An application of math to biology

Stanislaw Ulam also had an interest in the application of math to biology. One example that may have relevance is the subfield of cellular automata founded by von Neumann and Ulam. As an example of this class of problems imagine dividing a plane into a several small squares like a checkerboard with several objects placed in nearby squares. Then specify rules for the appearance of new objects or the disappearance of old objects in new squares depending on whether adjacent squares are occupied or not. With each application of the rules to all the squares, the pattern of occupied squares evolves with time. Depending on the initial configuration and the rules of growth, some computer generated cellular automata evolve into patterns of snowflakes or crystals and others seem to have an everchanging motion as if they are alive. In some cases, colonies of self replicating patterns expand to fill the available space like the growth of coral or bacteria in a petri dish.

Reference: Adventures of a mathematician, S. Ulam, 1991 edition.

Cheers, cheers, cheers,

Nalin Pithwa

DSP Filter to generate Fibonacci numbers

The title is slight misnomer; but I am presenting below is a closed form expression for the nth term of the Fibonacci sequence.

Reference: Digital Signal Processing by Proakis and Manolakis, Sixth Edition.

The one-sided z-transform is a very efficient tool for the solution of difference equations with nonzero intial conditions. It achieves that by reducing the difference equation relating the two time-domain signals to an equivalent algebraic equation relating their one-sided z-transforms. This equation can be easily solved to obtain the transform of the desired signal. The signal in the time domain is obtained by inverting the resulting z-transform. For instance:

Example: The well-known Fibonacci sequence of integers is obtained by computing each term as the sum of the two-previous ones. The first few terms of the sequence are: 1,1,2,3,5, 8, \ldots

Determine a closed form expression for the nth term of the Fibonacci sequence.

Solution: Let y(n) be the nth term of the Fibonacci sequence. Clearly, y(n) satisfies the difference equation:

y(n) = y(n-1) + y(n-2)…..Equation A

with initial conditions

y(0)=y(-1)+y(-2)=1…..B

y(1)=y(0)+y(-1)=1….C

From the above, y(-1)=0 and y(-2)=1. Thus, we have to determine y(n) for n \geq 0, which satisfies equation A with initial conditions y(-1)=0 and y(-2)=1.

By taking the one-sided z-transform of the equation A and using the shifting property, we obtain:

Y^{+}(z) = (z^{-1}Y^{+}(z)+y(-1))+(z^{-1}Y^{+}(z)+y(-2)+y(-1)z^{-1})

or Y^{+}(z) = \frac{1}{1-z^{-1}-z^{2}} = \frac{z^{2}}{z^{2}-z-1}….equation D

where we have used the fact that y(-1)=0 and y(-2)=1.

We can invert Y^{+}(z) by the partial fraction expansion method. The poles of Y^{+}(z) are:

p_{1} = \frac{1+\sqrt{5}}{2}, and p_{2} = \frac{1-\sqrt{5}}{2}

and the corresponding coefficients are A_{1} = \frac{p_{1}}{\sqrt{5}} and A_{2}=\frac{-p_{1}}{\sqrt{5}}. Therefore,

y(n) = (\frac{1+\sqrt{5}}{2\sqrt{5}})(\frac{1+\sqrt{5}}{2})^{n}u(n) - \frac{1-\sqrt{5}}{2\sqrt{5}}(\frac{1-\sqrt{5}}{2})^{n}u(n)

or, equivalently,

y(n) = \frac{1}{\sqrt{5}}(\frac{1}{2})^{n+1}((1+\sqrt{5})^{n+1}-(1-\sqrt{5})^{n+1})u(n).

One smallish comment: We can implement a difference equation of a filter as either FIR or IIR. Of course, based on the physical nature of the signal/processing, one or the other might be preferable.

Regards,

Nalin Pithwa.