Mathematical Modelling Functions In R – Part 2

Ok, let’s start, where I have left out Mathematical Modelling Functions In R – Part 1 (please read, in case you didn’t read first), start right there.

Polynomial functions: are functions in which x appears several times, each time raised to a different power. They are useful for describing curves with humps, inflexions or local maxima.

See graphs below..

 

polynomialfunctionplot2
polynomialfunctionplot1

 

polynomialfunctionplot3

 

Inverse polynomials are an important class of functions which are suitable for setting up
generalized linear models with gamma errors and inverse link functions:

1/y = a+b*x+c*x^2 +d*x^3 +…+z*x^n

Various shapes of function are produced, depending on the order of the polynomial (the
maximum power) and the signs of the parameters:

 

See Below
polynomialfunctionplot5

polynomialfunctionplot6

polynomialfunctionplot7
There are two ways of parameterizing the Michaelis–Menten equation:

In the first case, the asymptotic value of y is a/b and in the second it is 1/d.

y= a*x/1+b*x and y= x/c+d*x

Gamma function: The gamma function t is an extension of the factorial function, t! to positive real numbers:

Γt=∫x^(t−1)*e^(−x)dx      0<x<∞

It looks like this:

Note that t is equal to 1 at both t=1 and t=2. For integer values of t t+1=t! and.

Asymptotic functions: Much the most commonly used asymptotic function is

y= a*x/1+b*x

which has a different name in almost every scientific discipline. For example, in biochemistry it is called Michaelis–Menten, and shows reaction rate as a function of enzyme concentration; in ecology, it is called Holling’s disc equation and shows predator feeding rate as a function of prey density. The other common function is the asymptotic exponential

y=a*(1−e^(−b*x))

This, too, is a two-parameter model, and in many cases, the two functions would describe data equally well.
Let’s the behaviour at the limits of our two asymptotic functions, starting with the asymptotic exponential.
For x=0 we have

y=a*(1−e−^(b×0))=a*(1−e^0)=a*(1−1)=a×0=0

so the graph goes through the origin. At the other extreme, for x=∞, we have

y=a*(1−e^(−b×∞))=a*(1−e^(−∞))=a*(1−0)=a*1=a

had the look just like this
asymptoticfunction
which demonstrates that the relationship is asymptotic and that the asymptotic value of y is a.  For the Michaelis–Menten equation, determining the behaviour at the limits is somewhat more difficult, because for x=∞ we end up with y=∞/∞ which you might imagine is always going to be 1 no matter what the values of a and b. In fact, there is a special mathematical rule for this case, called l’Hospital’s rule: when you get a ratio of infinity to infinity, you work out the ratio of the derivatives to obtain the behaviour at the limit.

For x=0 the limit is easy:

y= a×0/1+b×0 = 0/1+0= 0/1=0

For x=∞ we get

y=∞/1+∞=∞/∞

The numerator is a*x so its derivative with respect to x is a. The denominator is 1+b*x so its derivative with respect to x is 0 +b = b.

So the ratio of the derivatives is a/b, and this is the asymptotic value of the Michaelis–Menten equation.
Parameter estimation in asymptotic functions There is no way of linearizing the asymptotic exponential model,

so we must resort to nonlinear least squares (NLS) to estimate parameter values for it (p. 662). One of the advantages of the Michaelis–Menten function is that it is easy to linearize. We use the reciprocal
transformation

1/y= (1+b*x)ax

which, at first glance, isn’t a big help. But we can separate the terms on the right because they have a common denominator. Then we can cancel the xs, like this:

1/y = (1/a*x)+b*x/a*x = 1/ax+ b/a

If we simplify the above equation by putting y=1/y, x=1/x, A=1/a, and C=b/a, we see that

Y =AX +C

which is linear: C is the intercept and A is the slope. So to estimate the values of a and b
from data, we would transform both x and y to reciprocals, plot a graph of 1/y against 1/x,
carry out a linear regression, then back-transform, to get:

a= 1/A
b=a*C

Suppose, the graph passed through the two points (x1,y1) as (0.2, 44.44) and (x2,y2) as (0.6,70.59). How do we work out the values of the parameters a and b?

slope A = (y2-y1)/(x2-x1)

First, we calculate the four reciprocals. The slope of the linearized function, A

 A = (70.59-44.44)/(0.6 – 0.2) = 0.002500781

so a = 1/A = 1/0.0025 = 400. Now we rearrange the equation and use one of the points
(say x =0.2 y=44.44) to get the value of b:

b= (1/x)*[(a*x/y)−1] = (1/0.2)[(400×02/4444)−1] = 4

The remaining function we will discuss in a later part.

If you have any doubts please put a comment below and we will try to address those.

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