Basic halving in a function

Question

I'm writing a function and per my experimenting I've found that my output needs to increment by 1 every time the input's distance to 1 is cut in half.

INPUT    OUTPUT
0.0      1
0.5      2
0.75     3
0.875    4
0.9375   5

So on and so forth. I know this can't be too complicated and I figure it has to be log or sqrt but it's just not clicking.

How can I solve this little math riddle ?

** EDIT **

Many good comments and answers so thanks all. The working c# came out like so

(int)((1-VAR_1) - (Mathf.Log(1-VAR_2))/Math.Log(2));

Answer 1

Just notice that if you reverse the roles of input and output you can see that

$$n\to f(n)=1-\frac{1}{2^{n-1}}$$ Now to switch back the inputs-outputs you need to invert this function, so the function you would be seeking is

$$m\to g(m)=1-\frac{\log(1-m)}{\log 2} $$

Answer 2

$$\text{input}=1-\left(\frac12\right)^{\text{output}-1}$$

$$\text{output}=1-\frac{\ln(1-\text{input})}{\ln(2)}$$

Answer 3

If you were to swap input/outputs you you'd have

$1\to 0$
$2 \to \frac 12$
$3\to \frac 34$
$4\to \frac 78$

And for any $k\to m$ we'd have $k+1$ to $1 - \frac {1-m}2$.

It's easy to so the pattern that the denominator is always $2^{k-1}$ and the numerator is $ ({2^{k-1} -1})$. and we have $k \to \frac {2^{k-1}-1}{2^{k-1}} = 1 - \frac 1{2^{k-1}}$.

So $k$ is our output and $1-\frac 1{2^{k-1}}$ is our input we have

So we have input expressed in terms of output. We want output expressed in terms of input.

Let $k$ be the output and $j$ be the input.

We have $j= 1-\frac 1{2^{k-1}}$. We need to reverse that to get $k = \text{something to do with }j$.

So.... solve for $k$.

$j = 1-\frac 1{2^{k-1}}$

$j-1 = -\frac 1{2^{k-1}}$

$\frac 1{2^{k-1}}=1-j$

$2^{k-1} = \frac 1{1-j}$

$\log_2 2^{k-1} = \log_2 \frac 1{1-j}$

$k-1 = \log_2\frac 1{1-j}$

$k = 1+ \log_2\frac 1{1-j} = 1-\log_2 (1-j)$.

Not that the requires $0 \le j < 1$.

If To do an example if $j = \frac {31}{32}$ we get

$k = 1-\log_2 (1-\frac {31}{32})=1-\log_2(1-\frac {31}{32})=1-\log_2(\frac 1{32})=1-\log_2 2^{-5} = 1-(-5) = 6$. Just as we expected!

And if we have an arbitrary value. So $j=\frac 23$ (meaning we'd expect $j$ to be between $2$ and $3$ [as $\frac 12 < \frac 23 < \frac 34$]) we have

$k = 1-\log_2(1-\frac 23)=1-\log_2 \frac 13 = 1-\log_2 3^{-1} = 1+\log_2 3 = 1+ \frac {\log_{10} 3}{\log_{10}2} = 1+\frac {0.47712125471966243729502790325512}{0.47712125471966243729502790325512} = 1+ 1.5849625007211561814537389439478=2.5849625007211561814537389439478$