Transcript 11.1: The constant e and continuous compound interest

11.1: The Constant e and
Continuous Compound Interest
Review (Mat 115)
• Just like π, e is an irrational number which can
not be represented exactly by any finite
decimal fraction.
x
• However, it can be approximated by 1  1 
e

x
for a sufficiently large x
The Constant e
Reminder:
Use your calculator, e = 2.718 281 828 459 …
DEFINITION OF THE NUMBER e
n
 1
1/ s
e  lim 1    lim 1  s 
n
 n  s0
Check the limit using table and graph
Review
• Simple interest: A = P + Prt
• Compound Interest: A = P(1 + r)t
nt
or
 r
A  P 1 
 n
with n = 1 (interest is compounded annually
– once per year)
• Other compounding periods:
A: future
value
P: principal
r: interest
rate
t: number of
years
semiannually(2), quarterly(4), monthly(12),
weekly(52), daily(365), hourly(8760)…
• Continuous Compounding: (see page 589 for the proof)
A = Pert
Example 1: Generous Grandma
Your Grandma puts $1,000 in a bank for you, at 5% interest. Calculate the amount
after 20 years.
Simple interest:
A = 1000 (1 + 0.0520) = $2,000.00
Compounded annually:
A = 1000 (1 + .05)20 =$2,653.30
Compounded daily:
 .05  (365 )( 20)
A  10001 
 $2,718.09

 365
Compounded continuously:
A = 1000 e(.05)(20) = $2,718.28
Example 2: IRA
After graduating from Barnett College, Sam Spartan landed
a great job with Springettsbury Manufacturing, Inc. His
first year he bought a $3,000 Roth IRA and invested it in a
stock sensitive mutual fund that grows at 12% a year,
compounded continuously. He plans to retire in 35 years.
a. What will be its value at the end of the time period?
A = Pert = 3000 e(.12)(35) =$200,058.99
b. The second year he repeated the purchase of an identical
Roth IRA. What will be its value in 34 years?
A = Pert = 3000 e(.12)(34) =$177,436.41
Example 3
What amount (to the nearest cent) will an account have
after 5 years if $100 is invested at an annual nominal rate
of 8% compounded annually? Semiannually? continuously?
•
•
•
Compounded annually
Compounded semiannually
1*5
 0.08 
A  1001

1


 0.08 
A  1001

2 

 146.93
2*5
 148.02
Compounded continuously
A = Pert = 100e(.08*5)
= 149.18
Example 4
If $5000 is invested in a Union Savings Bank 4-year CD
that earns 5.61% compounded continuously, graph the
amount in the account relative to time for a period of 4
years.
Use your graphing calculator:
Press y=
Type in 5000e^(x*0.0561)
Press ZOOM, scroll down,
then press ZoomFit
You will see the graph
•To find out the amount after
4 years
Press 2ND, TRACE, 1:VALUE
Then type in 4, ENTER
Example 5
How long will it take an investment of $10000
to grow to $15000 if it is invested at 9% compounded
continuously?
Formula:
A =P ert
15000 = 10000 e .09t
1.5 = e .09t
Ln (1.5) = ln (e .09t)
Ln (1.5) = .09 t
So t = ln(1.5) / .09
t = 4.51
It will take about 4.51 years
Example 6
How long will it take money to triple if it is
invested at 5.5% compounded continuously?
Formula:
A =P ert
3P = P e .055t
3 = e .055t
Ln 3
= ln (e .055t)
Ln 3 = .055t
So t = ln3 / .055
t = 19.97
It will take about 19.97 years
• Review on how to solve exponential
equations that involves e if needed
(materials in MAT 115)