Fórum de Matemática
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If \(f(x)\) is a real valued function discontinuous at all integral points lying in \(\left[0,n\right]\;\; n\in\mathbb{N}\) and if

(f(x))^2 = 1 forall \(0 \leq x \leq n\) then number of functions \(f(x)\) are

I have calculate it for n=2 and Getting 6 no. of function which is discontinuous in [0,2]

but how can i generalise it

Help required

Thanks

Hint:

Given a set \(A\subset\mathbb{R}\), consider the set \(S(A)\) contain any real valued function \(f\) discontinuous at all integral points lying in \(A\) such that \((f(x))^2 = 1\) for all \(x\in A\). You want to know how many elements has the set \(S([0,n])\) (i.e. the cardinality of \(S([0,n])\)).
Show that:
1-\(\sharp S([0,n])=\sharp S([0,n[)\) (i.e. there is one and only one way to extent a function in \(S([0,n[)\) to a function in \(S([0,n])\));
2-\(\sharp S([0,n+1])=3\sharp S([0,n[)\) (i.e. there are three and only three independent ways to extent a function in \(S([0,n[)\) to a function in \(S([0,n+1])\)).

Then you can conclude that \(\sharp S([0,n])=2\times 3^{n-1}\)

Notation: \([0,n[=\{x\in\mathbb{R}: 0\leq x < n\}\)

Thanks ruicarpenter got it

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