Arcade Fire said:I take it that the '12' is a typo, and should actually be a '2'?
FrostedNipple said:oo yer, if it is a 12 that isnt actually going to work!!!!
calnen said:That confused me as wellIf it's 'x to the power of 12' then it becomes rather more complicated.
Arcade Fire said:First thing that strikes me is that x=1 is an obvious solution.
calnen said:That confused me as wellIf it's 'x to the power of 12' then it becomes rather more complicated.
Angus-Higgins said:If you think of it as a straight line graph would that help?
(I mean that if you plot (1,1000) and then (12,255000)).
Would that work? (I am not thinking terribly deeply here).
Angus Higgins
What you've posted there isn't linear growth, it's geometric (or exponential) growth. An example of linear growth would be 1000 in month 1, 2000 in month 2, 3000 in month 3 etc.Topgun said:doh. it is not a quadratic.
Basically I'm starting a fictional .com business earning £1000 in the first month with a total of £255,000 earnt by the end of month 12. How much would I have to make each month asuming the growth in linear?
I got this far
a + ar + ar^2 + ar^3.... ar^11= Sum
a = 1
Sum = 255
r = ?
n = 12
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I think the flaw comes in the line where you start with "Assume x=254..." and end with "...and therefore x=1".SpeedFreak said:254 = 255x - x^12
x(255-x^11)= 254
therefore x= 254 and
255-x^11=254 ==x^11=1 therefore we can say x=1
x=1 and x =254 unless i am flawed in my thinking, it's been a few years since i've had to deal with these.
Topgun said:I'm not sure I follow. are you talking about graphing cumulative income? Because by the 12th month I want to have made 255k all together.
SpeedFreak said:254 = 255x - x^12
x(255-x^11)= 254
therefore x= 254 and
255-x^11=254 ==x^11=1 therefore we can say x=1
x=1 and x =254 unless i am flawed in my thinking, it's been a few years since i've had to deal with these.