here is an example f(n)=1X2n-1 Carson from Farmington, UT 1/16/2013 Comments for me, the question is not stated clearly enough --- what are the values of "n" --- could you possibly re-phrase the question? if the values of n are 0, -1, -2, -3, etc., then we would have the following: I wonder if the following would help: f(0) = 2^(0-1) = 2^(-1) = 1/2 = 0.5 f(-1) = 2^(-1-1) = 2^(-2) = 1/(2^2) = 1/4 = 0.25 f(-2) = 2^(-2-1) = 2^(-3) = 1/(2^3) = 1/8 = 0.125 etc., etc., etc. - Herb K. 1/17/2013 add comment
for me, the question is not stated clearly enough --- what are the values of "n" --- could you possibly re-phrase the question? if the values of n are 0, -1, -2, -3, etc., then we would have the following: I wonder if the following would help: f(0) = 2^(0-1) = 2^(-1) = 1/2 = 0.5 f(-1) = 2^(-1-1) = 2^(-2) = 1/(2^2) = 1/4 = 0.25 f(-2) = 2^(-2-1) = 2^(-3) = 1/(2^3) = 1/8 = 0.125 etc., etc., etc. - Herb K. 1/17/2013
Because any number multiplied by 1 is itself...1 X n = n, you could restate this function as f(n)=2(n-1). Therefore f(0)=2-1=1/2=.5. f(1)=20=1 f(2)=21=2 etc. Ben H. 2/14/2013 add comment
Comments
for me, the question is not stated clearly enough --- what are the values of "n" --- could you possibly re-phrase the question?
if the values of n are 0, -1, -2, -3, etc., then we would have the following:
I wonder if the following would help: f(0) = 2^(0-1) = 2^(-1) = 1/2 = 0.5
f(-1) = 2^(-1-1) = 2^(-2) = 1/(2^2) = 1/4 = 0.25
f(-2) = 2^(-2-1) = 2^(-3) = 1/(2^3) = 1/8 = 0.125
etc., etc., etc.
- Herb K. 1/17/2013