create a list fulfill special requirement by specifying only the
- To: mathgroup at smc.vnet.net
- Subject: [mg130630] create a list fulfill special requirement by specifying only the
- From: Joug Raw <jougraw at gmail.com>
- Date: Fri, 26 Apr 2013 23:10:58 -0400 (EDT)
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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?