## Why did Hopcroft and Karp write $M_0, M_1, M_2, \cdots, M_i, \cdots$? (Hopcroft – Karp Algorithm)

I am reading “An $$n^{\frac{5}{2}}$$ Algorithm for Maximum Matchings in Bipartite Graphs” by Hopcroft and Karp.

Please see the image below.

Let $$s$$ be the cardinality of a maximum matching.
I think any of $$M_0, M_1, M_2, \cdots, M_s$$ is a matching and $$M_s$$ is a maximum matching.
So, I think $$P_s$$ doesn’t exist.
But the authors wrote $$M_0, M_1, M_2, \cdots, M_i, \cdots$$ and $$|P_0|, |P_1|, \cdots, |P_i|, \cdots$$.

Why?

Maybe I am confused.

## Show that $1^{1987} + 2^{1987} + \cdots + n^{1987}$ never divides $n+2$

Show that for any natural number $$n$$, the result of $$1^{1987}+2^{1987} + \cdots + n^{1987}$$ cannot be divided by $$n+2$$ without a remainder.

## Problem

Solve for $$L$$, where $$P_n \cdots P_1 = I – XLX^T$$

where $$\Vert x_i \Vert_2=1$$, $$P_i := I – 2x_i x_i^T$$, and $$X_{m \times n} = [x_1 |\cdots | x_n]$$.

## Try

Note that $$L$$ is a lower triangular matrix. Denoting $$(i,j)$$ component of $$L$$ as $$L_{ij}$$, let us find $$L_{ij}$$‘s explicitly. We have

\begin{align} P – I &= (I – 2x_nx_n^T) \cdots (I – 2x_1x_1^T) – I\ &= – L_{11} x_1x_1^T – L_{21}x_2x_1^T – \cdots -L_{nn}x_nx_n^T \end{align}

where we can note that $$-L{ij}$$ is the coefficient for $$x_ix_j^T$$ in $$(I – 2x_nx_n^T) \cdots (I – 2x_1x_1^T)$$.

Let us observe more carefully. For $$L_{kk}$$‘s, we have

$$L_{kk} = 2$$

since these terms only come from $$I \cdots I (-2x_kx_k^T) I \cdots I$$. Next, we note that

$$L_{(k+1)k} = -2^2$$

and, next,

$$-L_{(k+2)k} = 2^2 – 2^3(v_{k+2}^Tv_{k+1})(v_{k+1}^Tv_k)$$

but I cannot see any simpler rule for $$L_{ij}$$‘s… Anyone to help me with finding all the $$L_{ij}$$s?

## If $\sum c_n$ is convergent where $c_n = a_1b_n + \cdots +a_nb_1$, then $\sum c_n = \sum a_n\sum b_n$

$$\sum a_n$$ and $$\sum b_n$$ is convergent and $$c_n = a_1b_n + \cdots + a_nb_1$$. Prove that if $$\sum c_n$$ is convergent then $$\sum c_n = \sum a_n \sum b_n$$.

This question has been answered using Abel’s Lemma in this post. But I am looking forward to a more basic solution which only uses basic definition and Cauchy’s convergence test.

## Estimate $\frac 1 {a+b} + \frac 1 {(a+b)(a+2b)} + \cdots + \frac 1 {(a+b)\cdots(a+kb)} + \cdots$

Suppose $$a>1,b>0$$ are real numbers. Consider the summation of the infinite series: $$S=\frac 1 {a+b} + \frac 1 {(a+b)(a+2b)} + \cdots + \frac 1 {(a+b)\cdots(a+kb)} + \cdots$$ How can I give a tight estimation on the summation? Apparently, one can get the upper bound: $$S\le \sum_{k=1} ^\infty \frac 1 {(a+b)^k}=\frac 1 {a+b-1}$$ But it is not tight enough. For example, fix $$a\rightarrow 1$$, and $$b=0.001$$, then $$S=38.969939$$, it seems that $$S=O(\sqrt{1/b})$$. Another example: $$a=1$$, and $$b=0.00001$$,$$S=395.039235$$.

## How to write a formula, $S$, so that the set $P_1, \cdots, P_k$ is not consistent iff $S$ is valid?

A set of propositional formulas $$P_1, \cdots, P_k$$ is consistent iff there is an environment in which they are all true.

Write a formula, $$S$$, so that the set $$P_1, \cdots, P_k$$ is not consistent iff $$S$$ is valid.

## $\bigcap I_\gamma$ if $I_0 \supset \cdots \supset\I_{\gamma} \cdots$ are unbounded

Let be $$\kappa$$ a regular cardinal and $$I_0 \supset \cdots \supset I_{\gamma} \cdots$$ are unbounded subsets of $$\kappa$$ for $$\gamma < \lambda <\kappa$$ where $$\lambda$$ is limit. I want to show $$\bigcap I_\gamma$$ is unbounded. Is it true? It’s easy show it’s not empty for regularity of $$\kappa$$.

## On constructing free action of the cyclic group $\Bbb Z/3\Bbb Z$ on $S^n \times \cdots \times S^n$($n$ is odd)

Can we construct a free action of the cyclic group $$\Bbb Z/3\Bbb Z$$ on $$S^n \times \cdots \times S^n$$($$n$$ is odd) without multiplying any coordinate by $$e^{2\pi i/3}$$? In other words, do I Need to multiply at least one coordinate by a 3rd root of unity to get a free action of $$\Bbb Z/3\Bbb Z$$? I have tried to construct such examples but could not find any.

Thank you so much in advance.

## Hypothesis $\cdots$

for $$n, p\in\mathbb{N}(p-prime\>number)$$ there exist $$m\in\mathbb{N}$$ for $$\p^m\equiv m\pmod n$$