我已经计算并检查过了,但似乎找不到问题所在。
\begin{flalign*}
u_0(x,\ t)\ =&\ x\left[t\ +\ \frac{t^3}{3}\right],\\
u_1(x,\ t)\ =&\ x\left[t\ -\ \frac{2t^5}{15}\ -\ \frac{t^7}{63}\ -\ \frac{t^{2-\alpha}}{\Gamma(3\ -\ \alpha)}\ -\ \frac{2t^{4-\alpha}}{\Gamma(5\ -\ \alpha)}\right],\\
u_3(x,\ t)\ =&\ x\left[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)}\ +\ \left(\frac{2}{\Gamma(3\ -\ \alpha)}\ -\ \frac{2}{\Gamma(4\ -\ \alpha)}\right)\frac{t^{4-\alpha}}{(4\ -\ \alpha)}\\
&+\ \left(\frac{4}{\Gamma(5\ -\ \alpha)}\ +\ \frac{16}{\Gamma(6\ -\ \alpha)}\right)\ \frac{t^{6-\alpha}}{(6\ -\ \alpha)}\\
&+\ \left(\frac{80}{\Gamma(8\ -\ \alpha)}\ -\ \frac{4}{15\Gamma(3\ -\ \alpha)}\right)\frac{t^{8-\alpha}}{(8\ -\ \alpha)}\\
&-\ \left(\frac{8}{15\Gamma(5\ -\ \alpha)}\ +\ \frac{2}{63\Gamma(3\ -\ \alpha)}\right)\frac{t^{10-\alpha}}{(10\ -\ \alpha)}\ -\ \frac{4t^{12-\alpha}}{63(12\ -\ \alpha)\Gamma(5\ -\ \alpha)}\\
&+\ \left(\frac{2}{\Gamma(5\ -\ 2\alpha)}\ -\ \frac{1}{\Gamma(3\ -\ \alpha)^2}\right)\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}\right],\\
\vdots
\end{flalign*}
答案1
例如,正如我在评论中指出的那样,可以将 all 替换\left
为\biggl
,将 all\right
替换为\biggr
。还有不同的尺寸:\big
,\Big
等,您可以在网站上搜索。
您的问题可能会被标记为重复,因为该问题已被问过很多次,例如这里:使用对齐环境进行方程式拆分时出错
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\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_3(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*}
\end{document}