Finding Distance Function with Increasing Outputs

[onako]

For determination of the distance from a certain point, a function should be used. Namely, the input values 'x' are from N (1,2,3,4,5...), and the corresponding outputs 'y' from R, but such that the sequence 'y' (relative to 'x') is increasing;

$y(1)<y(2)<y(3)<y(4)<...$ \\
but to satisfy \\
y(1)>[y(2)-y(1)]>[y(3)-y(2)]>[y(4)-y(3)]>...

The function that satisfies this is sqrt(x), but I'm interested in other possible functions that satisfy the above. The 'derivatives' of sqrt(x) should also work.

Thanks

 

[mathman]

Any function of the form x^a, where 0 < a < 1. Also log(x).

Derivatives will not work. They are decreasing functions.

 

[onako]

I put 'derivatives' with '' emphasis, to differentiate it from real ones, but this is a language obstacle. What is meant is x^a, 0<a<1. I would like to hear more suggestions on these kind of functions. I supposed to test their behaviour and to select the most appropriate one.
Note that the condition
y(1)>[y(2)-y(1)]>[y(3)-y(2)]>[y(4)-y(3)]>...

is better to be
y(1)=[y(2)-y(1)]=[y(3)-y(2)]>... (or, for the first few, approximately equal might be appropriate, but
to satisfy the first criterion from my first post).

Any help is highly appreciated.

 

[mathman]

You could be more specific as to what you have in mind by "most appropriate one".

 

[onako]

The most appropriate one would be the one satisfying the following:

1) y(1)<y(2)<y(3)<y(4)<...
2) y(1)=[y(2)-y(1)]=[y(3)-y(2)]>... first few '=' signs read as approximately equal. Possibly the number of
'few' should be determined by a parameter. Then [y(3)-y(2)]>[y(4)-y(3)]>... continues as usual.
Thanks

 

[mathman]

Take any of the examples I provided to start after the "first few". Get the tangent line at some point near the beginning and use values along the tangent line for the first few. Finally add (or subtract) a constant to all terms to get y(1)=[y(2)-y(1)].

 

[onako]

Thanks.
How could I then ensure the requirements are satisfied given a parameter 'k' corresponding to 'first few'. In other words, if 'first few' is 40, and I use 'log(x+1)', how to proceed?

 

[mathman]

Place the tangent line at x=40 and then add a constant to the whole thing so that the value at 1 is what you want.

 

[onako]

Thanks.