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How can we prove $\bigtriangledown = \bigtriangleup E^{-1}$?

where,$\bigtriangleup \rightarrow \text{ Forward difference operator }$ $\bigtriangledown \rightarrow \text{ Backward difference operator }$ $E \rightarrow \text{ Shift operator }$ I tried to utilize the method of seperation of symbol but not quite there...

Give some hints for this proof.

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up vote 4 down vote accepted

How about writing $\bigtriangleup(f(x))=f(x+1)-f(x)$, $E^{-1}(f(x))=\ldots$, so $\bigtriangleup E^{-1}(f(x))=$?

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Got it! One thing in my book $\bigtriangleup(f(x))$ is defined as $f(x+h)-f(x)$ where the consecutive values of $x$ differs by $h$ I have seen the proof of this derivation, I do agree that in problem solving I mean in while composing the difference table we do use $\bigtriangleup(f(x))=f(x+1)-f(x)$ but aren't we should use $f(x+h)-f(x)$ while proving equalities like this one? –  Quixotic Feb 7 '11 at 15:32
    
It depends upon what definition you are using for $\bigtriangleup$. In analyzing polynomials they often use +1. In calculus I have seen h, because we think of h as small. The proof is the same either way as long as you are consistent. –  Ross Millikan Feb 7 '11 at 15:35
    
Hm thanks. –  Quixotic Feb 7 '11 at 15:46
    
I have also seen the forward difference with a subscript $\bigtriangleup_h$ to show you what the size of the interval is. –  Ross Millikan Feb 7 '11 at 16:41
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