Consider the function f: N --> R defined by f(0) = 2 f(n+1) = f(n) + n - 2 Compute f(1), f(2), f(3). I know how to do this, but I was just wondering what is "n" - I mean f(n) for f(1) is 2, but is n just 1 for f(1) or 0?
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Originally Posted by brumby_3 Consider the function f: N --> R defined by f(0) = 2 f(n+1) = f(n) + n - 2 Compute f(1), f(2), f(3). I know how to do this, but I was just wondering what is "n" - I mean f(n) for f(1) is 2, but is n just 1 for f(1) or 0? $\displaystyle f(1)=f(0+1)=f(0)+0-2~\&~f(2)=f(1+1)=f(1)+1-2$
Originally Posted by Plato $\displaystyle f(1)=f(0+1)=f(0)+0-2~\&~f(2)=f(1+1)=f(1)+1-2$ Thank you Plato!
For $\displaystyle f(n+1) = f(n) + n - 2$ make $\displaystyle n=0$ you get $\displaystyle f(0+1) = f(0) + 0 - 2 \implies f(1) = 2 + 0 - 2 = 0$ Now make $\displaystyle n=1$
Thanks pickslides, I've got it
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