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Prove that for every $x$, we have $\Delta[f(x)+g(x)]=\Delta f(x)+ \Delta g(x)$. Thanks in advance.

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What is $\Delta(f(x))$ for you? –  Sigur Nov 11 '12 at 13:43
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The laplacian $\Delta$ is a linear differential operator. Don't you understand my answer? Well, try to explain what you are asking, please. –  Siminore Nov 11 '12 at 14:26
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u can solve this by using forward differences w.k.t D=delta (for my convinent) Df(x)=f(x+h)-f(x) LHS let a(x)=f(x)+g(x) Da(x)=a(x+h)-a(x) now substitute for a(x) D[f(x)+g(x)]=f(x+h)+g(x+h)-(f(x)+g(x))...................1 now take rhs of equ Df(x)=f(x+h)-f(x)......i Dg(x)=g(x+h)-g(x)......ii add i &ii lhs=rhs

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Since Notyathing said "derivative question", it's likely $\Delta$ is the Laplacian (or perhaps some other differential operator). It does not represent general differences as in your answer. –  Mark S. Nov 11 '12 at 15:50
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