Complicated condition inside piecewise function

Question

I want to define the piecewise function

$$f(x)=\begin{cases}-1,&x\ge1\\\frac1n,&\frac1{n+1}\le x<\frac1n,\; n\in\Bbb N_{>0}\\0,&x\le 0\end{cases}$$

in this form, but I don't know if this is possible (I know that we can define this function using a ceil function, but I'm trying to see if it is possible to write the function in the above form).

The problem is that I don't know if it is possible to define the condition

$$\frac1{n+1}\le x<\frac1n,\; n\in\Bbb N_{>0}$$

Thank you in advance

Answer 1

f[x_] := Piecewise[{{-1, x >= 1}, {0, 
    x <= 0}, {1/Quiet@Reduce[1/(n + 1) <= x < 1/n, n, Integers][[2]], 
    0 < x < 1}}]

f[0.4]

1/2

a = RandomReal[1, 10]

{0.946522, 0.753738, 0.520006, 0.50627, 0.684279, 0.57814, 0.470449, 0.860181, 0.639912, 0.0999967}

f /@ a

{1, 1, 1, 1, 1, 1, 1/2, 1, 1, 1/10}

Answer 2

Another way is as follows.

f = Function[x, Piecewise[{{-1, x > 1}, {0, x <= 0}, {1/Floor[1/x], x > 0 && x <= 1}}]]

Answer 3

this I think is a little closer to the actual statement:

f[x_?NumericQ] := Piecewise[{{-1, x >= 1}, {0, x < 0},
   Module[{n,nn},
       {1/nn,
       (nn = n /. First@FindInstance[ 1/n >= x > 1/(n + 1) , n, Integers, 1]) >  0 }]}]

note the order of evaluation allows us to use the result of the logical test in the returned expression.