1- a -0.11 1000 S, = (1.11)"| 1000 1000 + -0.11 - -0.11 1000 1110 = (1.11)"| +0.11 +0.11' ence, S40 = (1.11)º (10 090.090 909..) (9 090.909 090...) = (65.000 867...)(10 090.090 909..) – (9 090.909 090...) = 655 917.842... e $646 826. - (9 090.909 090..)
1- a -0.11 1000 S, = (1.11)"| 1000 1000 + -0.11 - -0.11 1000 1110 = (1.11)"| +0.11 +0.11' ence, S40 = (1.11)º (10 090.090 909..) (9 090.909 090...) = (65.000 867...)(10 090.090 909..) – (9 090.909 090...) = 655 917.842... e $646 826. - (9 090.909 090..)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Topic Video
Question
In the compound interest example on P.344. Calculate S20 and T240. Which is the better investment after 20 years?
![Example 8.2.3: Compound Interest
Suppose you are offered two retirement savings plans. In Plan
A, you start with $1,000, and each year (on the anniversary of
the plan), you are paid 11% simple interest, and you add
$1,000. In Plan B, you start with $100, and each month, you
are paid one-twelfth of 10% simple (annual) interest, and you
add $100. Which plan will be larger after 40
years?
// Can we apply a recurrence equation?
Consider Plan A and let S , denote the number of dollars in
the plan after (exactly) n years of operation. Then So = $1,000
and
Sn+1 = Sn + interest on S, +$1000
= S, + 11% of Sn
= S, (1 + 0.11)
+$1000
+$1000.
1000
In this RE, a = 1.11, c = 1000, so-
1 – a
and
-0.11'
%3D
1000
+
-0.11
1000
S, = (1.11)" | 1000
-0.11
1110
1000
= (1.11)"
+0.11
+0.11'
Hence, S40 = (1.11)40 (10 090.090 909. . )
- (9 090.909 090...)
(65.000 867..)(10 090.090 909...) – (9 090.909 090 .)
- (9 090.909 090...)
= 655 917.842...
= $646 826.
// Can that be right? You put in $40,000 and take out >
$600,000 in interest.
Now consider Plan B and let T, denote the number of
dollars in the plan after (exactly) n months of operation. Then
To = $100 and
Tn+1 = Tn +interest on Tn
+$100
%3D
= T, + (1/12) of 10% of T, +$100
= T„[1 + 0.1/12]
n
+$100.
с
100
In this RE,
a = 12.1/12,
c = 100, so
= -12000 and
-0.1/12
- a
T, = (12.1/12)" [100 + 12000] – 12000.
Hence, after 40 × 12 months,
T480 = (12.1/12)480 (12100)
= (1.008 333...)80 (12100) – (12000)
= (53.700 663 ...)(12100)
= 649 778.023 4...
e $637 778.
– (12000)
-
– (12000)
- (12000)
%D
Therefore, Plan A has a slightly larger value after 40 years.
The Most Important Ideas in This Section.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd55d239b-1399-4942-bcb8-cf1bf4c668fb%2Febb47619-6969-437c-999a-0b242445da7e%2Foxkxpw_processed.png&w=3840&q=75)
Transcribed Image Text:Example 8.2.3: Compound Interest
Suppose you are offered two retirement savings plans. In Plan
A, you start with $1,000, and each year (on the anniversary of
the plan), you are paid 11% simple interest, and you add
$1,000. In Plan B, you start with $100, and each month, you
are paid one-twelfth of 10% simple (annual) interest, and you
add $100. Which plan will be larger after 40
years?
// Can we apply a recurrence equation?
Consider Plan A and let S , denote the number of dollars in
the plan after (exactly) n years of operation. Then So = $1,000
and
Sn+1 = Sn + interest on S, +$1000
= S, + 11% of Sn
= S, (1 + 0.11)
+$1000
+$1000.
1000
In this RE, a = 1.11, c = 1000, so-
1 – a
and
-0.11'
%3D
1000
+
-0.11
1000
S, = (1.11)" | 1000
-0.11
1110
1000
= (1.11)"
+0.11
+0.11'
Hence, S40 = (1.11)40 (10 090.090 909. . )
- (9 090.909 090...)
(65.000 867..)(10 090.090 909...) – (9 090.909 090 .)
- (9 090.909 090...)
= 655 917.842...
= $646 826.
// Can that be right? You put in $40,000 and take out >
$600,000 in interest.
Now consider Plan B and let T, denote the number of
dollars in the plan after (exactly) n months of operation. Then
To = $100 and
Tn+1 = Tn +interest on Tn
+$100
%3D
= T, + (1/12) of 10% of T, +$100
= T„[1 + 0.1/12]
n
+$100.
с
100
In this RE,
a = 12.1/12,
c = 100, so
= -12000 and
-0.1/12
- a
T, = (12.1/12)" [100 + 12000] – 12000.
Hence, after 40 × 12 months,
T480 = (12.1/12)480 (12100)
= (1.008 333...)80 (12100) – (12000)
= (53.700 663 ...)(12100)
= 649 778.023 4...
e $637 778.
– (12000)
-
– (12000)
- (12000)
%D
Therefore, Plan A has a slightly larger value after 40 years.
The Most Important Ideas in This Section.
![Now consider Plan B and let T, denote the number of
dollars in the plan after (exactly) n months of operation. Then
To = $100 and
Tn+1
= T, + interest on Tn
= T, + (1/12) of 10% of T, +$100
= T„ [1+ 0.1/12]
+$100
n
n
+$100.
C
100
In this RE,
a = 12.1/12,
c = 100, so
= -12000 and
-0.1/12
T, = (12.1/12)" [100 + 12000] – 12000.
1- a
Hence, after 40 × 12 months,
– (12000)
= (1.008 333.)480 (12100) – (12000)
(53.700 663.. ) (12100) - (12000)
– (12000)
T480 = (12.1/12)480 (12100)
|
%3|
-
%3D
649 778.023 4...
e $637 778.
Therefore, Plan A has a slightly larger value after 40 years.
The Most Important Ideas in This Section.
A first-order linear recurrence equation relates consecutive
entries in a sequence by an equation of the form
Sn+1 = aS n + c
for V n e N.
The general solution is given in two parts:
if a = 1,
Sn = A + nc
for V n e N;
if a + 1,
Sn = a"A +
for Vn e N.
A particular solution is obtained by determining a
specific, numerical value for A. In fact, a particular solution is
determined by a specific, numerical value J for any
(particular) entry, S -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd55d239b-1399-4942-bcb8-cf1bf4c668fb%2Febb47619-6969-437c-999a-0b242445da7e%2F22803r9_processed.png&w=3840&q=75)
Transcribed Image Text:Now consider Plan B and let T, denote the number of
dollars in the plan after (exactly) n months of operation. Then
To = $100 and
Tn+1
= T, + interest on Tn
= T, + (1/12) of 10% of T, +$100
= T„ [1+ 0.1/12]
+$100
n
n
+$100.
C
100
In this RE,
a = 12.1/12,
c = 100, so
= -12000 and
-0.1/12
T, = (12.1/12)" [100 + 12000] – 12000.
1- a
Hence, after 40 × 12 months,
– (12000)
= (1.008 333.)480 (12100) – (12000)
(53.700 663.. ) (12100) - (12000)
– (12000)
T480 = (12.1/12)480 (12100)
|
%3|
-
%3D
649 778.023 4...
e $637 778.
Therefore, Plan A has a slightly larger value after 40 years.
The Most Important Ideas in This Section.
A first-order linear recurrence equation relates consecutive
entries in a sequence by an equation of the form
Sn+1 = aS n + c
for V n e N.
The general solution is given in two parts:
if a = 1,
Sn = A + nc
for V n e N;
if a + 1,
Sn = a"A +
for Vn e N.
A particular solution is obtained by determining a
specific, numerical value for A. In fact, a particular solution is
determined by a specific, numerical value J for any
(particular) entry, S -
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