Enter the next 3 terms from the sequence formed from the following recurrence relation.
T
n
+
1
=
T
n
−
3
,
T
1
=
25
Answers
Answer:
sorry my dear friend I am so weak in maths why don't know the answer
Answer:
Suppose we want to solve a recurrence relation expressed as a combination of the two previous terms, such as
a
n
=
a
n
−
1
+
6
a
n
−
2
.
In other words, we want to find a function of
n
which satisfies
a
n
−
a
n
−
1
−
6
a
n
−
2
=
0
.
Now iteration is too complicated, but think just for a second what would happen if we did iterate. In each step, we would, among other things, multiply a previous iteration by 6. So our closed formula would include
6
multiplied some number of times. Thus it is reasonable to guess the solution will contain parts that look geometric. Perhaps the solution will take the form
r
n
for some constant
r
.
The nice thing is, we know how to check whether a formula is actually a solution to a recurrence relation: plug it in. What happens if we plug in
r
n
into the recursion above? We get
r
n
−
r
n
−
1
−
6
r
n
−
2
=
0.
Now solve for
r
:
r
n
−
2
(
r
2
−
r
−
6
)
=
0
,
so by factoring,
r
=
−
2
or
r
=
3
(or
r
=
0
,
although this does not help us). This tells us that
a
n
=
(
−
2
)
n
is a solution to the recurrence relation, as is
a
n
=
3
n
.
Which one is correct? They both are, unless we specify initial conditions. Notice we could also have
a
n
=
(
−
2
)
n
+
3
n
.
Or
a
n
=
7
(
−
2
)
n
+
4
⋅
3
n
.
In fact, for any
a
and
b
,
a
n
=
a
(
−
2
)
n
+
b
3
n
is a solution (try plugging this into the recurrence relation). To find the values of
a
and
b
,
use the initial conditions.
This points us in the direction of a more general technique for solving recurrence relations. Notice we will always be able to factor out the
r
n
−
2
as we did above. So we really only care about the other part. We call this other part the characteristic equation for the recurrence relation. We are interested in finding the roots of the characteristic equation, which are called (surprise) the characteristic roots.
Characteristic Roots
Given a recurrence relation
a
n
+
α
a
n
−
1
+
β
a
n
−
2
=
0
,
the characteristic polynomial is
x
2
+
α
x
+
β
giving the characteristic equation:
x
2
+
α
x
+
β
=
0.
If
r
1
and
r
2
are two distinct roots of the characteristic polynomial (i.e, solutions to the characteristic equation), then the solution to the recurrence relation is
a
n
=
a
r
n
1
+
b
r
n
2
,
where
a
and
b
are constants determined by the initial conditions.
Example2.4.6 Solve the recurrence relation
a
n
=
7
a
n
−
1
−
10
a
n
−
2
with
a
0
=
2
and
a
1
=
3
.
Perhaps the most famous recurrence relation is
F
n
=
F
n
−
1
+
F
n
−
2
,
which together with the initial conditions
F
0
=
0
and
F
1
=
1
defines the Fibonacci sequence. But notice that this is precisely the type of recurrence relation on which we can use the characteristic root technique. When you do, the only thing that changes is that the characteristic equation does not factor, so you need to use the quadratic formula to find the characteristic roots. In fact, doing so gives the