alignat 中的评论

Question

我怎样才能使 $f(x^0 + d^0) = -b(x^0 + d^0)$ 移至 $f(x^0) = -ax^0$ 以下？

\documentclass[a4paper,fontsize=13pt]{scrartcl}

\usepackage[english, main=ngerman]{babel}
\usepackage[a4paper,left=3cm,right=3cm,top=2.5cm,bottom=2.5cm]{geometry}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{mathtools}
\usepackage{multirow}
\usepackage{graphicx}
\usepackage{xfrac}
\usepackage{enumitem}
\usepackage{aligned-overset}



\begin{document}

\begin{alignat*}{2}
\rho_{0}
    &= \frac{f(x^{0}) - f(x^{0} + d^{0})}{f(x^{0}) - \widetilde{q}_{0}(d^{0})}  \\
    &= \frac{f(x^{0}) - f(x^{0} + d^{0})}{-\widetilde{\phi}(x^{0}, \Delta_{0} ; d^{0})}   \\
    \overset{\star}&{=} \frac{-a x^{0}+b(x^{0}+d^{0})}{a d^{0}}  
        &&\qquad\qquad \text{\%} \quad
           f(x^{0})= -a x^{0} \text{ because } x^{0} < 0,
           f(x^{0} + d^{0}) = - b(x^{0} + d^{0})  \text{ because } x^{0}+d^{0} \in (0,1) \\
    &= \frac{-a x^{0}+b(x^{0}+\Delta_{0})}{a \Delta_{0}}
        &&\qquad\qquad \text{\%} \quad  d^{0}=\Delta_{0}  \\
    &= ... \\
    &= \left( \frac{b}{a} - 1 \right) \frac{x^0}{\Delta_0} + \frac{b}{a} \\
    &= \left( \frac{b}{a} - 1 \right) \frac{x^0}{\frac{\beta_{1} \beta_{2}-1}{\beta_{1}} x^{0}} + \frac{b}{a}
        &&\qquad\qquad \text{\%} \quad  \Delta_{0} = \tfrac{\beta_{1} \beta_{2}-1}{\beta_{1}} x^{0}  \\
    &= \left( \frac{b}{a} - 1 \right) \frac{\beta_{1}}{\beta_{1} \beta_{2}-1} + \frac{b}{a} \\
    &= \theta \leq \eta_1 \\
\end{alignat*}

\end{document}

Answer 1

您可以使用 来实现这一点array。请注意，您的字体大小对于这个带有注释的方程式来说太大，无法容纳在边距内。因此，我更改了一些间距，但仍然宽了近 16 pt，也许您应该重新考虑字体大小，至少对于这个方程式来说是这样。

\documentclass[a4paper,fontsize=13pt]{scrartcl}

\usepackage[english, main=ngerman]{babel}
\usepackage[a4paper, hmargin=3cm, vmargin=2.5cm]{geometry}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{mathtools}
\usepackage{multirow}
\usepackage{graphicx}
\usepackage{xfrac}
\usepackage{enumitem}
\usepackage{aligned-overset}

\begin{document}

\begin{alignat*}{2}
\rho_{0}
    &= \frac{f(x^{0}) - f(x^{0} + d^{0})}{f(x^{0}) - \widetilde{q}_{0}(d^{0})} \\
    &= \frac{f(x^{0}) - f(x^{0} + d^{0})}{-\widetilde{\phi}(x^{0}, \Delta_{0} ; d^{0})} \\
    \overset{\star}&{=} \frac{-a x^{0}+b(x^{0}+d^{0})}{a d^{0}}
        & & \text{\%}\;\begin{array}{|l@{}}
           f(x^{0})= -a x^{0}\; \text{ because } x^{0} < 0, \\
            f(x^{0} + d^{0}) = - b(x^{0} + d^{0}) \;\text{ because } x^{0}+d^{0} \in (0,1)
           \end{array}\\
    &= \frac{-a x^{0}+b(x^{0}+\Delta_{0})}{a \Delta_{0}}
        & & \text{\%} \; d^{0}=\Delta_{0} \\
    &= ... \\
    &= \left( \frac{b}{a} - 1 \right) \frac{x^0}{\Delta_0} + \frac{b}{a} \\
    &= \left( \frac{b}{a} - 1 \right) \frac{x^0}{\frac{\beta_{1} \beta_{2}-1}{\beta_{1}} x^{0}} + \frac{b}{a}
        &\hskip 1.5em&\text{\%} \; \Delta_{0} = \tfrac{\beta_{1} \beta_{2}-1}{\beta_{1}} x^{0} \\
    &= \left( \frac{b}{a} - 1 \right) \frac{\beta_{1}}{\beta_{1} \beta_{2}-1} + \frac{b}{a} \\
    &= \theta \leq \eta_1 \\
\end{alignat*}

\end{document}

Answer 2

您可以简单地在方程式中开始新行，然后“跳过前列”。

例如：

\documentclass[a4paper,fontsize=13pt]{scrartcl}

\usepackage[english, main=ngerman]{babel}
\usepackage[a4paper,left=3cm,right=3cm,top=2.5cm,bottom=2.5cm]{geometry}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{mathtools}
\usepackage{multirow}
\usepackage{graphicx}
\usepackage{xfrac}
\usepackage{enumitem}
\usepackage{aligned-overset}



\begin{document}

\begin{alignat*}{2}
\rho_{0}
    &= \frac{f(x^{0}) - f(x^{0} + d^{0})}{f(x^{0}) - \widetilde{q}_{0}(d^{0})}  \\
    &= \frac{f(x^{0}) - f(x^{0} + d^{0})}{-\widetilde{\phi}(x^{0}, \Delta_{0} ; d^{0})}   \\
    \overset{\star}&{=} \frac{-a x^{0}+b(x^{0}+d^{0})}{a d^{0}}  
        &&\qquad\qquad \text{\%} \quad
           f(x^{0})= -a x^{0} \text{ because } x^{0} < 0, \\
    &&&   \qquad\qquad \text{\%} \quad  f(x^{0} + d^{0}) = - b(x^{0} + d^{0})  \text{ because } x^{0}+d^{0} \in (0,1) \\
    &= \frac{-a x^{0}+b(x^{0}+\Delta_{0})}{a \Delta_{0}}
        &&\qquad\qquad \text{\%} \quad  d^{0}=\Delta_{0}  \\
    &= ... \\
    &= \left( \frac{b}{a} - 1 \right) \frac{x^0}{\Delta_0} + \frac{b}{a} \\
    &= \left( \frac{b}{a} - 1 \right) \frac{x^0}{\frac{\beta_{1} \beta_{2}-1}{\beta_{1}} x^{0}} + \frac{b}{a}
        &&\qquad\qquad \text{\%} \quad  \Delta_{0} = \tfrac{\beta_{1} \beta_{2}-1}{\beta_{1}} x^{0}  \\
    &= \left( \frac{b}{a} - 1 \right) \frac{\beta_{1}}{\beta_{1} \beta_{2}-1} + \frac{b}{a} \\
    &= \theta \leq \eta_1 \\
\end{alignat*}

\end{document}

将如下所示：