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如何缩小 $)$ 和 $^2$ 的间距?

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APPSYZY Posted 2022-9-13 02:26 |Read mode
比如 $\displaystyle\lim_{n\to\infty}\left(\dfrac{\displaystyle\sum^n_{k=1}\left(\dfrac{C^k_n}{k}\right)^2}{\left(\displaystyle\sum^n_{k=1}\dfrac{C^k_n}{k}\right)^2}\right)\sqrt n.$

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hbghlyj Posted 2022-9-13 02:36
为了保持上标2的高度,找到相同高度的括号放在\phantom里,然后用\kern调整字距:
$\displaystyle\lim_{n\to\infty}\left(\dfrac{\displaystyle\sum^n_{k=1}\left(\dfrac{C^k_n}{k}\right)\kern-1em\phantom{\bigg)}^2}{\left(\displaystyle\sum^n_{k=1}\dfrac{C^k_n}{k}\right)\kern-1em\phantom{\Bigg)}^2}\right)\sqrt n.$
这里用的
  1. \kern1em
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转换为MathML:
  1. <mspace width="-1em"></mspace>
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或者写成
  1. <mspace width="-1em"/>
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 Author| APPSYZY Posted 2022-9-13 10:59
hbghlyj 发表于 2022-9-13 02:36
为了保持上标2的高度,找到相同高度的括号放在\phantom里,然后用\kern调整字距:
$\displaystyle\lim_{n\to\i ...
谢谢

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kuing Posted 2022-9-13 13:25
hbghlyj 发表于 2022-9-13 02:36
为了保持上标2的高度,找到相同高度的括号放在\phantom里,然后用\kern调整字距:
$\displaystyle\lim_{n\to\i ...
  1. $\displaystyle\lim_{n\to\infty}\left(\dfrac{\displaystyle\sum^n_{k=1}\left(\dfrac{C^k_n}{k}\right)\kern-1em\phantom{\bigg)}^2}{\left(\displaystyle\sum^n_{k=1}\dfrac{C^k_n}{k}\right)\kern-1em\phantom{\Bigg)}^2}\right)\sqrt n.$
Copy the Code
你的这段代码在真 LaTeX 里效果并不正确,不过我也不理解它为什么会不正确。

经测试,需要对 \phantom 外再套一层 { } 才正确,就是 \phantom{\bigg)}^2 得改成 {\phantom{\bigg)}}^2,完整代码:
  1. $\displaystyle\lim_{n\to\infty}\left(\dfrac{\displaystyle\sum^n_{k=1}\left(\dfrac{C^k_n}{k}\right)\kern-1em{\phantom{\bigg)}}^2}{\left(\displaystyle\sum^n_{k=1}\dfrac{C^k_n}{k}\right)\kern-1em{\phantom{\Bigg)}}^2}\right)\sqrt n.$
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真是有点奇怪呢。

另外 -1em 似乎还不够明显,-1.2em 还差不多,分母还应该更大一些可以 -1.3em。

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kuing Posted 2022-9-13 13:37
其实手工调距离也不需要这么麻烦啊,直接将 ^2 改成 ^{\kern-3pt2} 不就行了嘛?
$\displaystyle\lim_{n\to\infty}\left(\dfrac{\displaystyle\sum^n_{k=1}\left(\dfrac{C^k_n}{k}\right)^{\kern-3pt2}}{\left(\displaystyle\sum^n_{k=1}\dfrac{C^k_n}{k}\right)^{\kern-3.5pt2}}\right)\sqrt n.$
  1. $\displaystyle\lim_{n\to\infty}\left(\dfrac{\displaystyle\sum^n_{k=1}\left(\dfrac{C^k_n}{k}\right)^{\kern-3pt2}}{\left(\displaystyle\sum^n_{k=1}\dfrac{C^k_n}{k}\right)^{\kern-3.5pt2}}\right)\sqrt n.$
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hbghlyj Posted 2022-9-13 17:50
话说...这个极限怎么求呢

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 Author| APPSYZY Posted 2022-9-13 18:13
kuing 发表于 2022-9-13 13:37
其实手工调距离也不需要这么麻烦啊,直接将 ^2 改成 ^{\kern-3pt2} 不就行了嘛?
$\displaystyle\lim_{n\to ...
这个间距更紧更好看了

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