\begin{frame}{Example 1 cont.}
\begin{minipage}[c][.5\textheight][c]{\linewidth}
Consequently, the first few components of the homotopy perturbation solution for Eq. (4.1) are derived as follows
\begin{flalign*}
u_0(x, t) =& x\biggl[t + \frac{t^3}{3}\biggr],\\
u_1(x, t) =& x\biggl[t - \frac{2t^5}{15} - \frac{t^7}{63} - \frac{t^{2-\alpha}}{\Gamma(3 - \alpha)} - \frac{2t^{4-\alpha}}{\Gamma(5 - \alpha)}\biggr],\\
u_2(x, t) =& x\biggl[t - \frac{t^3}{3} - \frac{2t^5}{15} + \frac{t^7}{45} + \frac{2t^9}{567} - \frac{4t^{11}}{2475} - \frac{4t^{13}}{12285} - \frac{t^{15}}{59535} - \frac{2t^{2-\alpha}}{\Gamma(3 - \alpha)}\\
&+ \frac{t^{3-2\alpha}}{\Gamma(4 - 2\alpha)} + \biggl(\frac{2}{\Gamma(3 - \alpha)} - \frac{2}{\Gamma(4 - \alpha)}\biggr)\frac{t^{4-\alpha}}{(4 - \alpha)}\\
&+ \biggl(\frac{4}{\Gamma(5 - \alpha)} + \frac{16}{\Gamma(6 - \alpha)}\biggr) \frac{t^{6-\alpha}}{(6 - \alpha)}\\
&+ \biggl(\frac{80}{\Gamma(8 - \alpha)} - \frac{4}{15\Gamma(3 - \alpha)}\biggr)\frac{t^{8-\alpha}}{(8 - \alpha)}\\
&- \biggl(\frac{8}{15\Gamma(5 - \alpha)} + \frac{2}{63\Gamma(3 - \alpha)}\biggr)\frac{t^{10-\alpha}}{(10 - \alpha)} - \frac{4t^{12-\alpha}}{63(12 - \alpha)\Gamma(5 - \alpha)}\\
&+ \biggl(\frac{2}{\Gamma(5 - 2\alpha)} - \frac{1}{\Gamma(3 - \alpha)^2}\biggr)\frac{t^{5-2\alpha}}{5 - 2\alpha} - \frac{4t^{7-2\alpha}}{(7 - 2\alpha)\Gamma(3 - \alpha)\Gamma(5 - \alpha)}\\
&- \frac{4t^{9-2\alpha}}{(9 - 2\alpha)\Gamma(5 - \alpha)^2}\biggr],\\
\vdots
\end{flalign*}
and so on.
\end{minipage}
\end{frame}
答案1
我手动将其强制放在一张幻灯片上。坦率地说，只有非常靠近屏幕或非常大的屏幕的观众才能看到。
\documentclass{beamer}
\begin{document}
\begin{frame}{Example 1 cont.}\par\vspace{1em}
\begin{minipage}[c][.5\textheight][c]{1.1\linewidth}
\tiny Consequently, the first few components of the homotopy perturbation solution for Eq. (4.1) are derived as follows\par
\vspace{-2em}
\begin{flalign*}
u_0(x, t) =& x\biggl[t + \frac{t^3}{3}\biggr],\\
u_1(x, t) =& x\biggl[t - \frac{2t^5}{15} - \frac{t^7}{63} - \frac{t^{2-\alpha}}{\Gamma(3 - \alpha)} - \frac{2t^{4-\alpha}}{\Gamma(5 - \alpha)}\biggr],\\
u_2(x, t) =& x\biggl[t - \frac{t^3}{3} - \frac{2t^5}{15} + \frac{t^7}{45} + \frac{2t^9}{567} - \frac{4t^{11}}{2475} - \frac{4t^{13}}{12285} - \frac{t^{15}}{59535} - \frac{2t^{2-\alpha}}{\Gamma(3 - \alpha)}\\
&+ \frac{t^{3-2\alpha}}{\Gamma(4 - 2\alpha)} + \biggl(\frac{2}{\Gamma(3 - \alpha)} - \frac{2}{\Gamma(4 - \alpha)}\biggr)\frac{t^{4-\alpha}}{(4 - \alpha)}\\
&+ \biggl(\frac{4}{\Gamma(5 - \alpha)} + \frac{16}{\Gamma(6 - \alpha)}\biggr) \frac{t^{6-\alpha}}{(6 - \alpha)}\\
&+ \biggl(\frac{80}{\Gamma(8 - \alpha)} - \frac{4}{15\Gamma(3 - \alpha)}\biggr)\frac{t^{8-\alpha}}{(8 - \alpha)}\\
&- \biggl(\frac{8}{15\Gamma(5 - \alpha)} + \frac{2}{63\Gamma(3 - \alpha)}\biggr)\frac{t^{10-\alpha}}{(10 - \alpha)} - \frac{4t^{12-\alpha}}{63(12 - \alpha)\Gamma(5 - \alpha)}\\
&+ \biggl(\frac{2}{\Gamma(5 - 2\alpha)} - \frac{1}{\Gamma(3 - \alpha)^2}\biggr)\frac{t^{5-2\alpha}}{5 - 2\alpha} - \frac{4t^{7-2\alpha}}{(7 - 2\alpha)\Gamma(3 - \alpha)\Gamma(5 - \alpha)}\\
&- \frac{4t^{9-2\alpha}}{(9 - 2\alpha)\Gamma(5 - \alpha)^2}\biggr],\\
\vdots
\end{flalign*} \par
\vspace{-1em}
and so on.
\end{minipage}
\end{frame}
\end{document}
