含[x]的递推

hbghlyj

@tian27546西西 发表于2011-07-04 22:38 恩的 因为是给群里训练的 顺手也发到这

红色编号的题还没找到答案: 1、设 $a_0=1$, 当 $n \geq 0$ 时, 有 $a_{n+1}=3 a_n+\left[\sqrt{5} a_n\right]$, 求 $a_n$ 2、设 $a_1=6$, 且 $a_{n+1}=\left[\frac{5}{4} a_n+\frac{3}{4} \sqrt{a_n^2-2}\right]$, 求 $a_n$ 3、设 $a_1=4$, $a_{n+1}=a_n+\left[\frac{2 a_n-(n+3)}{n+1}\right]+1, n=1,2, \cdots$, 求 $a_n$ 4、设 $A$ 是2011阶实方阵, 且 $A^r=0$, $r$ 是自然数, 求 $\operatorname{Rank}(A)$ 的最大值. 5、设 $\Omega$ 为由 $x=0, y=0, z(x+y)=1$ 与 $x+y+z=4$ 围成的区域, 求证: $$ \iiint_\Omega\frac{\rmd x\rmd y\rmd z}{\sqrt{x^2+y^2+z^2}} \leq 2^{\frac{7}{4}} 3^{\frac{1}{2}} $$ 6、求以原点为顶点，且包含三条坐标轴的正半轴的圆锥面方程. 7、设 $E \subset \mathbb{R}^1, m(E)>0$, 则存在 Lebesgue不可测集 $A \subset E$. 8、设 $F$是特征为 0 的域, 则域的有限扩张 $\frac{E}{F}$ 必是单扩张. 9、若 $p$ 和 $q$ 是不同的素数, 证: $[\mathbb Q(\sqrt{p}, \sqrt{q}):\mathbb Q]=4$, 并且 $\mathbb Q(\sqrt{p}, \sqrt{q})=\mathbb Q(\sqrt{p}+\sqrt{q})$ 10、$C$ 是从原点到点 $z=e^{i a}$ 的直线段, 问 $a\;(-\pi< a \leq \pi)$ 取哪些值时, 积分 $I=\int_C e^{-\frac1z} d z$ 存在.

hbghlyj

hbghlyj 发表于 2023-4-28 22:15
3、设 $a_1=4$, $a_{n+1}=a_n+\left[\frac{2 a_n-(n+3)}{n+1}\right]+1, n=1,2, \cdots$, 求 $a_n$

代入$a_n=\frac{(n+1)(n+2)}2+1$
右边的$\left[\frac{2 a_n-(n+3)}{n+1}\right]$化简得$n+1$
于是$a_{n+1}=a_n+n+2$
由归纳法得证

hbghlyj

hbghlyj 发表于 2023-4-28 22:15
6、试求以原点为顶点，且包含三条坐标轴的正半轴的圆锥面方程

经过点$(1,0,0),(0,1,0),(0,0,1)\implies$母线与轴$(1,1,1)$的夹角为$\cos^{-1}(\frac13),$\[\left(x-\frac{x+y+z}{3}\right)^2 + \left(y-\frac{x+y+z}{3}\right)^2 + \left(z-\frac{x+y+z}{3}\right)^2 = \frac{2}{3}(x^2+y^2+z^2)\]展开得一个3项的方程\[xy+yz+zx=0\]这是Steiner's Circumellipse的重心坐标方程 Steiner's circumellipse of $\triangle ABC$ is described in the barycentric coordinates associated with the triangle by the following equation\[xy+yz+zx=0.\]

hbghlyj

hbghlyj 发表于 2023-4-28 22:15
4、设 $A$ 是2011阶实方阵, 且 $A^r=0$, $r$ 是自然数, 求 $\operatorname{Rank}(A)$ 的最大值

当$r=1$, 得$A=0$, 显然$\operatorname{Rank}(A)=0$.
因为2011是质数, 由下方红字得, 当$r>1$, $\operatorname{Rank}(A)$最大值为$2010-[2011/r]$

math.stackexchange.com/questions/863063/

Suppose $A$ is an element of $M_n(\mathbb{C})$. By Jordan form, if $A^k=0$, we have: (i) all eigenvalue are $0$; (ii) each Jordan block of $A$ has at most $k-1$ entries of $1$. To obtain the maximal value of $\operatorname{rank}(A)$, we consider the case of minimal number of Jordan blocks. The minimal number of Jordan blocks is equal to $\frac{n}{k}$ if $k$ divides $n$ and $\left\lfloor\frac{n}{k}\right\rfloor+1$ if $k$ doesn't divide $n$ (where $\lfloor x\rfloor$ is the unique integer such that $x\leqslant\lfloor x\rfloor<x+1$). Hence the maximal rank is equal to $n-\frac{n}{k}$ if $k$ divides $n$, and $n-\left\lfloor\frac{n}{k}\right\rfloor-1$ if $k$ doesn't divide $n$.

The equality is obtainable by choosing$$A=\begin{pmatrix}D_1&\dots&\dots&\dots&0\\0&\dots&\dots&\dots&0\\\dots&\dots&\dots&\dots&\dots\\0&\dots&\dots&D_k&0\\0&\dots&\dots&\dots&E\end{pmatrix}$$where $D_i=\begin{pmatrix}0&1&\dots&0\\0&0&\dots&0\\0&0&\dots&1\\0&0&\dots&0\end{pmatrix}$ of size $k$ and $E=\begin{pmatrix}0&1&\dots&0\\0&0&\dots&0\\0&0&\dots&1\\0&0&\dots&0\end{pmatrix}$ of size $n\bmod k$.

hbghlyj

hbghlyj 发表于 2023-4-28 22:15
8、设 $F$是特征为 0 的域, 则域的有限扩张 $\frac{E}{F}$ 必是单扩张.

en.wikipedia.org/wiki/Field_extension

In characteristic 0, every finite extension is a simple extension. This is the primitive element theorem, which does not hold true for fields of non-zero characteristic.

kuing.cjhb.site/forum.php?mod=viewthread&tid=11341

hbghlyj

hbghlyj 发表于 2023-4-28 22:15
7、设 $E \subset \mathbb{R}^1, m(E)>0$, 则存在 Lebesgue不可测集 $A \subset E$.

Every non null measurable set contains a non-measurable subset MSE
You can just imitate the usual construction - let $ A $ be a measurable subset of $ \mathbb R $, and choose a bounded interval $ I $ such that $ A \cap I = X $ has nonzero measure. By translating and scaling, we may wlog assume $ I = [0, 1] $. Let $ V $ be a set of representatives in $ I $ for the cosets of $ \mathbb R / \mathbb Q $ intersecting $ X $ (obviously $ V $ may be chosen to be a subset of $ X $, and thus of $ A $), and note that choosing an enumeration $ q_k $ of the rationals in $ [-1, 1] $ and defining $ V_k = V + q_k $ gives that
$$ \mu(X) \leq \sum_{k=1}^{\infty} \mu(V_k) = \sum_{k=1}^{\infty} \mu(V) \leq\mu([-1,2]) =3 $$
which is a contradiction, since $ \mu(X) \neq 0 $.

hbghlyj

hbghlyj 发表于 2023-4-28 22:15
9、若 $p$ 和 $q$ 是不同的素数, 证: $[\mathbb Q(\sqrt{p}, \sqrt{q}):\mathbb Q]=4$, 并且 $\mathbb Q(\sqrt{p}, \sqrt{q})=\mathbb Q(\sqrt{p}+\sqrt{q})$

MSE
LEMMA $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\ $ if  $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\:b}\ $  all are not in $\rm\:K\:$ and $\rm\: 2 \ne 0\:$ in $\rm\:K\:.$

Proof $\ \ $  Let  $\rm\ L = K(\sqrt{b})\:.\:$ Then $\rm\:  [L:K] = 2\:$  via  $\rm\:\sqrt{b}  \not\in K\:,\:$  so it is sufficient to prove $\rm\: [L(\sqrt{a}):L] = 2\:.\:$ It fails only if  $\rm\:\sqrt{a} \in L = K(\sqrt{b})\ $ and then $\rm\ \sqrt{a}\ =\  r + s\ \sqrt{b}\ $  for $\rm\ r,s\in K\:.\:$ But that is impossible since squaring yields $\rm(1):\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\  \sqrt{b}\:,\: $ which contradicts hypotheses as follows:  

$\rm\qquad\qquad r\:s \ne 0\ \Rightarrow\ \ \ \sqrt{b}\ \in\  K\ $ by solving $(1)$ for $\rm\sqrt{b}\:,\:$ using  $\rm\:2 \ne 0$  

$\rm\qquad\qquad\  s = 0\ \ \Rightarrow\ \  \ \sqrt{a}\ \in\  K\ \ $  via  $\rm\ \sqrt{a}\ =\ r \in K$

$\rm\qquad\qquad\  r = 0\ \ \Rightarrow\ \  \sqrt{a\:b}\in K\ \ $  via  $\rm\ \sqrt{a}\ =\ s\ \sqrt{b}\:,\: \ $times $\rm\:\sqrt{b}\quad\quad$ QED

Using the above as the inductive step one easily proves the following result of Besicovic.

THEOREM Let $\rm\:Q\:$ be a field with $2 \ne 0\:,\:$ and $\rm\ L = Q(S)\ $ be an extension of $\rm\:Q\:$ generated by $\rm n$ square roots  $\rm\ S = \{ \sqrt{a}, \sqrt{b},\ldots \}$ of elts  $\rm\ a,\:b,\:\ldots \in  Q\:.\:$
If every nonempty subset of $\rm\:S\:$ has product not in $\rm\:Q\:$ then each successive adjunction  $\rm\ Q(\sqrt{a}),\  Q(\sqrt{a},\:\sqrt{b}),\:\ldots$ doubles the degree over $\rm\:Q\:,\:$ so, in total, $\rm\: [L:Q] \ =\ 2^n.\:$  Hence the $\rm2^n$ subproducts of the product of $\rm\:S\:$ comprise a basis of $\rm L$ over $\rm\:Q\:.$

ljh25252

hbghlyj 发表于 2023-4-29 06:37
1、oeis.org/A018903
如何证明$S_n=\sum_{i=0}^na_i$
其中$a_{n+1}=4a_n+S_n$

可以直接求通项
$$
\begin{aligned}
a_{n+1}&=3a_n+[\sqrt{5}a_n]\\
a_{n+1}-3a_n&=[\sqrt{5}a_n]\\
a_{n+1}-3a_n&\leq\sqrt{5}a_n<a_{n+1}-3a_n+1\\
3a_n-a_{n+1}&\geq-\sqrt{5}a_n>3a_n-1-a_{n+1}\\
3a_n+\sqrt{5}a_n&\geq a_{n+1}>3a_n+\sqrt{5}a_n-1\\
4a_n&\geq(3-\sqrt{5})a_{n+1}>4a_n-1\\
3a_{n+1}-4a_n&\leq\sqrt{5}a_{n+1}<3a_{n+1}-4a_n+1\\
[\sqrt{5}a_{n+1}]&=3a_{n+1}-4a_n\\
a_{n+2}&=3a_{n+1}+[\sqrt{5}a_{n+1}]=3a_{n+1}+3a_{n+1}-4a_{n}=6a_{n+1}-4a_{n}
\end{aligned}
$$

后面就是线性递推了。

hbghlyj

hbghlyj 发表于 2023-4-28 22:15
10、$C$ 是从原点到点 $z=e^{i a}$ 的直线段, 问 $a\;(-\pi< a \leq \pi)$ 取哪些值时, 积分 $I=\int_C e^{-\frac1z} d z$ 存在.

$I=\int_C e^{-\frac1z} d z$ 存在的条件是$\abs a\le\frac\pi2$吗

hbghlyj

hbghlyj 发表于 2023-4-28 22:15
5、设 $\Omega$ 为由 $x=0, y=0, z(x+y)=1$ 与 $x+y+z=4$ 围成的区域, 求证:
$$
\iiint_\Omega\frac{\rmd x\rmd y\rmd z}{\sqrt{x^2+y^2+z^2}} \leq 2^{\frac{7}{4}} 3^{\frac{1}{2}}
$$

区域$\Omega$

Asymptote HTML

如何证明呢?