11.1: the constant e and continuous compound interest

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11.1: The Constant e and Continuous Compound Interest

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Page 1: 11.1: The Constant e and Continuous Compound Interest

11.1: The Constant e and Continuous Compound Interest

Page 2: 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.

• However, it can be approximated by

for a sufficiently large x

x

x

11

ee

e

Page 3: 11.1: The Constant e and Continuous Compound Interest

The Constant e

Reminder:

Use your calculator, e = 2.718 281 828 459 …

DEFINITION OF THE NUMBER e

s

s

n

ns

ne /1

01lim

11lim

Page 4: 11.1: The Constant e and Continuous Compound Interest

Check the limit using table and graph

Page 5: 11.1: The Constant e and Continuous Compound Interest

Review

• Simple interest: A = P + Prt = P(1 + rt)• Compound Interest: A = P(1 + r)t

or

with n = 1 (interest is compounded annually – once per year)

• Other compounding periods: semiannually(2), quarterly(4), monthly(12), weekly(52), daily(365), hourly(8760)…

• Continuous Compounding: (see page 589 for the proof)

A = Pert

A: future value

P: principal

r: interest rate

t: number of years

nt

n

rPA

1

Page 6: 11.1: The Constant e and Continuous Compound Interest

Example 1: Generous GrandmaYour Grandma puts $1,000 in a bank for you, at 5% interest. Calculate the amount after 20 years.

Simple interest:

A = 1000 (1 + 0.0520) = $2,000.00

Compounded annually:

A = 1000 (1 + .05)20 =$2,653.30

Compounded daily:

Compounded continuously:

A = 1000 e(.05)(20) = $2,718.28

10.718,2$365

05.11000 )20)(365(

A

Page 7: 11.1: The Constant e and Continuous Compound Interest

Example 2: IRAAfter 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

Page 8: 11.1: The Constant e and Continuous Compound Interest

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

• Compounded continuously A = Pert = 100e(.08*5)

= 149.18

93.1461

08.01100

5*1

A

02.1482

08.01100

5*2

A

Page 9: 11.1: The Constant e and Continuous Compound Interest

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

Page 10: 11.1: The Constant e and Continuous Compound Interest

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

Page 11: 11.1: The Constant e and Continuous Compound Interest

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

Page 12: 11.1: The Constant e and Continuous Compound Interest

• Review on how to solve exponential equations that involves e if needed (materials in MAT 115)