Example: Let us utilize Mathematica to generate the first 15 values of the
difference equation,
a[n+1] = a[n] + a[n-1],
with the initial values a[0] = a[1] = 1.
To find the first 15 values of a[n], we shall first input the initial values
In[1]:= a[0] = 1
Out[1]= 1
In[2]:= a[1] = 1
Out[2]= 1
It is tempting at this point to input the difference equation in some manner,
but to do so would create a great difficulty due to the fact that a difference
equation is defined recursively. Mathematica would automatically begin to
generate values of a[n] until the internal limits of the system would force it
to stop. Thus we must apply some restrictions on the values of n before we can
input the difference equation.
One of the most convenient ways in which we can derive the first 15 values
of this recursively defined function a[n], is to restrict the possible values
that n can attain. Usually when one thinks of generating the values of a
function at n = 1, 2, ..., 15, one thinks of some sort of looping devise. In
our case we will incorporate a Table command:
In[3]:= Table[a[n+1]=a[n]+a[n-1],{n,1,14}]
This generates the values of a[n+1] as n ranges from 1 to 14 (the values a[2],
a[3], ..., a[15]), and places everything in a list as follows:
Out[3]= {2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987}
These values, along with our initial values, are then the values of a[n] as n
ranges from 0 to 15.
Example
: Consider the difference equation
, a[n+1] = 5*a[n] - 4*a[n-1],
with initial values a[0] = 2 and a[1] = 3. We shall use Mathematica to
obtain the values of a[n] for the values of n = 2, 3, ..., 15.
To define a[n] as a recurrent function in Mathematica while restricting
the value of n, we shall utilize a Table command. We do this as follows:
In[1]:= a[0]=2
Out[1]= 2
In[2]:= a[1]=3
Out[2]= 3
In[3]:= Table[a[n+1]=5*a[n]-4*a[n-1],{n,1,14}]
Out[3]= {7, 23, 87, 343, 1367, 5463, 21847, 87383, 349527, 1398103, 5592407,
> 22369623, 89478487, 357913943}
Thus we see that the size of a[n] grows fairly rapidly with n.
Example: The difference equation
( y[n] )
y[n+1] = y[n] ( 4 - ----- )
( 300 )
is a form of the discrete logistic equation, describing the size y of a
population at time t =n+ 1 units. Suppose that y[0] = 450. Let us use
Mathematica to obtain the values of y[n] for n = 1, 2, ..., 15.
Once again, to do this we shall use a Table for the purpose of
restricting the values of n. We obtain the solution of this problem as follows:
In[4]:= y[0]=450.0
Out[4]= 450.
In[5]:= Table[y[n+1]=y[n]*(4-y[n]/300),{n,0,14}]
Out[5]= {1125., 281.25, 861.328, 972.359, 737.83, 1136.68, 239.928, 767.828,
> 1106.11, 346.168, 985.231, 705.324, 1163.02, 143.351, 504.906}
Note that if we had input the initial value as an integer 450 instead as a real
number 450.0, then the output would have been in terms of fractions of integers
whose sizes would grow extremely large rather quickly.
Last modified: Mon May 24 01:24:39 EDT