Are we allowed, by the definition of induction, to assume that $f_{1}(x)$ still has the same formula, even though we switched $n$?

by ArielK   Last Updated November 18, 2018 11:20 AM

I'm pretty sure i'm supposed to demonstate this by induction but i think there is a problem with the way the proposition is stated that impossibilitates that, I need to prove that

For all natural number $n$, $f_{0}(x) = \frac{x}{x+1} \& f_{n+1}(x)=(f_{0}(f_{n}(x))) \implies f_{n}(x) = \frac{x}{(n+1)x + 1}$

So if we first verify it for n=0 then the explicit formula for $f_{n}$ is given. But then, if we try to verify it for any other $n$, let's say $n=1$, we would have the formula for $f_{0}(x)$ as always, and we would also have that $f_{2}(x) = f_{0}(f_{1}(x))$ but how is that suppose to lead us to conclude the explicit formula for $f_{1}$ if we don't know what $f_{2}(x)$ looks like?

Did I get induction wrong?



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