Not quite the Fibonacci sequence

If instead of the recursion that we use to get the Fibonacci sequence:

f(n + 2) = f(n + 1) + f(n)

what happens if we change the sign of the coefficient of f(n) to get the recursion

f(n + 2) = f(n + 1) - f(n)

instead?

With the initial conditions

f(0) = 0

and

f(1) = 1

we get

f(0) = 0

f(1) = 1

f(2) = 1

f(3) = 0

f(4) = -1

f(5) = -1

f(6) = 0

f(7) = 1

at which point it's obvious that we'll just keep cycling through the pattern 0, 1, 1, 0, -1, -1 again. This probably means that the corresponding function of a real variable f(x) probably has an interesting graph. What does this look like?

We have that

f(x) = (axbx) / (ab)

where

a = (1 + i √3) / 2

and

b = (1 – i √3) / 2

which means that we also have that

ab = 1

abi √3

and

(a / b) = (-1 + i √3) / 2 = exp ( 2 π i / 3 )

so that

|a / b| = 1

and

arg (a / b) = 2 π / 3

and that 

log (a / b) = log( |a / b| ) + i arg (a / b) = 2 π i / 3

We can substitute these into the expression from yesterday's post:

f(n) = - [ 2 i (ab) n/2 / (ab) ] sin [ (i n / 2) log (a / b) ]

to find that

f(n) = (2 / √3) sin ( n π / 3)

so that the graph of f(x) just looks like this:

Image001

That's not quite as interesting as the graphs that we got from the Fibonacci sequence, is it?

Leave a Reply

Your email address will not be published. Required fields are marked *