Population growth using e^x versus using 2^x

huskyItalian · 2007-12-24 22:30:34

I know that population growth, for example, DOUBLES every time period. Since this is the case, couldn't we use the equation y=2^x to describe DOUBLING of the population rather than the equation e^x? Lets say a population of bacteria doubles everyday. Lets say starting population is 2 bacteria. Using the formula y=2^x, you get the following population growth: day 1= 2 bacteria cells; day 2 has 4 bacteria, day 3 has 8 bacteria (and each day it doubles: 2, 4, 8, 16, 32, 64.....). This SEEMS so perfect for the situation, BUT every text book recommends the equation e^x to describe population growth. Why is this? Could both equations be correct?

JaneFairfax · 2007-12-24 22:39:15

What textbook tells you e[sup]x[/sup] describes growth by doubling? y = e[sup]x[/sup] is an equation of exponential growth, but not by doubling.

luca-deltodesco · 2007-12-24 23:04:03

you can use eulers numbers as the base of the exponent if you want using the identity

mathsyperson · 2007-12-25 03:52:53

What luca said. In this case, 2^x is the best way to go, but in general exponential growth is described by e^(αx), where α is some constant that defines how fast the growth is.

George,Y · 2007-12-25 23:34:11

2^(ax) makes more sense.
But mathematicians are lazy, they don't count difference over each multiplication, they simply imagine the differential.
And e^(bx) is most the convenient in doing differential.

George,Y · 2007-12-25 23:43:21

1 D
2 1
4 2
8 4
16 8

Taking difference of 2^k is fairly easy, but it needs specifying steps, which seems hard. In the end, they abandon 2^k and resort to fit a line with the function e^(bx), which makes less sense. However, you can fit a line with 2^(ax) and the relationship between a and b is

b/a=ln2

Or b/a=log(2/e) , whatever log,
or b log(e)=a log(2)

whichever you find easy to memory.

Ricky · 2007-12-26 09:21:24

But mathematicians are lazy, they don't count difference over each multiplication, they simply imagine the differential.

This is starting to get ridiculous, George.

I know that population growth, for example, DOUBLES every time period.

The above statement is incorrect. Population growth does not double every time period. It can, but it doesn't have to. When modeling growth, the assumption is that the change in population is directly proportional to the population size. This gives a differential equation, the solution to which is C*e^(kt).

This is a much more accurate model than a power series. For you see, the power series is derived from the assumption that every generation comes at exactly the same time. Let's say a generation is 20 years. Power series calculates what would happen if everyone gave birth at the 20 year mark.

However, this is not the case. You start a P(0), but within the first year, people are having kids. This adds to your population before you get to the 20 mark, where they have kids again. As you can see, this leads to the natural model of compounding continuously, which is what the differential equation gives us.

It's rather beautiful.

George,Y · 2007-12-26 09:25:52

But there is absolutely no instant growth rate in the real world in this case, don't you agree?

Ricky · 2007-12-26 09:34:07

George,Y wrote:

But there is absolutely no instant growth rate in the real world in this case, don't you agree?

This model does not take into account premature deaths, deaths after a certain lifespan, nor infertility. Along with these, this model does not take into account that people can only give birth after 13 years of growth and development.

But this has nothing to do with laziness George. huskyItalian is probably either a middle/high school student, or taking freshman calculus. The model does not take these factors into account because it would complicate things beyond belief.

George,Y · 2007-12-26 10:47:08

"Along with these, this model does not take into account that people can only give birth after 13 years of growth and development."

Yes, that's what 2^(bx) model fitting is superior on.

Given fitting models, e^(ax) and 2^(bx) require nearly the same amount of effort to take, after all you have to run a regression. And why is e^(ax) preferred? The only reason I can think of is it is easy to differentiate, which makes no sense in this case however.

Last edited by George,Y (2007-12-26 10:47:40)

Ricky · 2007-12-26 14:00:31

The only reason I can think of is it is easy to differentiate, which makes no sense in this case however.

I already described how the exponential function compounds the population growth continuously, which is a more accurate representation of population growth. You did not object to this, so I will assume you agree with me on this point. But then how can you say that you see no reason to use the exponential function?

The power function has it's applications and may be useful when doing some population growth problems. However, I have not seen any where it would be. So try this sample problem, would you?

For population growth problems, the input is what percent of the population produces offspring and the starting population. So George, given that 60% of the starting population produces offspring and there are 100 people to start with, can you produce a power function which models this? If we assume that everyone produces children at exactly the same moment (which is a restriction that the power function forces us to use), then the function should look like the following:

f(0) = 100
f(1) = 160
f(2) = 256
f(3) = 418.6
...

Please produce this function as a power function, f(x) = 2^(bx) + 99.

George,Y · 2007-12-27 08:47:20

I object continuous growth.
Clearly, no multiplication is on a continuous basis, it is on an discrete basis.

First and foremost, f(3) = 418.6 is wrong. Either 418 cells or 419 cells and you cannot arbitarily define what a 0.6 cell is. Even if you do that, I disagree. And because the number of cells is Discrete!

And further, it is easy to fit in A[sub]0[/sub]2[sup]bx[/sup] to this model.
f(x)=100*2^(xlog[sub]2[/sub]1.6)

Besides, Ricky I don't know how you come up with the +99 term, which doesn't make sense at all. Even in e function, it's displayed as Ce^(kT) without a "+". You come up with this term just to puzzle me huh? How about this game. You plot in f(x)=e^(ax)+99?

And you know too well in the real world you cannot get the data like this, which just be perfectly fit by a model. Instead, we got to have data that just cannot be fit by either model because models created by us are just simplification. Like this
100 156 270 500
And it makes no difference you fit in the curve with Ce[sup]at[/sup] or C2[sup]bt[/sup]
And I guarantee if you run a regression for each of them, you end up with
a-estimate=b-estimate/ln2
Like the results in Post 6.

Ricky · 2007-12-27 11:21:15

Clearly, no multiplication is on a continuous basis, it is on an discrete basis.

It's a model.

First and foremost, f(3) = 418.6 is wrong. Either 418 cells or 419 cells and you cannot arbitarily define what a 0.6 cell is. Even if you do that, I disagree. And because the number of cells is Discrete!

It's a model.

Besides, Ricky I don't know how you come up with the +99 term, which doesn't make sense at all. Even in e function, it's displayed as Ce^(kT) without a "+". You come up with this term just to puzzle me huh? How about this game. You plot in f(x)=e^(ax)+99?

Sorry, I was trying to simplify it for you. I did not know you wished to have another parameter in front of the base. Then try it with: f(t) = C*2^(kt) + D.

And you know too well in the real world you cannot get the data like this, which just be perfectly fit by a model. Instead, we got to have data that just cannot be fit by either model because models created by us are just simplification. Like this

It's a model.

George,Y · 2007-12-28 10:16:29

"It's a model."
Base 2 is a Better model, based on the reasons I have stated.

Then try it with: f(t) = C*2^(kt) + D.
I have already done it
f(x)=100*2^(xlog[sub]2[/sub]1.6)
C=100, k=log[sub]2[/sub], D=0
Doesn't it fit in the data?

Ricky · 2007-12-28 18:06:51

Base 2 is a Better model, based on the reasons I have stated.

No, it is a different model. It has shortcomings that I addressed, and showed why the exponential model does not have these shortcomings. That isn't to say that the exponential model is perfect or even good, but it does do better than the power model in some respects.

Oh, and George, maybe you were hoping I wouldn't notice, but log base 2 of 1.6 is not a rational number. You just used a number that you don't even believe exists! Furthermore, you might as well just say your function is:

f(t) = 100*1.6^t

Math Is Fun Forum

#1 2007-12-24 22:30:34

Population growth using e^x versus using 2^x

#2 2007-12-24 22:39:15

Re: Population growth using e^x versus using 2^x

#3 2007-12-24 23:04:03

Re: Population growth using e^x versus using 2^x

#4 2007-12-25 03:52:53

Re: Population growth using e^x versus using 2^x

#5 2007-12-25 23:34:11

Re: Population growth using e^x versus using 2^x

#6 2007-12-25 23:43:21

Re: Population growth using e^x versus using 2^x

#7 2007-12-26 09:21:24

Re: Population growth using e^x versus using 2^x

#8 2007-12-26 09:25:52

Re: Population growth using e^x versus using 2^x

#9 2007-12-26 09:34:07

Re: Population growth using e^x versus using 2^x

#10 2007-12-26 10:47:08

Re: Population growth using e^x versus using 2^x

#11 2007-12-26 14:00:31

Re: Population growth using e^x versus using 2^x

#12 2007-12-27 08:47:20

Re: Population growth using e^x versus using 2^x

#13 2007-12-27 11:21:15

Re: Population growth using e^x versus using 2^x

#14 2007-12-28 10:16:29

Re: Population growth using e^x versus using 2^x

#15 2007-12-28 18:06:51

Re: Population growth using e^x versus using 2^x

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