令u=x+y,v=x-y，则x=(u+v)/2y=(u-v)/2故f（x＋y,x－y）＝f(u,v)=[(u+v)/2]^2＋[(u-v)/2]^2＋e^[(u+v)/2*(u-v)/2]=（u²+v²）/2+e^[(u²-v²)/4]故f(x,y)=（x²+y²）/2+e^[(x²-y²)/4]
