Re: create a list fulfill special requirement by

*To*: mathgroup at smc.vnet.net*Subject*: [mg130641] Re: create a list fulfill special requirement by*From*: Ray Koopman <koopman at sfu.ca>*Date*: Sun, 28 Apr 2013 01:00:39 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net

f[a_,x_] := x/(x + a(1-x)), with a > 0, maps [0,1]->[0,1]. The slope at (0,0) is 1/a, the slope at (1,1) is a. Try f[a,Range[0,13]/13]*3.5+2.5 with a = 10 or so. On Fri, Apr 26, 2013 @ 08:10 PM, Joug Raw <jougraw at gmail.com> wrote: > > I want to generate a list of number bewteen 2.5 and 6 with 14 elements > totally. > > The number of the list should fulfill such requirement that it will be very > tight at the beginning then loos at the end. That means the numbers started > from 2.5 will be more closed then the number at around 6. Something like > {2.5, 2.52, 2.55, 2.60, 2.71, 2.80, 2.95, 3.1...4.2, 5, 6} > > I tried exponential as well as power function for generating this list with > the following code: > > 1. list = With[{a = 3}, > Array[a^# &, 14, {x, y}] /. > Join[Solve[a^x == 2.5, x][[-1]], Solve[a^y == 6., y][[-1]]]] > > which fit the list points with exponential function > > 2. list = With[{a = 5}, > Array[#^a &, 14, {x, y}] /. > Join[Solve[x^a == 2.5, x][[-1]], Solve[y^a == 6., y][[-1]]]] > > which fit the list points with power function > > The 1st one gives me output > {2.5, 2.67416, 2.86045, 3.05972, 3.27286, 3.50086, 3.74474, 4.00561, \ > 4.28466, 4.58314, 4.90242, 5.24393, 5.60924, 6.} > > The 2nd one gives me output > {2.5, 2.6895, 2.89031, 3.10295, 3.32792, 3.56575, 3.81699, 4.0822, \ > 4.36195, 4.65683, 4.96744, 5.29441, 5.63838, 6.} > > They are indeed tight at the end closed to 2.5 and lose at the end near 6, * > BUT* it is not enough for me, what I need is VERY tight and VERY lose at > the two ends such as {2.5, 2.52, 2.6, 2.69... > and ...3.0, 4.2, 6.0} and in between the transition should be gradually and > smoothly done. > > I tired to adjust the number "a" in the exponential function and the power > function in my codes. Even if I gave it a very big value for a, this dose > not really change so much the tight and the lose of the numbers at both > ends. > > What function should I use to generate the list as I wished? > > And, how to make my codes a bit smarter? The code I used I think is bit > more complected. Any suggestion that I could bypass the "With" and the "/." > in my code which still allow me to define directly the two ends{2.5, 6} and > size of "a" in a more straightforward and easier way?