Digital Signal Processing Reference
In-Depth Information
longer periods of time. Let's see if we can determine a general formula for the
balance of this account. Starting with the first month we find that
y
0( =
100
(5.12)
y
(
=
1
01
⋅
y
(
0
+
100
=
100
⋅
(
+
1
01
)
(5.13)
2
y
(
2
=
1
01
⋅
y
(
+
100
=
100
⋅
(
+
1
01
+
1
01
)
(5.14)
or in general
n
∑
=
k
y
(
n
)
=
100
⋅
1
01
k
0
(5.15)
Although this expression is compact, it still would require the calculation of
the sum of
n
+ 1 terms in order to determine the value. What we really need is a
closed form solution. (Of course I wouldn't have mentioned such a thing if one
didn't exist.) We can define what is referred to as a finite geometric sum, as
shown below:
n
∑
=
k
FGS
=
a
k
0
(5.16)
Then, with some ingenious mathematics, we can define a difference that
cancels most of the terms:
n
n
∑
∑
k
k
n
+
1
FGS
−
a
⋅
FGS
=
a
−
a
⋅
a
=
1
−
a
k
=
0
k
=
0
(5.17)
Finally, we can determine the value of the FGS, as shown below. (The value
of FGS when
a
= 1 is determined directly from (5.16):
a
n
+
1
⎧
1
−
⎨
,
for
a
≠
1
(5.18)
FGS
=
1
−
a
⎩
n
+
1
for
a
=
1
