答案1
- 以您的聲譽,您應該知道您應該提供 MWE。
- 如果你已經知道它們的名字,為什麼你要問如何在 LaTeX 中複製
\zeta
和符號呢?\sum
如果此程式碼無法解決您的問題,請編輯您的問題並更具體:
\documentclass{book}
\usepackage{amsmath, amssymb}
\begin{document}
\[
-\frac{{\zeta}^{\prime}(s)}{\zeta(s)} = \sum_{m\geq 1}\frac{\Lambda(n)}{{n}^{s}}
\]
\end{document}
編輯:
正如Mico所說,這是打包後的結果newtxmath
:
\documentclass{book}
\usepackage{newtxmath}
\DeclareMathOperator*{\mysum}{\text{\raisebox{-2pt}{\scalebox{2}{$\Sigma$}}}}
\begin{document}
\[
-\frac{{\zeta}^{\prime}(s)}{\zeta(s)} = \sum_{m\geq 1}\frac{\Lambda(n)}{{n}^{s}}
\]
\end{document}
最終,您可以創建自己的數學運算符。
第二次編輯
正如您正確指出的那樣,我的第一個解決方案(\mysum
)在不處於顯示樣式時不會變小。
我創建了另一個命令 ( \mynewsum
),它根據運算符的大小進行縮放\sum
。
如果您喜歡我的第一個顯示樣式解決方案,您可以混合使用前兩個解決方案(請參閱 參考資料\myfinesum
)。
\documentclass{book}
\usepackage{amsmath, amssymb}
\usepackage{array}
\usepackage{booktabs}
\renewcommand*{\arraystretch}{3}
\usepackage{graphicx}
\usepackage{scalerel}
\DeclareMathOperator*{\mysum}{\raisebox{-2pt}{\scalebox{2}{$\Sigma$}}}
\DeclareMathOperator*{\mynewsum}{\scalerel*{\Sigma}{\sum}}
\DeclareMathOperator*{\myfinesum}{%
\mathchoice
{\raisebox{-2pt}{\scalebox{2}{$\Sigma$}}}%
{\scalerel*{\Sigma}{\sum}}%
{\scalerel*{\Sigma}{\sum}}%
{\scalerel*{\Sigma}{\sum}}
}
\begin{document}
\noindent
\begin{tabular}{l>{$\displaystyle}c<{$}>{$\textstyle}c<{$}>{$\scriptstyle}c<{$}>{$\scriptscriptstyle}c<{$}}
\toprule
&
\text{Display style}
&
\text{Text style}
&
\textstyle\text{Script style}
&
\textstyle\text{Scriptscript style}
\\[10pt]
\midrule
\textbackslash\texttt{mysum}
&
-\frac{{\zeta}^{\prime}(s)}{\zeta(s)} = \mysum_{m\geq 1}\frac{\Lambda(n)}{{n}^{s}}
&
-\frac{{\zeta}^{\prime}(s)}{\zeta(s)} = \mysum_{m\geq 1}\frac{\Lambda(n)}{{n}^{s}}
&
-\frac{{\zeta}^{\prime}(s)}{\zeta(s)} = \mysum_{m\geq 1}\frac{\Lambda(n)}{{n}^{s}}
&
-\frac{{\zeta}^{\prime}(s)}{\zeta(s)} = \mysum_{m\geq 1}\frac{\Lambda(n)}{{n}^{s}}
\\[10pt]
\textbackslash\texttt{mynewsum}
&
-\frac{{\zeta}^{\prime}(s)}{\zeta(s)} = \mynewsum_{m\geq 1}\frac{\Lambda(n)}{{n}^{s}}
&
-\frac{{\zeta}^{\prime}(s)}{\zeta(s)} = \mynewsum_{m\geq 1}\frac{\Lambda(n)}{{n}^{s}}
&
-\frac{{\zeta}^{\prime}(s)}{\zeta(s)} = \mynewsum_{m\geq 1}\frac{\Lambda(n)}{{n}^{s}}
&
-\frac{{\zeta}^{\prime}(s)}{\zeta(s)} = \mynewsum_{m\geq 1}\frac{\Lambda(n)}{{n}^{s}}
\\[10pt]
\textbackslash\texttt{myfinesum}
&
-\frac{{\zeta}^{\prime}(s)}{\zeta(s)} = \myfinesum_{m\geq 1}\frac{\Lambda(n)}{{n}^{s}}
&
-\frac{{\zeta}^{\prime}(s)}{\zeta(s)} = \myfinesum_{m\geq 1}\frac{\Lambda(n)}{{n}^{s}}
&
-\frac{{\zeta}^{\prime}(s)}{\zeta(s)} = \myfinesum_{m\geq 1}\frac{\Lambda(n)}{{n}^{s}}
&
-\frac{{\zeta}^{\prime}(s)}{\zeta(s)} = \myfinesum_{m\geq 1}\frac{\Lambda(n)}{{n}^{s}}
\\[10pt]
\bottomrule
\end{tabular}
\end{document}