如何替换聚合表达式

发布于 2024-10-17 02:32:25 字数 153 浏览 0 评论 0原文

例如，我象征性地

1/n*Sum[ee[k] + 1, {k, j, n}]^2

将 Sum[ee[k], {k, j+1, n}] 替换为 x。我该怎么做？感谢您的帮助！

原文

For example, I have symbollically

1/n*Sum[ee[k] + 1, {k, j, n}]^2

And I want to substitute Sum[ee[k], {k, j+1, n}] to be x. How can I do this? May thanks for your help!

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谁的年少不轻狂 2024-10-24 02:32:25

您可以使用递推关系求和。例如：

f[j] := f[j + 1] + (ee[j] + 1);

1/N f[j]^2 /. f[j + 1] -> x

Out

(1 + x + ee[j])^2/N

Edit

根据您发布的几个问题，我认为您在某种程度上误解了 Replace[] 命令的作用。它不是基于“代数”，而是基于“模式”。它不理解也不使用比已经定义的代数变换（由您或 Mma 本身定义）更多的代数变换。

例如：

x/. (x-1)->y

不会匹配任何内容。但

(x-1) /. x->y-1

会给你 (y-2) 因为模式 x 是匹配的。

此外：

x = 3;
(x - 1) /. x -> y - 1

会给你 2 因为 x 在可能的匹配之前被评估，并且模式中的 x 也被评估（只需粘贴，执行并查看符号颜色）。

You may use the recurrence relation for the sum. For example:

f[j] := f[j + 1] + (ee[j] + 1);

1/N f[j]^2 /. f[j + 1] -> x

Out

(1 + x + ee[j])^2/N

Edit

Based on several questions you posted, I think you are somehow misinterpreting what the Replace[] command does. It is not "algebraic" based, but "pattern" based. It doesn't understand nor use more algebraic transformations than those already defined (by you or by Mma itself).

For example:

x/. (x-1)->y

will not match anything. But

(x-1) /. x->y-1

Will give you (y-2) because the pattern x is matched.

Moreover:

x = 3;
(x - 1) /. x -> y - 1

will give you 2 because x is evaluated before the possible match, and the x in the pattern is also evaluated (just paste, execute and look at the symbol color).

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