2

I've counted and checked but can't seem to find the problem.

\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*}
Werner
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newbie
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    Thie third row has a \left, but no \right. Things are out of sync. If it was your intent to split \left and \right across lines of the flalign, that is not permitted. Instead use \bigg-style left/right arguments, instead of \left and \right. See http://tex.stackexchange.com/questions/231696/error-while-equation-splitting-with-align-environment – Steven B. Segletes May 05 '15 at 19:22

1 Answers1

3

For example, as noted in my comment, one could replace all \left with \biggl and all \right with \biggr. There are different sizes, as well: \big, \Big, etc. that you can search for on the site.

Your question will likely be marked duplicative, because the question has been asked many times, for example, here: Error while equation splitting with align environment

\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}

output

Community
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Steven B. Segletes
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  • @StevenB.Segletes thank you so much sir. it's 3a.m already and i really appreciate your help :') – newbie May 05 '15 at 19:36
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    @newbie You are welcome. Welcome to the site. The real way to say thanks to an answer that best addresses your question is to click the check mark to the left of that answer to "accept" it. For other answers that help, you can (once you get enough "reputation") also click the up-arrow icon to the left of the answer. – Steven B. Segletes May 05 '15 at 19:40
  • @newbie Also, you can combine both (upvoting and accepting) they are not exclusive. – Manuel May 05 '15 at 19:42