Suppose we have the function f
that satisfies the recurrence formula f of n + 1
equals 2 times f of n + 3.
Know that this is not enough information
to define the function since we don't have
our initial condition.
For each of the initial conditions below,
find the value of f of 4.
To determine the value of f of 4,
we first have to determine f of 1, f of 2, and f of 3.
First, we were given the initial condition
f of 0 equals 0.
Because we are given f of 0,
we can now determine f of 1.
Where for f of 1, notice n is equal to 0
since f of the quantity 0 + 1 is f of 1,
which is equal to 2 times f of 0
since n is equal to 0 + 3.
We are given f of 0 is equal to 0
and therefore f of 1 is equal to 2 times 0 + 3,
which is equal to 3.
So now we know f of 1 is equal to 3,
we can determine f of 2.
For f of 2, notice n is equal to 1
since f of the quantity 1 + 1 is f of 2,
which is equal to 2 times f of 1, again,
because n is equal to 1, + 3.
We now know f of 1 is equal to 3,
which gives us 2 times 3 + 3,
which is 6 + 3, which is equal to 9.
We now know f of 2 is equal to 9.
Now that we know f of 2, we can determine f of 3
where for f of 3 notice n is equal to 2
since f of the quantity 2 + 1 is f of 3,
which is equal to 2 times f of 2 + 3.
2 times f of 2 + 3 is 2 times 9 + 3,
which is 18 + 3, which is 21.
We now know f of 3 is equal to 21.
And now finally we can determine f of 4.
So notice the drawback for a recursive function
is that we have to have the previous function values
for images of previous elements from the domain
to determine function values.
So to determine f of 4, notice n is equal to 3
since f of the quantity 3 + 1
is f of 4, which is equal to 2 times f of 3 + 3.
And again, f of 3 is equal to 21,
giving us 2 times 21 + 3,
which is 42 + 3, which equals 45.
And finally, we now know that f of 4 is equal to 45.
And now we're asked to go through the same process
where now f of 0 is equal to 3.
Starting over again, for f of 1
n is equal to 0 since f of 0 + 1 is f of 1,
which is equal to 2 times f of 0 + 3,
which is equal to 2 times 3 + 3,
which is equal to 6 + 3 or 9.
We now know f of 1 is equal to 9.
Well, since we're using the same recursive formula
and since f of 1 is equal to 9
and from the previous problem, f of 2 is equal to 9,
then therefore in our case, now f of 2
should be equal to 21.
f of 3 should be equal to 45, and so on,
but we'll go ahead and go through the same process.
For f of 2, n is equal to 1
since f of 1 + 1 is f of 2,
which gives us 2 times f of 1 + 3,
which is 2 times 9 + 3, which is equal to 21.
f of 2 is equal to 21.
For f of 3, n is equal to 2
since f of the quantity 2 + 1 is f of 3,
which gives us 2 times f of 2 + 3,
which is equal to 2 times 21 + 3,
which you know is equal to 45.
f of 3 is equal to 45.
Now that we know f of 3, we can determine f of 4.
for f of 4, notice n is equal to 3
since f of 3 + 1 is equal to f of 4,
which is equal to 2 times f of 3 + 3,
which is 2 times 45 + 3.
2 times 45 is 90, 90 + 3 is 93.
When the initial condition is f of 0 equals 3,
f of 4 is equal to 93.
I hope you found this helpful.