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带有分数的失控论证

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带有分数的失控论证

我试图写出以下等式：

\begin{equation}
    \text{$N_R$} \text{x} (\text{$\Delta\sigma_R$})^m = 2 \text{x} {10^6} \text{x} (\text{$\Delta\sigma_C$})^m    with  m = 3 for \text{N} \leq 5 \text{x} {10^6}
    \text{$N_R$} \text{x} (\text{$\Delta\tau_R$})^m = 2 \text{x} {10^6} \text{x} (\text{$\Delta\tau_C$})^m    with  m = 5 for \text{N} \leq 5 \text{x} {10^6}
    where \text{$\Delta\sigma_D$} = \text{$\frac{2}{5}^{\frac{1}{3}$} \text{x} \text{$\Delta\sigma_C$} = 0.7368 \text{x} \text{$\Delta\sigma_C$} is the constant amplitude fatigue limit
    \end{equation}

但出现错误：

Runaway argument?
{$\frac{2}{5}^{\frac{1}{3}$} \text{x} \text{$\Delta\sigma_C$} =\ETC.
! Paragraph ended before \text@ was complete.
<to be read again>
             \par

请帮忙……

答案1

您可能正在寻找以下内容，但我真的不确定。

\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{align}
    N_R \times (\Delta\sigma_R)^m &= 2 \times {10^6} \times (\Delta\sigma_C)^m    
&&\text{with~}  m = 3 \text{~for~} N\leq 5 \times {10^6}\\
N_R \times (\Delta\tau_R)^m &= 2 \times {10^6} \times (\Delta\tau_C)^m    
&&\text{with~}  m = 5 \text{~for~} N \leq 5 \times {10^6}
\end{align}
where $\Delta\sigma_D = \left(\frac{2}{5}\right)^{\frac{1}{3}} \times \Delta\sigma_C = 0.7368 \times \Delta\sigma_C $
is the constant amplitude fatigue limit
\end{document}

答案2

我不太确定你真正想要得到什么，但这可能是另一种解决方案。

\documentclass{article}
\usepackage{amsmath}
\usepackage{siunitx}
\begin{document}
\begin{equation}
 N_R \times (\Delta\sigma_R)^m = \num{2e6}\times(\Delta\sigma_C)^m
\end{equation}

with $m = 3 \quad\forall\quad N\leq\num{5e6}$

\begin{equation}
 N_R \times (\Delta\tau_R)^m = \num{2e6}\times(\Delta\tau_C)^m
\end{equation}

with $m = 5 \quad\forall\quad N\leq\num{5e6}$

where

\begin{equation}
 \Delta\sigma_D = \Bigl(\frac{2}{5}\Bigr)^{\frac{1}{3}} \times
 \Delta\sigma_C = 0.7368 \times \Delta\sigma_C
\end{equation}

is the constant amplitude fatigue limit

\end{document}

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