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## Main Question or Discussion Point

**Proving - numerical analysis (separation of symbols)**

How do I prove this?

f(x+n) = f(x+n-1) + /\f(x+n-2) + ... + /\^(n-2) f(x+1) + /\^(n-1) f(x) +

/\^n f(x-1)

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- Thread starter irony of truth
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How do I prove this?

f(x+n) = f(x+n-1) + /\f(x+n-2) + ... + /\^(n-2) f(x+1) + /\^(n-1) f(x) +

/\^n f(x-1)

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I stumbled upon this problem as well... i can't really see how these two are equivalent without their corresponding binomial coefficients when f(x+n) = E^n f(x) = (1+/\)^n f(x) and (1+/\ )^n is expanded ^^;irony of truth said:Which of these must be true?

f(x+n) = f(x+n-1) + /\f(x+n-2) + ... + /\^(n-2) f(x+1) + /\^(n-1) f(x) +

/\^n f(x-1),

OR

f(x+n) = f(x+n-1) + /\f(x+n-2) + ... + /\^(n-2) f(x+1) + /\^(n-1) f(x) +

/\^n f(x)

?

This was the question given in my homework... I just doubt it because my

professor could have miswritten our assignment.

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