I am asked to explain why the function of the exponential growth is always above the function of the logistic growth.
I am given the formula
y'[x] = r y[x] (1 - y[x]/b)
Still going with the same values of r and starter as used in the plots above, look at the differential equations
y'[x] = r y[x]
with y[0] = starter
and the solution of
y'[x] = r y[x] (1 - y[x]/b)
with y[0] = starter
with 0 < starter < b
and explain the statement:
No matter what positive b you go with, the logistic solution of the second differential equation has no choice but to run below the exponential solution of the first differential equation.
My answer looks like this:
When we take the logistic differential equation
y'[x] = r y[x] (1 - y[x]/b)
with y[0] small relative to b , then as x advances from 0 , the solution of this differential equation runs close to the exponential solution of
y'[x] = r y[x] .
If y[0] is small relative to b , then logistic growth mimics exponential growth at the start.
That is why some people call it a "controlled exponential growth"
with r and b > 0 and 0 < y[0] < b
must go into global scale with
Underscript[lim, x → Infinity] y[x] = b .
As a result, logistic growth is like exponential growth because if y[0] is small relative to b , then as x advances from 0, y[x] grows like an exponential function until it starts to move into its global scale.
But this answer does not satisfy my mentor.
This is all conceptually true, but it doesn't specify exactly why this would be the case. Consider, what portion of the logistic equation is the limiting portion. Why does it limit it in this way specifically? You are saying correct concepts, but you need to specifically explain what is happening to the derivative when the term (1-y[x]/b) has the specifics given. For instance, what happens to the (1-y[x]/b) term specifically when y[x]->b? What happens to the derivative in that case specifically that means that y[x] can't go beyond it.
Can somebody please help to understand this?
I am given the formula
y'[x] = r y[x] (1 - y[x]/b)
Still going with the same values of r and starter as used in the plots above, look at the differential equations
y'[x] = r y[x]
with y[0] = starter
and the solution of
y'[x] = r y[x] (1 - y[x]/b)
with y[0] = starter
with 0 < starter < b
and explain the statement:
No matter what positive b you go with, the logistic solution of the second differential equation has no choice but to run below the exponential solution of the first differential equation.
My answer looks like this:
When we take the logistic differential equation
y'[x] = r y[x] (1 - y[x]/b)
with y[0] small relative to b , then as x advances from 0 , the solution of this differential equation runs close to the exponential solution of
y'[x] = r y[x] .
If y[0] is small relative to b , then logistic growth mimics exponential growth at the start.
That is why some people call it a "controlled exponential growth"
with r and b > 0 and 0 < y[0] < b
must go into global scale with
Underscript[lim, x → Infinity] y[x] = b .
As a result, logistic growth is like exponential growth because if y[0] is small relative to b , then as x advances from 0, y[x] grows like an exponential function until it starts to move into its global scale.
But this answer does not satisfy my mentor.
This is all conceptually true, but it doesn't specify exactly why this would be the case. Consider, what portion of the logistic equation is the limiting portion. Why does it limit it in this way specifically? You are saying correct concepts, but you need to specifically explain what is happening to the derivative when the term (1-y[x]/b) has the specifics given. For instance, what happens to the (1-y[x]/b) term specifically when y[x]->b? What happens to the derivative in that case specifically that means that y[x] can't go beyond it.
Can somebody please help to understand this?
