一题简单的数学题

streamleaf

我知道你没说它 diverges to infinity. 我只是引用 cesaro 的论点。

你的意思是应该说 cesaro sum of the series 是 0.5, 而不是 series 的 sum 是 0 .5 对吗？

madara_tachi

回复 21# streamleaf


是的, 我不擅长用华语解释.

Quote from wikipedia.

1) Not our case
a) If the series converges in the usual sense to a sum A, then the series is also Cesàro summable and has Cesàro sum A.
b) However, a series which positively diverges to an infinite value, will not be summable to a finite value in any case.

2) 0ur case
The significance of Cesàro summation is that a series which does not converge may still have a well-defined Cesàro sum.

---------

In short, the Cesàro summation is different from the original summation. Looking at the formula, we know that what it does is tracking the average of the partial sum.

Let's consider the summation formula of a convergent geometric series.

If |r| < 1, then a + ar + ar^2 + ar^3 + ......  = a / (1 - r)

The easiest proof is probably like the one below.

Let S = a + ar + ar^2 + ar^3 + ....            -(1)
then rS = ar + ar^2 + ar^3 + ....          -(2)
(1) - (2)
S(1-r) = a
S = a / (1-r)

However, this is true only if |r| < 1, which you can test by using
ratio test on the power series.
We have to be careful when we add infinite number of terms. In the example above, when you let S = a + ar +ar^2 + ar^3 + .....  , you really are assuming that it does converge to a value, hence you can represent it by S (or whatever). However, as we know, the series only converges if |r| < 1.  (So what baba000 did isn't appropriate)

1/2 can also be obtained from the formula of the convergent geometric series, 1 - 1 + 1 + ... = 1 / (1 - (-1)) = 1/2
where a = 1 and r = -1. We know we can't do this because |r| = |-1| = 1 is not less than 1, so we can't use this formula. However, there may be some interpretation of this sum, like the one you mentioned (Grandi's Series).

So I would say it is the Cesàro sum (in this case, 1/2) that has some relation and applicable to the area that you talked about.

streamleaf

赞同你的说法，convergent geometric series 不适合用在这问题。

但如果用 divergent geometric series 呢？

我比较倾向于这样的算法，相信 baba0000 也用这个：（我抄来的）

Treating Grandi's series as a divergent geometric series we may use the same algebraic methods that evaluate convergent geometric series to obtain a third value.

S = 1 &#8722; 1 + 1 &#8722; 1 + …, so

1 &#8722; S = 1 &#8722; (1 &#8722; 1 + 1 &#8722; 1 + …) = 1 - 1 + 1 - 1 + … = S,

resulting in S = 1&#8260;2. The same conclusion results from calculating &#8722;S, subtracting the result from S, and solving 2S = 1.

The above manipulations do not consider what the sum of a series actually means. Still, to the extent that it is important to be able to bracket series at will, and that it is more important to be able to perform arithmetic with them, one can arrive at two conclusions:

• The series 1 &#8722; 1 + 1 &#8722; 1 + … has no sum
• ...but its sum should be 1&#8260;2

可是回到一开始的问题，到底能不能说 1-1+1-1+.... = 1/2 呢？
红色的字很扣人心弦，到最后可能都没有一个定论。

我得空时去把 Devlin 的书找出来看看....      

baba0000 和 madara_tachi 你们是数学系的吗？

baba0000

我不是数学系的。大学教育已离我超过十载之年。只是目前的工作需要一些数学统计的东西，而且本身对这方面也有兴趣，所以有时间就自己研究一下。

biohazard

看到都头晕了

kelfaru

在另一个论坛也看到这题,但我真是完完整整被炸到

他给的答案竟然是>>> "独一无二"

ratloverice

(0,1)

龟龟龟

说infinity的朋友要看清楚题目。说0或1的朋友的答案虽然不完全对，但也应该看出这series在不同的情况下，会 ...
baba0000 发表于 17-4-2010 12:15 PM

   
x = 1 - 1 + 1 - 1 + 1 - 1 + ......

1 - x = 1 - (1 - 1 + 1 - 1 + 1 - 1 + ......）

1 - x = 1 - 1 + 1 - 1 + 1 - 1 + ......

1 - (1 - x) = 1 - (1 - 1 + 1 - 1 + 1 - 1 + ......)

x = 1 - 1 + 1 - 1 + 1 - 1 + ......

x = (1 - 1) + (1 - 1) + (1 - 1) + ......

x = 0 + 0 + 0 + 0 + 0 + 0.....

x = 0

管你什么 cesaro，这就是我的论点

Kitty_1978

好好玩哦。。。

streamleaf

在另一个论坛也看到这题,但我真是完完整整被炸到

他给的答案竟然是>>> "独一无二":shak ...
kelfaru 发表于 28-6-2010 09:43 PM

哈哈。。。。 好幽默！

streamleaf

x = 1 - 1 + 1 - 1 + 1 - 1 + ......

1 - x = 1 - (1 - 1 + 1 - 1 + 1 - 1 + ......）

1 ...
龟龟龟 发表于 29-6-2010 04:03 PM

也可以写成
x = 1 - 1 + 1 - 1 + 1 .... = 1 + (-1+1) + (-1+1) + .....
   = 1 + 0 + 0 ...... = 1


典型的 telescoping series.......

飘逸手语

还有其他"简单"的题目吗? 既然这题已经解到了,很想挑战下一个题目