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\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}
  • Looks like shrinking the font size to \scriptsize won't even make the formulas fit the page. That's a lot of math to fit into one slide. Even if you manage to do that, I'm not sure how much the audience can swallow this much material. You may want to consider breaking them into two or more slides. – Herr K. Dec 7 '15 at 5:05
0

I manually forced it to fit on a single slide. Bluntly this is will only be visible to an audience either very close to the screen or on a very large screen.

\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}

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