OK math people, I knew this 20 years ago but I haven’t the faintest idea now.
What do you call this kind of curve, and what equation would yield a set like this:
x y
1 32
2 48
3 56
4 60
5 62
6 63
Or in other words, the first iteration is 2^5, next is 2^5 + 2^4, next is 2^5 + 2^4 + 2^3, etc.
I never was much of a math person. All I can say is that it looks a lot like the logarithmic growh curves we used to see in biology class.
PS - this isn’t homework, what I’m doing is writing a letter to the sunrise alarm clock manufacturer to describe how the illumination cycle of their product really ought to work. I want to say that the voltage should increase ______ally instead of arithmetically, but I can’t come up with the right adjective.
Gah, I swapped the X and the Y. So I guess if anybody is very very quick on the trigger with the answer, I will have to ask for the inverse of that function.
Well, I don’t know that it’s a common series, at least not one with a name I know of. It’s a convergant series.
And damn you for making me try to recall my Dif. Eq. class. Anyways, it’s hard to illustrate a series on the board so I won;t try and write the proper equation.
[sub]hows that for a cop out?[/sub]
Yeah, absolutely. You’ve written down 100000[sub]2[/sub], then 110000[sub]2[/sub], then 111000[sub]2[/sub], etc. You can carry on flipping fractional bits on if you like, and you converge at 111111.11111…[sub]2[/sub], otherwise known as 64. Not logarithmic, since in any base, log(inf) = inf.
Thats not the equation you want. Hyperbolas asymptote at a diagonal line instead of flattening out. A logarithmic function gets you the general shape of the graph you want and looks like this. It does go up until infinity but I believe thats the shape you want.
It’s an exponential series, albeit a decaying one that’s been inverted. You could write it as
y(x) = (sum from 1 to x) (64 * 2[sup]-x[/sup])
which can in turn be evaluated in closed form to be y(x) = 64(1 - 2[sup]-x[/sup]). The 2[sup]-x[/sup] bit means that it’s exponential.
I would personally describe this as y “asymptoting to” 64, since although this curve is technically exponential, “exponential” often implies growth, not decay.
As it happens, what the human eye usually perceives as equal increments in brightness actually turn out to be exponential increments. In other words, if we think that “A is as much brighter than B as B is brighter than C”, it turns out that in terms of light energy output, A/B = B/C. Just a fun fact you might be interested in.
I missed the reply from MikeS - it looks similar to what I came up with, they might resolve to the same thing. I’ll confirm this when I have more time, have to leave for work now.
Not so. The asympotes of a hyperbola may be diagonal, but nearly flat, diagonal, but nearly vertical, or indeed, actually horizontal or vertical. Simplest example: y=1/x has asymptotes at x=0 and y=0 - vertical and horizontal.
