I am trying to show that the 9 point is equal to $$\Delta^2u+\frac{1}{12}h^2(u_{xxxx}+2u_{xxyy}+u_{yyyy})+O(h^4)$$ like it says in my book. But I am not getting that. I can't figure out where I went wrong... I was hoping you could help me out. Thanks.
Note: Where it says \cancel, that is a term that I eliminated.
\noindent PROBLEM 2\ \noindent a) \ Show that the 9-point Laplacian (3.17) has the truncation error given in Section 3.5 as $)(h^4)$.\
The 9-point Laplacian is $$\Delta_9^hu_{i,j}=\frac{1}{6h^2}[4u_{i-1,j}+4u_{i+1,j}+4u_{i,j-1}+4u_{i,j+1}+u_{i-1,j-1}+u_{i-1,j+1}+u_{i+1,j-1}+u_{i+1,j+1}-20u_{ij}]$$
If we find the Taylor series expansion, we get:\ \begin{align*} u_{i-1,j} &= u - hu_x + \frac{h^2}{2}u_{xx} - \frac{h^3}{6}u_{xxx} + \frac{h^4}{24}u_{xxxx} + O(h^5)\\ u_{i+1,j} &= u + hu_x + \frac{h^2}{2}u_{xx} + \frac{h^3}{6}u_{xxx} + \frac{h^4}{24}u_{xxxx} + O(h^5)\\ u_{i,j-1} &= u - hu_y + \frac{h^2}{2}u_{yy} - \frac{h^3}{6}u_{yyy} + \frac{h^4}{24}u_{yyyy} + O(h^5)\\ u_{i,j+1} &= u + hu_y + \frac{h^2}{2}u_{yy} + \frac{h^3}{6}u_{yyy} + \frac{h^4}{24}u_{yyyy} + O(h^5)\\ u_{i-1,j-1} &= u - h(u_x + u_y) + \frac{h^2}{2}(u_{xx} + 2u_{xy} + u_{yy}) - \frac{h^3}{6}(u_{xxx} + 3u_{xxy} + 3u_{xyy} + u_{yyy}) + \frac{h^4}{24}(u_{xxxx} + 4u_{xxxy} + 6u_{xxyy} + 4u_{xyyy} + u_{yyyy}) + O(h^5)\\ u_{i-1,j+1} &= u - h(u_x - u_y) + \frac{h^2}{2}(u_{xx} - 2u_{xy} + u_{yy}) - \frac{h^3}{6}(u_{xxx} - 3u_{xxy} + 3u_{xyy} - u_{yyy}) + \frac{h^4}{24}(u_{xxxx} - 4u_{xxxy} + 6u_{xxyy} - 4u_{xyyy} + u_{yyyy}) + O(h^5)\\ u_{i+1,j-1} &= u + h(u_x - u_y) + \frac{h^2}{2}(u_{xx} - 2u_{xy} + u_{yy}) + \frac{h^3}{6}(u_{xxx} - 3u_{xxy} + 3u_{xyy} - u_{yyy}) + \frac{h^4}{24}(u_{xxxx} - 4u_{xxxy} + 6u_{xxyy} - 4u_{xyyy} + u_{yyyy}) + O(h^5)\\ u_{i+1,j+1} &= u + h(u_x + u_y) + \frac{h^2}{2}(u_{xx} + 2u_{xy} + u_{yy}) + \frac{h^3}{6}(u_{xxx} + 3u_{xxy} + 3u_{xyy} + u_{yyy}) + \frac{h^4}{24}(u_{xxxx} + 4u_{xxxy} + 6u_{xxyy} + 4u_{xyyy} + u_{yyyy}) + O(h^5)\\ \end{align*}\
Notice that $u_{i-1,j} + u_{i+1,j} + u_{i,j-1} + u_{i,j+1}$ \begin{align*} &= u - \cancel{hu_x} + \frac{h^2}{2}u_{xx} - \cancel{\frac{h^3}{6}u_{xxx}} + \frac{h^4}{24}u_{xxxx} + O(h^5)\\ &+ u + \cancel{hu_x} + \frac{h^2}{2}u_{xx} + \cancel{\frac{h^3}{6}u_{xxx}} + \frac{h^4}{24}u_{xxxx} + O(h^5)\\ &+ u - \cancel{hu_y} + \frac{h^2}{2}u_{yy} - \cancel{\frac{h^3}{6}u_{yyy}} + \frac{h^4}{24}u_{yyyy} + O(h^5)\\ &+ u + \cancel{hu_y} + \frac{h^2}{2}u_{yy} + \cancel{\frac{h^3}{6}u_{yyy}} + \frac{h^4}{24}u_{yyyy} + O(h^5)\\ \end{align*} $$=4u+h^2(u_{xx}+u_{yy})+\frac{1}{12}h^4(u_{xxxx}+u_{yyyy})+O(h^5)$$ So, $4[u_{i-1,j} + u_{i+1,j} + u_{i,j-1} + u_{i,j+1}]=$ $$=16u+4h^2(u_{xx}+u_{yy})+\frac{1}{3}h^4(u_{xxxx}+u_{yyyy})+O(h^5)$$\
Next, we will find $u_{i-1,j-1} + u_{i-1,j+1}$ \begin{align*} &= u - h(u_x + u_y) + \frac{h^2}{2}(u_{xx} + \cancel{2u_{xy}} + u_{yy}) - \frac{h^3}{6}(u_{xxx} + \cancel{3u_{xxy}} + 3u_{xyy} + \cancel{u_{yyy}}) + \frac{h^4}{24}(u_{xxxx} + \cancel{4u_{xxxy}} + 6u_{xxyy} + \cancel{4u_{xyyy}} + u_{yyyy}) + O(h^5)\\ &+ u - h(u_x - u_y) + \frac{h^2}{2}(u_{xx} \cancel{- 2u_{xy}} + u_{yy}) - \frac{h^3}{6}(u_{xxx} \cancel{- 3u_{xxy}} + 3u_{xyy} \cancel{- u_{yyy}}) + \frac{h^4}{24}(u_{xxxx} \cancel{- 4u_{xxxy}} + 6u_{xxyy} \cancel{- 4u_{xyyy}} + u_{yyyy}) + O(h^5)\\ \end{align*}
$$=2u-2hu_x+h^2(u_{xx}+u_{yy})-\frac{h^3}{3}(u_{xxx}+3u_{xyy})+\frac{h^4}{12}(u_{xxxx}+6u_{xxyy}+u_{yyyy})+O(h^5)$$
Then, similarly, $u_{i+1,j-1} + u_{i+1,j+1}$ \begin{align*} &= u + h(u_x - u_y) + \frac{h^2}{2}(u_{xx} - 2u_{xy} + u_{yy}) + \frac{h^3}{6}(u_{xxx} - 3u_{xxy} + 3u_{xyy} - u_{yyy}) + \frac{h^4}{24}(u_{xxxx} - 4u_{xxxy} + 6u_{xxyy} - 4u_{xyyy} + u_{yyyy}) + O(h^5)\\ &+ u + h(u_x + u_y) + \frac{h^2}{2}(u_{xx} + 2u_{xy} + u_{yy}) + \frac{h^3}{6}(u_{xxx} + 3u_{xxy} + 3u_{xyy} + u_{yyy}) + \frac{h^4}{24}(u_{xxxx} + 4u_{xxxy} + 6u_{xxyy} + 4u_{xyyy} + u_{yyyy}) + O(h^5)\\ \end{align*}
$$=2u+2hu_x+h^2(u_{xx}+u_{yy})+\frac{h^3}{3}(u_{xxx}+3u_{xyy})+\frac{h^4}{12}(u_{xxxx}+6u_{xxyy}+u_{yyyy})+O(h^5)$$
Finally, we compute $u_{i-1,j-1} + u_{i-1,j+1}+u_{i+1,j-1} + u_{i+1,j+1}$
\begin{align*} &=2u\cancel{-2hu_x}+h^2(u_{xx}+u_{yy})\cancel{-\frac{h^3}{3}(u_{xxx}+3u_{xyy})}+\frac{h^4}{12}(u_{xxxx}+6u_{xxyy}+u_{yyyy})+O(h^5)\\ &+2u+\cancel{2hu_x}+h^2(u_{xx}+u_{yy})+\cancel{\frac{h^3}{3}(u_{xxx}+3u_{xyy})}+\frac{h^4}{12}(u_{xxxx}+6u_{xxyy}+u_{yyyy})+O(h^5)\\ \end{align*}
$$=4u+2h^2(u_{xx}+u_{yy})+\frac{1}{6}h^4(u_{xxxx}+6u_{xxyy}+u_{yyyy})+O(h^5)$$
Then, $$\Delta_9^hu_{i,j}=\frac{1}{6h^2}[16u+4h^2(u_{xx}+u_{yy})+\frac{1}{3}h^4(u_{xxxx}+u_{yyyy})+4u+2h^2(u_{xx}+u_{yy})+\frac{1}{6}h^4(u_{xxxx}+6u_{xxyy}+u_{yyyy})-20u+O(h^5)]$$ $$=\frac{1}{6h^2}[\cancel{16u}+4h^2(u_{xx}+u_{yy})+\frac{1}{3}h^4(u_{xxxx}+u_{yyyy})+\cancel{4u}+2h^2(u_{xx}+u_{yy})+\frac{1}{6}h^4(u_{xxxx}+6u_{xxyy}+u_{yyyy})\cancel{-20u}+O(h^5)]$$ $$=\frac{1}{6h^2}[6h^2(u_{xx}+u_{yy})+\frac{1}{3}h^4(u_{xxxx}+u_{yyyy})+\frac{1}{6}h^4(u_{xxxx}+6u_{xxyy}+u_{yyyy})+O(h^5)$$ $$=(u_{xx}+u_{yy})+\frac{1}{18}h^2(u_{xxxx}+u_{yyyy})+\frac{1}{36}h^2(u_{xxxx}+6u_{xxyy}+u_{yyyy})+O(h^3)$$
Perfect! $$\begin{align}\frac1{18}h^2\left(u_{xxxx}+u_{yyyy}\right)+\frac1{36}h^2\left(u_{xxxx}+6u_{xxyy}+u_{yyyy}\right)&=\frac1{36}h^2\left(2u_{xxxx}+2u_{yyyy}+u_{xxxx}+6u_{xxyy}+u_{yyyy}\right)\\ &=\frac1{36}h^2\left(3u_{xxxx}+6u_{xxyy}+3u_{yyyy}\right)\\ &=\frac1{12}h^2\left(u_{xxxx}+2u_{xxyy}+u_{yyyy}\right)\end{align}$$ So you nailed it. Well, except that you didn't notice that all the odd-order terms canceled when you added $u_{i-1,j}+u_{i+1,j}$, $u_{i,j-1}+u_{i,j+1}$, $u_{i-1,j-1}+u_{i+1,j+1}$, and $u_{i-1,j+1}+u_{i+1,j-1}$ so your error terms should have been $O\left(h^6\right)$ and then when you divided by $h^2$ your eventual error would have been $O\left(h^4\right)$.