Forgot password?
 Register account
View 2559|Reply 12

[数列] 数列存在问题

[Copy link]

1

Threads

1

Posts

11

Credits

Credits
11

Show all posts

nanjingmath Posted 2013-9-26 17:24 |Read mode
本人高二,有幸发现了论坛,希望以后能帮渣渣解决最近碰到的数学问题。

下面这个问题思考了很久。
设$a_{1},a_{2},\cdots,a_{N}$是非负数,且不全为0,证明:存在一个整数列
$1=n_{0}<n_{1}<\cdots<n_{k}=N+1$,且满足
$$n_{1}a_{n_{0}}+n_{2}a_{n_{2}}+\cdots+n_{k}a_{n_{k-1}}<3(a_{1}+a_{2}+\cdots+a_{N})$$

来源:tieba.baidu.com/p/2614176109

686

Threads

110K

Posts

910K

Credits

Credits
91229
QQ

Show all posts

kuing Posted 2013-9-26 17:26
下标有没有打错?

1

Threads

1

Posts

11

Credits

Credits
11

Show all posts

 Author| nanjingmath Posted 2013-9-26 17:32
按照上面打的 好像没有打错。。请假在家复习。你懂得。大神

6

Threads

245

Posts

2284

Credits

Credits
2284

Show all posts

睡神 Posted 2013-9-26 17:45
嗯嗯,大神是懂的,关键系我木系大神…
除了不懂,就是装懂

84

Threads

2339

Posts

110K

Credits

Credits
13091

Show all posts

其妙 Posted 2013-9-26 18:10
楼主高二这么厉害就懂latex啦?

67

Threads

407

Posts

3537

Credits

Credits
3537

Show all posts

Tesla35 Posted 2013-9-26 18:32
看到是西西的题,我还是洗洗睡吧

686

Threads

110K

Posts

910K

Credits

Credits
91229
QQ

Show all posts

kuing Posted 2013-9-26 18:33
瞬间变成水贴……
PS、其实我是想说第二项的下标跟原贴不一样……

84

Threads

2339

Posts

110K

Credits

Credits
13091

Show all posts

其妙 Posted 2013-9-27 15:37
回复 7# kuing
对,那个下标搞错了吧

770

Threads

4692

Posts

310K

Credits

Credits
35048

Show all posts

isee Posted 2013-9-27 17:23
楼主高二这么厉害就懂latex啦?
其妙 发表于 2013-9-26 18:10

这有何难,看看k写的《本论坛的 $\LaTeX$ 公式的基本输入[2013-9-1更新]》就会了

686

Threads

110K

Posts

910K

Credits

Credits
91229
QQ

Show all posts

kuing Posted 2013-9-27 17:45
好吧继续水……公式输入只是 LaTeX 的一小部分……不能叫懂 LaTeX ……其他的更难

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

hbghlyj Posted 2022-9-14 19:16
A[1].gif B[1].gif C[1].gif
KoMaL A.394
Let $a_1,a_2,\dots,a_N$ be nonnegative reals, not all 0. Prove that there exists a sequence $1=n_0<n_1<\dots<n_k=N+1$ of integers such that\[n_1a_{n_0}+n_2a_{n_1}+\dots+n_ka_{n_{k-1}}<3(a_1+a_2+\dots+a_N).\]
Solution.
Define $a_{N+2}+a_{N+3}=\dots=0$ as well and look for an infinite sequence $1=n_0<n_1<\dots$ of integers for which
\[\sum_{i=1}^\infty n_ia_{n_{i-1}} < 3\sum_{n=1}^\infty a_n.\]
Define the function $f:[1,\infty)\to R$ such that $f(x)=a_{[x]}$.

The required sequence is constructed randomly. Take a random variable $t\in[0,1]$ of uniform distribution. Set $n_0=1$ and $n_i=\left[2\cdot e^{i-1+t}\right]$ for $i=1,2,\dots$. Then\begin{aligned}
&E\left(\sum_{i=1}^\infty n_ia_{n_{i-1}}\right) \le E(n_1)\cdot a_0 + E \left(\sum_{i=2}^\infty2\cdot e^{i-1+t}f(2\cdot e^{i-2+t})\right)\\
=&E(n_1)\cdot a_0 + \sum_{i=2}^\infty\int_0^1 \big(2\cdot e^{i-1+t}f(2\cdot e^{i-2+t})\big)dt\\
=&E(n_1)\cdot a_0 + \int_0^\infty 2\cdot e^{u+1}f(2\cdot e^u)du\\=&E(n_1)\cdot a_0 + e\int_2^\infty f(x)dx\\
=&E(n_1)\cdot a_0 + e\sum_{n=2}^Na_n.
\end{aligned}The value of $n_1$ is 2,3,4 or 5 if $t<\ln\frac32$, $\ln\frac32\le t<\ln\frac42$, $\ln\frac42\le t<\ln\frac52$, or $\ln\frac52\le t<1$, respectively. Therefore
\[E(n_1)= \left(\ln\frac32\right)\cdot2 + \left(\ln\frac42-\ln\frac32\right)\cdot3 + \left(\ln\frac52-\ln\frac42\right)\cdot4 + \left(1-\ln\frac52\right)\cdot5 \approx 2.985<3\]
and
\[E\left(\sum_{i=1}^\infty n_ia_{n_{i-1}}\right) < 3a_1 + e(a_2+\dots+a_N).\]
There always exists a sequence $1=n_0<n_1<...<n_k=N+1$ such that
\[n_1a_{n_0}+n_2a_{n_1}+\dots+n_ka_{n_{k-1}}<3a_1+e(a_2+a_3+...+a_N).\]
math.stackexchange.com/questions/1060803

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

hbghlyj Posted 2022-9-14 19:31

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

hbghlyj Posted 2022-9-14 19:36
这个外链挂了.
引用外链时,记得到WayBack Machine存一下档
WayBack Machine有一个Chrome Extension,浏览网页时,只要顺手点一下就存档了.

Mobile version|Discuz Math Forum

2025-5-31 11:10 GMT+8

Powered by Discuz!

× Quick Reply To Top Edit