# Fast Simulation of Laser Heating Processes on Thin Metal Plates with FFT Using CPU/GPU Hardware
## Appendix D Scheme 4--Odd-Symmetry 2D FFT  
In this scheme, consider the real part \(D_{kl}\) (instead of the complex one) of Equation (A9) as follows:  
\[ \begin{split} D_{kl}&=\frac{1}{4}\mathcal{R}e\left[ \sum_{n=0}^{N-2}\mathcal{R}e\left[\sum_{m=0}^{M-2}\theta_{nm}e^{-\frac{2\pi k}{2 }k(m+1)}\right]e^{-\frac{2\pi l}{2N}l(n+1)}\right]\\ &-\frac{1}{4}\mathcal{R}e\left[\sum_{n=0}^{N-2}\mathcal{R}e\left[ \sum_{m=M+1}^{M-1}\theta_{(2M-m-1)n}e^{-\frac{2\pi l}{2N}lmn}\right]e^{-\frac {2\pi l}{2N}l(n+1)}\right]\\ &-\frac{1}{4}\mathcal{R}e\left[\sum_{n=N+1}^{2N-1}\mathcal{R}e \left[\sum_{m=0}^{M-2}\theta_{m(2N-n-1)}e^{-\frac{2\pi l}{2N}l(m+1)}\right]e^ {-\frac{2\pi l}{2N}ln}\right]\\ &+\frac{1}{4}\mathcal{R}e\left[\sum_{n=N+1}^{2N-1}\mathcal{R}e \left[\sum_{m=M+1}^{2M-1}\theta_{(2M-m-1)(2N-n-1)}e^{-\frac{2\pi}{2N}km} \right]e^{-\frac{2\pi l}{2N}ln}\right]\\ k&=0,1,\ldots M-1,\qquad l=0,1,\ldots,N-1\end{split}  \tag{10}\]  
which in fact consists of the cosine parts of the Fourier series:  
\[ \begin{split} D_{kl}&=\frac{1}{4}\sum_{n=0}^{N-2} \sum_{m=0}^{M-2}\theta_{mn}\cos\left(\frac{2\pi k(m+1)}{2M}\right)\cos\left( \frac{2\pi l(n+1)}{2N}\right)\\ &-\frac{1}{4}\sum_{n=0}^{N-2}\sum_{m=M+1}^{2M-1}\theta_{(2M-m-1) n}\cos\left(\frac{2\pi km}{2M}\right)\cos\left(\frac{2\pi l(n+1)}{2N}\right)\\ &-\frac{1}{4}\sum_{n=N+1}^{2N-1}\sum_{m=0}^{M-2}\theta_{(2N-n-1) }\cos\left(\frac{2\pi k(m+1)}{2M}\right)\cos\left(\frac{2\pi ln}{2N}\right)\\ &+\frac{1}{4}\sum_{n=N+1}^{2N-1}\sum_{m=M+1}^{2M-1}\theta_{(2M-m -1)(2N-n-1)}\cos\left(\frac{2\pi km}{2M}\right)\cos\left(\frac{2\pi ln}{2N} \right)\\ k&=0,1,\ldots M-1,\qquad l=0,1,\ldots,N-1\end{split}  \tag{11}\]  
Consider the second term of the previous expansion, and the change of the series variable \(m\gets 2M-m-1\). Since \(\cos(x)=\cos(-x)\) and \(\cos(x)=\cos(x+2\pi k)\) (with \(k\in\mathbb{N}_{+}\)), then the following equation holds:  
\[ \begin{split}&\frac{1}{4}\sum_{n=0}^{N-2}\sum_{m=M+1}^{2M-1} \theta_{(2M-m-1)n}\cos\left(\frac{2\pi km}{2M}\right)\cos\left(\frac{2\pi l(n +1)}{2N}\right)\\ &=\frac{1}{4}\sum_{n=0}^{N-2}\sum_{m=0}^{M-2}\theta_{mn}\cos \left(\frac{2\pi k(2M-1-m)}{2M}\right)\cos\left(\frac{2\pi l(n+1)}{2N}\right) \\ &=\frac{1}{4}\sum_{n=0}^{N-2}\sum_{m=0}^{M-2}\theta_{mn}\cos \left(\frac{2\pi k(m+1)}{2M}\right)\cos\left(\frac{2\pi l(n+1)}{2N}\right)\end{split}  \tag{12}\]  
Applying the same procedure to the fourth term in Equation (11), we obtain:  
\[ \begin{split}&\frac{1}{4}\sum_{n=N+1}^{2N-1}\sum_{m=M+1}^{2M-1} \theta_{(2M-m-1)(2N-n-1)}\cos\left(\frac{2\pi km}{2M}\right)\cos\left(\frac{2 \pi ln}{2N}\right)\\ &=\frac{1}{4}\sum_{n=N+1}^{2N-1}\sum_{m=0}^{M-2}\theta_{m(2N-n-1) }\cos\left(\frac{2\pi k(2M-1-m)}{2M}\right)\cos\left(\frac{2\pi ln}{2N}\right) \\ &=\frac{1}{4}\sum_{n=N+1}^{2N-1}\sum_{m=0}^{M-2}\theta_{m(2N-n-1) }\cos\left(\frac{2\pi k(m+1)}{2M}\right)\cos\left(\frac{2\pi ln}{2N}\right) \end{split}  \tag{13}\]Substituting Equations (A12) and (A13) into Equation (A11):  
\[ \begin{split} D_{kl}&=\frac{1}{4}\sum_{n=0}^{N-2}\sum_ {m=0}^{M-2}\theta_{mn}\cos\left(\frac{2\pi k(m+1)}{2M}\right)\cos\left(\frac{2 \pi l(n+1)}{2N}\right)\\ &-\frac{1}{4}\sum_{n=0}^{N-2}\sum_{m=0}^{M-2}\theta_{mn}\cos \left(\frac{2\pi k(m+1)}{2M}\right)\cos\left(\frac{2\pi l(n+1)}{2N}\right)\\ &-\frac{1}{4}\sum_{n=N+1}^{2N-1}\sum_{m=0}^{M-2}\theta_{m(2N-n-1) }\cos\left(\frac{2\pi k(m+1)}{2M}\right)\cos\left(\frac{2\pi ln}{2N}\right)\\ &+\frac{1}{4}\sum_{n=N+1}^{2N-1}\sum_{m=0}^{M-2}\theta_{m(2N-n-1) }\cos\left(\frac{2\pi k(m+1)}{2M}\right)\cos\left(\frac{2\pi ln}{2N}\right)\\ k&=0,1,\ldots M-1,\qquad l=0,1,\ldots,N-1\end{split} \] (A14)  
in which the first and second terms cancel out, as well as terms three and four, respectively. Therefore:  
\[\forall_{k,l}\qquad D_{kl}=0\] (A15)  
Finally, we add \(D_{kl}\) to Equation (A9):  
\[