\frac{1}{x+k}}\tag{5}\\
\end{align*}

Comment:

In (3) we factor out $(-1)^{2p}=1$ from the product. We also change the order of summation $k\to p-1-k$.

In (4) we substitute $j$ with $-j$ in the product and we expand numerator and denominator of the summands with $-1$.

In (5) we shift the index $j$ by one to start with $j=-p$. We also shift the index $k$ by $p$ to start with $k=-p$.

We can now sum up (2) and (5) and get
\begin{align*}
\color{blue}{f(x)+f(1-x)}&\color{blue}{=\prod_{j=-p}^{p-1}(x+j)\sum_{k=-p}^{p-1}
\frac{(-1)^{p+k}}{(p+k)!(p-1-k)!}\,\frac{1}{x+k}}\tag{6}
\end{align*}

The right-hand side of (6) looks like a partial fraction decomposition. Indeed, we consider a function
\begin{align*}
g(x)=\frac{1}{\prod_{j=-p}^{p-1}(x+j)}
\end{align*}
The function $g(x)$ is a rational function with $2p$ pairwise different simple poles. Therefore we can find a representation
\begin{align*}
g(x)=\frac{1}{\prod_{j=-p}^{p-1}(x+j)}=\sum_{k=-p}^{p-1}\frac{a_k}{x+k}\qquad\qquad a_k\in\mathbb{R}
\end{align*}

Multiplication with the denominator of $g(x)$ gives
\begin{align*}
\color{blue}{1=\prod_{j=-p}^{p-1}(x+j)\sum_{k=-p}^{p-1}\frac{a_k}{x+k}}\tag{7}
\end{align*}

We can now evaluate (7) at $x=-k_0, -p\leq k_0\leq p-1$. We obtain
\begin{align*}
1&=\left.\prod_{j=-p}^{p-1}(x+j)\sum_{k=-p}^{p-1}\frac{a_k}{x+k}\right|_{x=-k_0}\\
&=\prod_{\substack{j=-p\\j\neq k_0}}^{p-1}\left(-k_0+j\right)a_{k_0}\\
a_{k_0}&=\frac{1}{\prod_{\substack{j=-p\\j\neq k_0}}^{p-1}(j-k_0)}\\
&=\frac{1}{\prod_{j=-p}^{k_0-1}(j-k_0)}\,\frac{1}{\prod_{j=k_0+1}^{p-1}(j-k_0)}\\
&=\frac{(-1)^{p+k_0}}{(p+k_0)!}\,\frac{1}{\left(p-1-k_0\right)!}
\end{align*}
The coefficients $a_{k_0}$ can now be put into (7) and we obtain expression (6). We conclude
\begin{align*}
\color{blue}{f(x)+f(1-x)=1}
\end{align*}
and the claim follows.


问:
solve the equation $\tanh(\mu \pi)=\frac{2\mu}{a}$
I am trying to find what condition it must meet for the equation to have a real solution for $\mu$.

I understand that the function $\tanh(x)$ is bounded in the interval $ (- 1,1) $ if I make use of this I find the following:

$-1<\tanh(\mu \pi)<1$ then $-1< \frac{2\mu}{a}< 1$

where do i get that

$a>2\mu$

But I was reading the procedure for this and it tells me the following:

A simple sketch shows that this equation can only have a real solution for $\mu$ if
the slope of $f(\mu) = \tanh(\mu \pi)$ at $\mu = 0$ is greater than the slope of $g(\mu)=\frac{2µ}{a}$ at
the same place. The former slope is π and the latter $\frac{2}{a}$. Thus the condition for
negative eigenvalues of the original equation is $a > \frac{2}{\pi}$.

But the latter I cannot fully understand. Could you explain me?

答:
First of all, we always have the trivial solution $\mu = 0$. $\DeclareMathOperator{sech}{sech}$

But what about other solutions? We confine ourselves to considering solutions $\mu > 0$ (since if $\mu$ is a solution, so too is $-\mu$).

We note that for $\mu > 0$, $\tanh$ is a concave-down function. This can be easily verified by taking its second derivative. $\frac{d}{dx} \tanh(x) = \sech^2(x)$ and $\frac{d}{dx} \sech^2(x) = 2 \sech(x) (-\sech(x) \tanh(x)) = -2 \sech^2(x) \tanh(x) < 0$ when $x > 0$.

Therefore, the function $f(\mu) = \tanh(\pi \mu) - \frac{2 \mu}{a}$ is a concave down function.

Now suppose we have $f(\mu) = 0$ for some $\mu > 0$. Then by the Rolle's theorem, we can find some $x \in (0, \mu)$ such that $f'(\mu) = 0$. But $f'$ is a decreasing function, so this means $f'(0) < 0$. This means that $\pi - \frac{2}{a} > 0$; that is, $\pi > \frac{2}{a}$; that is, $a > \frac{2}{\pi}$.

Now on the other hand, if we have $a > \frac{2}{\pi}$, then we see that $f'(0) > 0$. So for sufficiently small $x > 0$, we have $f(x) > 0$. But we also see that for very large $x$, we have $f(x) < 0$. So by the intermediate value theorem, there must exist some solution $\mu$ of the equation $f(\mu) = 0$.

So there is a non-zero solution of $\tanh(\pi \mu) = \frac{2 \mu}{a}$ if, and only if, $\frac{2}{\pi} < a$.


问:
Proving $a_n = \frac{q^n}{n}$ diverges
So, I have recently been studying sequences and I'm having quite a bit of trouble with the following problem:

Let $a_n = \frac{q^n}{n}$ with $q > 1, n \in \mathbb{N}$. Find the limit of $a_n$ if it exists.

Now, I know that this is a form of a geometric sequence $(q^n)_{n \in \mathbb{N}}$, which diverge for $|q| > 1$ and plotting the graph of $a_n$ for some values of $q$ shows that $a_n$ will most likely also diverge towards $\infty$. However, I just can't quite prove that this is the case. I attempted to show the following to prove this diverges:

$$\forall K>0: \exists N=N(K) \in \mathbb{N}: \forall n > N: a_n > K$$

The most obvious way to show this would be to show that this sequence is increasing from some point on-wards for all $q > 1$ and then solving $\frac{q^n}{n} > K$ for n. However, this didn't help much since I don't know how to solve that equation and I also feel like there should be some easier or more obvious way to show $a_n$ diverges.

答: Since $\lim \frac{a_{n+1}}{a_n}=q \gt 1,$ choose $N$ such that $n \gt N \Rightarrow \frac{a_{n+1}}{a_n} \gt \frac {1+q}{2} \gt 1$.  From that point forward, the sequence must get and stay larger than a geometric sequence with ratio $\frac{1+q}{2}$, which you know diverges.  Since the sequence in question is eventually larger than a sequence that diverges to $\infty$, our sequence also must do so.

Title: Proving $a_n = \frac{q^n}{n}$ diverges


答:
Ignore the number line. That was there just in case "aleph-0" was supposed to represent a real number.

Cardinal subtraction and division are problematic to define, for many reasons, some of which are included here, others here.

Title: arithmetic operations on alephs in WolframAlpha

Body: Can someone please enlighten me if or how both of these facts can be true? I entered $\aleph_0-\aleph_0$ into WolframAlpha, which gave the result as undefined. But it also plotted the result on a number line at zero. Similarly, $\frac{\aleph_0}{\aleph_0}$ is 1 on the number line, but undefined. Is this just a bug? </p>


问:
Number of pieces produced by cutting a folded ribbon
A piece of ribbon 1 metre long is folded in half so that the two ends are on top of each other. This doubled piece of ribbon is then folded in half again. The folded piece of ribbon is then cut  right through its midway point (see my drawing below).

This produces 5 pieces of ribbon of lengths: $\frac{1}{8},\frac{1}{4},\frac{1}{4}, \frac{1}{4}, \frac{1}{8}$.

My question is what happens after $n$ folds?

Trying a few more examples has led me to guess that after $n$ folds: there are $2^n + 1$ pieces of ribbon, with two strands of length $\frac{1}{2^{n+1}}$ and the rest of length $\frac{1}{2^{n}}$. I am having difficulty creating a convincing proof of this.

I came across this problem on an Oxford University Thinking Skills Admissions test ( Question 38, page 23 ). If this is a well known problem, I would also be interested in a source to read about the question in more detail.

答:
HINT: I think you can convince yourself with this argument:

After $N$ folds, how many layers "thick" is the ribbon?

The cut will create $2$ open ends per layer.  So how many total open ends are there after the cut?  Don't forget the $2$ original open ends (the ends of the original ribbon).

No. of pieces is related to total no. of open ends is a very simple way.