4.3 continuous compound interests perta x
TRANSCRIPT
Periodic Compound Interest
Let P = principal
i = (periodic) interest rate,
N = number of periods
A = accumulation
Recall the period-compound PINA formula.
Periodic Compound Interest
Let P = principal
i = (periodic) interest rate,
N = number of periods
A = accumulation
then P(1 + i) N = A
Recall the period-compound PINA formula.
Periodic Compound Interest
Let P = principal
i = (periodic) interest rate,
N = number of periods
A = accumulation
then P(1 + i) N = A
Recall the period-compound PINA formula.
We use the following time line to see what is happening.
0 1 2 3 Nth period N–1
Let P = principal
i = (periodic) interest rate,
N = number of periods
A = accumulation
then P(1 + i) N = A
We use the following time line to see what is happening.
P
0 1 2 3 Nth period N–1
Rule: Multiply (1 + i) each period forward
Recall the period-compound PINA formula.
Periodic Compound Interest
Let P = principal
i = (periodic) interest rate,
N = number of periods
A = accumulation
then P(1 + i) N = A
We use the following time line to see what is happening.
P
0 1 2 3 Nth period N–1
Rule: Multiply (1 + i) each period forward
P(1 + i)
Recall the period-compound PINA formula.
Periodic Compound Interest
Let P = principal
i = (periodic) interest rate,
N = number of periods
A = accumulation
then P(1 + i) N = A
We use the following time line to see what is happening.
P
0 1 2 3 Nth period N–1
Rule: Multiply (1 + i) each period forward
P(1 + i) P(1 + i) 2
Recall the period-compound PINA formula.
Periodic Compound Interest
Let P = principal
i = (periodic) interest rate,
N = number of periods
A = accumulation
then P(1 + i) N = A
We use the following time line to see what is happening.
P
0 1 2 3 Nth period N–1
Rule: Multiply (1 + i) each period forward
P(1 + i) P(1 + i) 2 P(1 + i) 3
Recall the period-compound PINA formula.
Periodic Compound Interest
Let P = principal
i = (periodic) interest rate,
N = number of periods
A = accumulation
then P(1 + i) N = A
We use the following time line to see what is happening.
P
0 1 2 3 Nth period N–1
Rule: Multiply (1 + i) each period forward
P(1 + i) P(1 + i) 2 P(1 + i) 3 P(1 + i) N - 1
Recall the period-compound PINA formula.
Periodic Compound Interest
Let P = principal
i = (periodic) interest rate,
N = number of periods
A = accumulation
then P(1 + i) N = A
We use the following time line to see what is happening.
P
0 1 2 3 Nth period N–1
Rule: Multiply (1 + i) each period forward
P(1 + i) P(1 + i) 2 P(1 + i) 3 P(1 + i) N - 1
Recall the period-compound PINA formula.
P(1 + i) N
Periodic Compound Interest
Let P = principal
i = (periodic) interest rate,
N = number of periods
A = accumulation
then P(1 + i) N = A
We use the following time line to see what is happening.
P
0 1 2 3 Nth period N–1
Rule: Multiply (1 + i) each period forward
P(1 + i) P(1 + i) 2 P(1 + i) 3 P(1 + i) N - 1 P(1 + i) N = A
Example A. $1,000 is in an account that has a monthly interest
rate of 1%. How much will be there after 60 years?
Recall the period-compound PINA formula.
P(1 + i) N
Periodic Compound Interest
Let P = principal
i = (periodic) interest rate,
N = number of periods
A = accumulation
then P(1 + i) N = A
We use the following time line to see what is happening.
P
0 1 2 3 Nth period N–1
Rule: Multiply (1 + i) each period forward
P(1 + i) P(1 + i) 2 P(1 + i) 3 P(1 + i) N - 1
Example A. $1,000 is in an account that has a monthly interest
rate of 1%. How much will be there after 60 years?
We have P = $1,000, i = 1% = 0.01, N = 60 *12 = 720 months
so by PINA, there will be 1000(1 + 0.01) 720
Recall the period-compound PINA formula.
P(1 + i) N = AP(1 + i) N
Periodic Compound Interest
Let P = principal
i = (periodic) interest rate,
N = number of periods
A = accumulation
then P(1 + i) N = A
We use the following time line to see what is happening.
P
0 1 2 3 Nth period N–1
Rule: Multiply (1 + i) each period forward
P(1 + i) P(1 + i) 2 P(1 + i) 3 P(1 + i) N - 1
Example A. $1,000 is in an account that has a monthly interest
rate of 1%. How much will be there after 60 years?
We have P = $1,000, i = 1% = 0.01, N = 60 *12 = 720 months
so by PINA, there will be 1000(1 + 0.01) 720 = $1,292,376.71
after 60 years.
Recall the period-compound PINA formula.
P(1 + i) N = AP(1 + i) N
Periodic Compound Interest
The graphs shown here are the different returns with r = 20%
with different compounding frequencies.
Compounded return on $1,000 with annual interest rate r = 20% (Wikipedia)
Periodic Compound Interest
The graphs shown here are the different returns with r = 20%
with different compounding frequencies. We observe that
I. the more frequently we compound, the bigger the return
Compounded return on $1,000 with annual interest rate r = 20% (Wikipedia)
Periodic Compound Interest
The graphs shown here are the different returns with r = 20%
with different compounding frequencies. We observe that
I. the more frequently we compound, the bigger the return
II. but the returns do not go above the blue-line
the continuous compound return, which is the next topic.
Compounded return on $1,000 with annual interest rate r = 20% (Wikipedia)
Periodic Compound Interest
Example B. We deposited $1000 in an account with annual
compound interest rate r = 8%. How much will be there after
20 years if it's compounded 100 times a year? 1000 times a
year? 10000 times a year?
Continuous Compound Interest
Example B. We deposited $1000 in an account with annual
compound interest rate r = 8%. How much will be there after
20 years if it's compounded 100 times a year? 1000 times a
year? 10000 times a year?
P = 1000, r = 0.08, T = 20,
Continuous Compound Interest
Example B. We deposited $1000 in an account with annual
compound interest rate r = 8%. How much will be there after
20 years if it's compounded 100 times a year? 1000 times a
year? 10000 times a year?
P = 1000, r = 0.08, T = 20,
For 100 times a year, 1000.08i = = 0.0008,
Continuous Compound Interest
Example B. We deposited $1000 in an account with annual
compound interest rate r = 8%. How much will be there after
20 years if it's compounded 100 times a year? 1000 times a
year? 10000 times a year?
P = 1000, r = 0.08, T = 20,
For 100 times a year, 1000.08i = = 0.0008,
N = (20 years)(100 times per years) = 2000
Continuous Compound Interest
Example B. We deposited $1000 in an account with annual
compound interest rate r = 8%. How much will be there after
20 years if it's compounded 100 times a year? 1000 times a
year? 10000 times a year?
P = 1000, r = 0.08, T = 20,
For 100 times a year, 1000.08i = = 0.0008,
N = (20 years)(100 times per years) = 2000
Hence A = 1000(1 + 0.0008 )2000
Continuous Compound Interest
Example B. We deposited $1000 in an account with annual
compound interest rate r = 8%. How much will be there after
20 years if it's compounded 100 times a year? 1000 times a
year? 10000 times a year?
P = 1000, r = 0.08, T = 20,
For 100 times a year, 1000.08i = = 0.0008,
N = (20 years)(100 times per years) = 2000
Hence A = 1000(1 + 0.0008 )2000 4949.87 $
Continuous Compound Interest
Example B. We deposited $1000 in an account with annual
compound interest rate r = 8%. How much will be there after
20 years if it's compounded 100 times a year? 1000 times a
year? 10000 times a year?
P = 1000, r = 0.08, T = 20,
For 100 times a year, 1000.08i = = 0.0008,
N = (20 years)(100 times per years) = 2000
Hence A = 1000(1 + 0.0008 )2000 4949.87 $
For 1000 times a year, 10000.08i = = 0.00008,
Continuous Compound Interest
Example B. We deposited $1000 in an account with annual
compound interest rate r = 8%. How much will be there after
20 years if it's compounded 100 times a year? 1000 times a
year? 10000 times a year?
P = 1000, r = 0.08, T = 20,
For 100 times a year, 1000.08i = = 0.0008,
N = (20 years)(100 times per years) = 2000
Hence A = 1000(1 + 0.0008 )2000 4949.87 $
For 1000 times a year, 10000.08i = = 0.00008,
N = (20 years)(1000 times per years) = 20000
Continuous Compound Interest
Example B. We deposited $1000 in an account with annual
compound interest rate r = 8%. How much will be there after
20 years if it's compounded 100 times a year? 1000 times a
year? 10000 times a year?
P = 1000, r = 0.08, T = 20,
For 100 times a year, 1000.08i = = 0.0008,
N = (20 years)(100 times per years) = 2000
Hence A = 1000(1 + 0.0008 )2000 4949.87 $
For 1000 times a year, 10000.08i = = 0.00008,
N = (20 years)(1000 times per years) = 20000
Hence A = 1000(1 + 0.00008 )20000
Continuous Compound Interest
Example B. We deposited $1000 in an account with annual
compound interest rate r = 8%. How much will be there after
20 years if it's compounded 100 times a year? 1000 times a
year? 10000 times a year?
P = 1000, r = 0.08, T = 20,
For 100 times a year, 1000.08i = = 0.0008,
N = (20 years)(100 times per years) = 2000
Hence A = 1000(1 + 0.0008 )2000 4949.87 $
For 1000 times a year, 10000.08i = = 0.00008,
N = (20 years)(1000 times per years) = 20000
Hence A = 1000(1 + 0.00008 )20000 4952.72 $
Continuous Compound Interest
Example B. We deposited $1000 in an account with annual
compound interest rate r = 8%. How much will be there after
20 years if it's compounded 100 times a year? 1000 times a
year? 10000 times a year?
P = 1000, r = 0.08, T = 20,
For 100 times a year, 1000.08i = = 0.0008,
N = (20 years)(100 times per years) = 2000
Hence A = 1000(1 + 0.0008 )2000 4949.87 $
For 1000 times a year, 10000.08i = = 0.00008,
N = (20 years)(1000 times per years) = 20000
Hence A = 1000(1 + 0.00008 )20000 4952.72 $
For 10000 times a year, 100000.08i = = 0.000008,
Continuous Compound Interest
Example B. We deposited $1000 in an account with annual
compound interest rate r = 8%. How much will be there after
20 years if it's compounded 100 times a year? 1000 times a
year? 10000 times a year?
P = 1000, r = 0.08, T = 20,
For 100 times a year, 1000.08i = = 0.0008,
N = (20 years)(100 times per years) = 2000
Hence A = 1000(1 + 0.0008 )2000 4949.87 $
For 1000 times a year, 10000.08i = = 0.00008,
N = (20 years)(1000 times per years) = 20000
Hence A = 1000(1 + 0.00008 )20000 4952.72 $
For 10000 times a year, 100000.08i = = 0.000008,
N = (20 years)(10000 times per years) = 200000
Continuous Compound Interest
Example B. We deposited $1000 in an account with annual
compound interest rate r = 8%. How much will be there after
20 years if it's compounded 100 times a year? 1000 times a
year? 10000 times a year?
P = 1000, r = 0.08, T = 20,
For 100 times a year, 1000.08i = = 0.0008,
N = (20 years)(100 times per years) = 2000
Hence A = 1000(1 + 0.0008 )2000 4949.87 $
For 1000 times a year, 10000.08i = = 0.00008,
N = (20 years)(1000 times per years) = 20000
Hence A = 1000(1 + 0.00008 )20000 4952.72 $
For 10000 times a year, 100000.08i = = 0.000008,
N = (20 years)(10000 times per years) = 200000
Hence A = 1000(1 + 0.000008 )200000
Continuous Compound Interest
P = 1000, r = 0.08, T = 20,
For 100 times a year, 1000.08i = = 0.0008,
N = (20 years)(100 times per years) = 2000
Hence A = 1000(1 + 0.0008 )2000 4949.87 $
For 1000 times a year, 10000.08i = = 0.00008,
N = (20 years)(1000 times per years) = 20000
Hence A = 1000(1 + 0.00008 )20000 4952.72 $
For 10000 times a year, 100000.08i = = 0.000008,
N = (20 years)(10000 times per years) = 200000
Hence A = 1000(1 + 0.000008 )200000 4953.00 $
Continuous Compound Interest
Example B. We deposited $1000 in an account with annual
compound interest rate r = 8%. How much will be there after
20 years if it's compounded 100 times a year? 1000 times a
year? 10000 times a year?
We list the results below as the number compounded per year
f gets larger and larger.
Continuous Compound Interest
We list the results below as the number compounded per year
f gets larger and larger.
4 times a year 4875.44 $
Continuous Compound Interest
We list the results below as the number compounded per year
f gets larger and larger.
100 times a year 4949.87 $
4 times a year 4875.44 $
Continuous Compound Interest
We list the results below as the number compounded per year
f gets larger and larger.
10000 times a year 4953.00 $
1000 times a year 4952.72 $
100 times a year 4949.87 $
4 times a year 4875.44 $
Continuous Compound Interest
We list the results below as the number compounded per year
f gets larger and larger.
10000 times a year 4953.00 $
1000 times a year 4952.72 $
100 times a year 4949.87 $
4 times a year 4875.44 $
Continuous Compound Interest
We list the results below as the number compounded per year
f gets larger and larger.
10000 times a year 4953.00 $
1000 times a year 4952.72 $
100 times a year 4949.87 $
4 times a year 4875.44 $
4953.03 $
Continuous Compound Interest
We list the results below as the number compounded per year
f gets larger and larger.
10000 times a year 4953.00 $
1000 times a year 4952.72 $
100 times a year 4949.87 $
4 times a year 4875.44 $
4953.03 $
We call this amount the continuously compounded return.
Continuous Compound Interest
We list the results below as the number compounded per year
f gets larger and larger.
10000 times a year 4953.00 $
1000 times a year 4952.72 $
100 times a year 4949.87 $
4 times a year 4875.44 $
4953.03 $
We call this amount the continuously compounded return.
This way of compounding is called compounded continuously.
Continuous Compound Interest
We list the results below as the number compounded per year
f gets larger and larger.
10000 times a year 4953.00 $
1000 times a year 4952.72 $
100 times a year 4949.87 $
4 times a year 4875.44 $
4953.03 $
We call this amount the continuously compounded return.
This way of compounding is called compounded continuously.
The reason we want to compute interest this way is because
the formula for computing continously compound return is
easy to manipulate mathematically.
Continuous Compound Interest
Continuous Compound InterestFormula for Continuously Compounded Return (Perta)Formula for Continuously Compounded Return (Perta)
Let P = principal
r = annual interest rate (compound continuously)
t = number of years
A = accumulated value, then
Per*t = A where e 2.71828…
Continuous Compound Interest
There is no “f” because
it’s compounded continuously
Formula for Continuously Compounded Return (Perta)
Let P = principal
r = annual interest rate (compound continuously)
t = number of years
A = accumulated value, then
Per*t = A where e 2.71828…
Example C. We deposited $1000 in an account compounded
continuously.
a. if r = 8%, how much will be there after 20 years?
Continuous Compound InterestFormula for Continuously Compounded Return (Perta)
Let P = principal
r = annual interest rate (compound continuously)
t = number of years
A = accumulated value, then
Per*t = A where e 2.71828…
Example C. We deposited $1000 in an account compounded
continuously.
a. if r = 8%, how much will be there after 20 years?
P = 1000, r = 0.08, t = 20.
Continuous Compound InterestFormula for Continuously Compounded Return (Perta)
Let P = principal
r = annual interest rate (compound continuously)
t = number of years
A = accumulated value, then
Per*t = A where e 2.71828…
Example C. We deposited $1000 in an account compounded
continuously.
a. if r = 8%, how much will be there after 20 years?
P = 1000, r = 0.08, t = 20. So the continuously compounded
return is A = 1000*e0.08*20
Continuous Compound InterestFormula for Continuously Compounded Return (Perta)
Let P = principal
r = annual interest rate (compound continuously)
t = number of years
A = accumulated value, then
Per*t = A where e 2.71828…
Example C. We deposited $1000 in an account compounded
continuously.
a. if r = 8%, how much will be there after 20 years?
P = 1000, r = 0.08, t = 20. So the continuously compounded
return is A = 1000*e0.08*20 = 1000*e1.6
Continuous Compound InterestFormula for Continuously Compounded Return (Perta)
Let P = principal
r = annual interest rate (compound continuously)
t = number of years
A = accumulated value, then
Per*t = A where e 2.71828…
Example C. We deposited $1000 in an account compounded
continuously.
a. if r = 8%, how much will be there after 20 years?
P = 1000, r = 0.08, t = 20. So the continuously compounded
return is A = 1000*e0.08*20 = 1000*e1.6 4953.03$
Continuous Compound InterestFormula for Continuously Compounded Return (Perta)
Let P = principal
r = annual interest rate (compound continuously)
t = number of years
A = accumulated value, then
Per*t = A where e 2.71828…
Example C. We deposited $1000 in an account compounded
continuously.
a. if r = 8%, how much will be there after 20 years?
P = 1000, r = 0.08, t = 20. So the continuously compounded
return is A = 1000*e0.08*20 = 1000*e1.6 4953.03$
b. If r = 12%, how much will be there after 20 years?
Continuous Compound InterestFormula for Continuously Compounded Return (Perta)
Let P = principal
r = annual interest rate (compound continuously)
t = number of years
A = accumulated value, then
Per*t = A where e 2.71828…
Example C. We deposited $1000 in an account compounded
continuously.
a. if r = 8%, how much will be there after 20 years?
P = 1000, r = 0.08, t = 20. So the continuously compounded
return is A = 1000*e0.08*20 = 1000*e1.6 4953.03$
b. If r = 12%, how much will be there after 20 years?
r = 12%, A = 1000*e0.12*20
Continuous Compound InterestFormula for Continuously Compounded Return (Perta)
Let P = principal
r = annual interest rate (compound continuously)
t = number of years
A = accumulated value, then
Per*t = A where e 2.71828…
Example C. We deposited $1000 in an account compounded
continuously.
a. if r = 8%, how much will be there after 20 years?
P = 1000, r = 0.08, t = 20. So the continuously compounded
return is A = 1000*e0.08*20 = 1000*e1.6 4953.03$
b. If r = 12%, how much will be there after 20 years?
r = 12%, A = 1000*e0.12*20 = 1000e 2.4
Continuous Compound InterestFormula for Continuously Compounded Return (Perta)
Let P = principal
r = annual interest rate (compound continuously)
t = number of years
A = accumulated value, then
Per*t = A where e 2.71828…
Example C. We deposited $1000 in an account compounded
continuously.
a. if r = 8%, how much will be there after 20 years?
P = 1000, r = 0.08, t = 20. So the continuously compounded
return is A = 1000*e0.08*20 = 1000*e1.6 4953.03$
b. If r = 12%, how much will be there after 20 years?r = 12%, A = 1000*e0.12*20 = 1000e 2.4 11023.18$
Continuous Compound InterestFormula for Continuously Compounded Return (Perta)
Let P = principal
r = annual interest rate (compound continuously)
t = number of years
A = accumulated value, then
Per*t = A where e 2.71828…
Example C. We deposited $1000 in an account compounded
continuously.
a. if r = 8%, how much will be there after 20 years?
P = 1000, r = 0.08, t = 20. So the continuously compounded
return is A = 1000*e0.08*20 = 1000*e1.6 4953.03$
b. If r = 12%, how much will be there after 20 years?r = 12%, A = 1000*e0.12*20 = 1000e 2.4 11023.18$
c. If r = 16%, how much will be there after 20 years?
Continuous Compound InterestFormula for Continuously Compounded Return (Perta)
Let P = principal
r = annual interest rate (compound continuously)
t = number of years
A = accumulated value, then
Per*t = A where e 2.71828…
Example C. We deposited $1000 in an account compounded
continuously.
a. if r = 8%, how much will be there after 20 years?
P = 1000, r = 0.08, t = 20. So the continuously compounded
return is A = 1000*e0.08*20 = 1000*e1.6 4953.03$
b. If r = 12%, how much will be there after 20 years?r = 12%, A = 1000*e0.12*20 = 1000e 2.4 11023.18$
c. If r = 16%, how much will be there after 20 years?
r = 16%, A = 1000*e0.16*20
Continuous Compound InterestFormula for Continuously Compounded Return (Perta)
Let P = principal
r = annual interest rate (compound continuously)
t = number of years
A = accumulated value, then
Per*t = A where e 2.71828…
Example C. We deposited $1000 in an account compounded
continuously.
a. if r = 8%, how much will be there after 20 years?
P = 1000, r = 0.08, t = 20. So the continuously compounded
return is A = 1000*e0.08*20 = 1000*e1.6 4953.03$
b. If r = 12%, how much will be there after 20 years?r = 12%, A = 1000*e0.12*20 = 1000e 2.4 11023.18$
c. If r = 16%, how much will be there after 20 years?
r = 16%, A = 1000*e0.16*20 = 1000*e 3.2
Continuous Compound InterestFormula for Continuously Compounded Return (Perta)
Let P = principal
r = annual interest rate (compound continuously)
t = number of years
A = accumulated value, then
Per*t = A where e 2.71828…
Example C. We deposited $1000 in an account compounded
continuously.
a. if r = 8%, how much will be there after 20 years?
P = 1000, r = 0.08, t = 20. So the continuously compounded
return is A = 1000*e0.08*20 = 1000*e1.6 4953.03$
b. If r = 12%, how much will be there after 20 years?r = 12%, A = 1000*e0.12*20 = 1000e 2.4 11023.18$
c. If r = 16%, how much will be there after 20 years?r = 16%, A = 1000*e0.16*20 = 1000*e 3.2 24532.53$
Continuous Compound InterestFormula for Continuously Compounded Return (Perta)
Let P = principal
r = annual interest rate (compound continuously)
t = number of years
A = accumulated value, then
Per*t = A where e 2.71828…
Just as the number π, the number e 2.71828… occupies a
special place in mathematics.
Continuous Compound InterestAbout the Number e
Just as the number π, the number e 2.71828… occupies a
special place in mathematics. Where as π 3.14156… is a
geometric constant–the ratio of the circumference to the
diameter of a circle, e is derived from calculations.
Continuous Compound InterestAbout the Number e
Just as the number π, the number e 2.71828… occupies a
special place in mathematics. Where as π 3.14156… is a
geometric constant–the ratio of the circumference to the
diameter of a circle, e is derived from calculations.
For example, the following sequence of numbers zoom–in on
the number,
( )1,2 1 …( )4,
5 4
( )3,4 3
( )2,3 2
2.71828…
Continuous Compound InterestAbout the Number e
Just as the number π, the number e 2.71828… occupies a
special place in mathematics. Where as π 3.14156… is a
geometric constant–the ratio of the circumference to the
diameter of a circle, e is derived from calculations.
For example, the following sequence of numbers zoom–in on
the number,
( 2.71828…)the same as
( )1,2 1 …( )4,
5 4
( )3,4 3
( )2,3 2
2.71828…which is
Continuous Compound InterestAbout the Number e
Just as the number π, the number e 2.71828… occupies a
special place in mathematics. Where as π 3.14156… is a
geometric constant–the ratio of the circumference to the
diameter of a circle, e is derived from calculations.
For example, the following sequence of numbers zoom–in on
the number,
This number emerges often in the calculation of problems in
physical science, natural science, finance and in mathematics.
( 2.71828…)the same as
( )1,2 1 …( )4,
5 4
( )3,4 3
( )2,3 2
2.71828…which is
Continuous Compound InterestAbout the Number e
Just as the number π, the number e 2.71828… occupies a
special place in mathematics. Where as π 3.14156… is a
geometric constant–the ratio of the circumference to the
diameter of a circle, e is derived from calculations.
For example, the following sequence of numbers zoom–in on
the number,
http://en.wikipedia.org/wiki/E_%28mathematical_constant%29
This number emerges often in the calculation of problems in
physical science, natural science, finance and in mathematics.
Because of its importance, the irrational number 2.71828…
is named as “e” and it’s called the “natural” base number.
( 2.71828…)the same as
http://www.ndt-ed.org/EducationResources/Math/Math-e.htm
( )1,2 1 …( )4,
5 4
( )3,4 3
( )2,3 2
2.71828…which is
Continuous Compound InterestAbout the Number e
Continuous Compound InterestGrowth and Decay
In all the interest examples we have positive interest rate r,
hence the return A = Perx grows larger as time x gets larger.
Continuous Compound InterestGrowth and Decay
In all the interest examples we have positive interest rate r,
hence the return A = Perx grows larger as time x gets larger.
We call an expansion that may be modeled by A = Perx
with r > 0 as “ an exponential growth with growth rate r”.
Continuous Compound Interest
y = e1x has the growth rate of
r = 1 or 100%.
Growth and Decay
In all the interest examples we have positive interest rate r,
hence the return A = Perx grows larger as time x gets larger.
We call an expansion that may be modeled by A = Perx
with r > 0 as “ an exponential growth with growth rate r”.
For example,
Continuous Compound Interest
y = e1x has the growth rate of
r = 1 or 100%.
Exponential growth are rapid
expansions compared to other
expansions as shown here
by their graphs.
y = x3
y = 100x
y = ex
Growth and Decay
In all the interest examples we have positive interest rate r,
hence the return A = Perx grows larger as time x gets larger.
We call an expansion that may be modeled by A = Perx
with r > 0 as “ an exponential growth with growth rate r”.
For example,An Exponential Growth
Continuous Compound Interest
y = e1x has the growth rate of
r = 1 or 100%.
Exponential growth are rapid
expansions compared to other
expansions as shown here
by their graphs.
y = x3
y = 100x
y = ex
The world population may be
modeled with an exponential
growth with r ≈ 1.1 % or 0.011
or that A ≈ 6.5e0.011t in billions,
with 2011as t = 0.
Growth and Decay
In all the interest examples we have positive interest rate r,
hence the return A = Perx grows larger as time x gets larger.
We call an expansion that may be modeled by A = Perx
with r > 0 as “ an exponential growth with growth rate r”.
For example,An Exponential Growth
Continuous Compound InterestIf the rate r is negative, or that r < 0 then the return A = Perx
shrinks as time x gets larger.
Continuous Compound InterestIf the rate r is negative, or that r < 0 then the return A = Perx
shrinks as time x gets larger.
We call a contraction that may be modeled by A = Perx
with r < 0 as “an exponential decay at the rate | r |”.
Continuous Compound InterestIf the rate r is negative, or that r < 0 then the return A = Perx
shrinks as time x gets larger.
We call a contraction that may be modeled by A = Perx
with r < 0 as “an exponential decay at the rate | r |”.
For example,
y = e–1x has the decay or
contraction rate of I r I = 1 or 100%.
y = e–xAn Exponential Decay
Continuous Compound InterestIf the rate r is negative, or that r < 0 then the return A = Perx
shrinks as time x gets larger.
We call a contraction that may be modeled by A = Perx
with r < 0 as “an exponential decay at the rate | r |”.
For example,
y = e–1x has the decay or
contraction rate of I r I = 1 or 100%.
In finance, shrinking values is
called “depreciation” or
”devaluation”.
y = e–xAn Exponential Decay
Continuous Compound InterestIf the rate r is negative, or that r < 0 then the return A = Perx
shrinks as time x gets larger.
We call a contraction that may be modeled by A = Perx
with r < 0 as “an exponential decay at the rate | r |”.
For example,
y = e–1x has the decay or
contraction rate of I r I = 1 or 100%.
In finance, shrinking values is
called “depreciation” or
”devaluation”. For example,
a currency that is depreciating
at a rate of 4% annually may be
modeled by A = Pe –0.04x
where x is the number of yearss elapsed.
y = e–xAn Exponential Decay
Continuous Compound InterestIf the rate r is negative, or that r < 0 then the return A = Perx
shrinks as time x gets larger.
We call a contraction that may be modeled by A = Perx
with r < 0 as “an exponential decay at the rate | r |”.
For example,
y = e–1x has the decay or
contraction rate of I r I = 1 or 100%.
In finance, shrinking values is
called “depreciation” or
”devaluation”. For example,
a currency that is depreciating
at a rate of 4% annually may be
modeled by A = Pe –0.04x
where x is the number of yearss elapsed.
Hence if P = $1, after 5 years, its purchasing power is
1*e–0.04(5) = $0.82 or 82 cents.
y = e–xAn Exponential Decay
Continuous Compound InterestIf the rate r is negative, or that r < 0 then the return A = Perx
shrinks as time x gets larger.
We call a contraction that may be modeled by A = Perx
with r < 0 as “an exponential decay at the rate | r |”.
For example,
y = e–1x has the decay or
contraction rate of I r I = 1 or 100%.
In finance, shrinking values is
called “depreciation” or
”devaluation”. For example,
a currency that is depreciating
at a rate of 4% annually may be
modeled by A = Pe –0.04x
where x is the number of yearss elapsed.
Hence if P = $1, after 5 years, its purchasing power is
1*e–0.04(5) = $0.82 or 82 cents. For more information:
y = e–xAn Exponential Decay
http://math.ucsd.edu/~wgarner/math4c/textbook/chapter4/expgrowthdecay.htm
Doubling Time and the 72-Rule Given that r > 0, the return A = Perx grows as time x gets larger.
The time it takes to double the principal P so that A = 2P.
is called the doubling time.
Doubling Time and the 72-Rule Given that r > 0, the return A = Perx grows as time x gets larger.
The time it takes to double the principal P so that A = 2P.
is called the doubling time.
Solving for the doubling time we have
Perx = 2P or that
erx= 2
Doubling Time and the 72-Rule Given that r > 0, the return A = Perx grows as time x gets larger.
The time it takes to double the principal P so that A = 2P.
is called the doubling time.
Solving for the doubling time we have
Perx = 2P or that
erx= 2 so from the result of the next section,
rx = In(2)
Doubling Time and the 72-Rule Given that r > 0, the return A = Perx grows as time x gets larger.
The time it takes to double the principal P so that A = 2P.
is called the doubling time.
Solving for the doubling time we have
Perx = 2P or that
erx= 2 so from the result of the next section,
rx = In(2) and that the doubling time is
x = In(2)/r
Doubling Time and the 72-Rule Given that r > 0, the return A = Perx grows as time x gets larger.
The time it takes to double the principal P so that A = 2P.
is called the doubling time.
Solving for the doubling time we have
Perx = 2P or that
erx= 2 so from the result of the next section,
rx = In(2) and that the doubling time is
x = In(2)/r ≈ 0.72/r
Doubling Time and the 72-Rule Given that r > 0, the return A = Perx grows as time x gets larger.
The time it takes to double the principal P so that A = 2P.
is called the doubling time.
Solving for the doubling time we have
Perx = 2P or that
erx= 2 so from the result of the next section,
rx = In(2) and that the doubling time is
x = In(2)/r ≈ 0.72/r
The last formula for the doubling time x is the 72-Rule, i.e.
x ≈ 0.72/r
Doubling Time and the 72-Rule Given that r > 0, the return A = Perx grows as time x gets larger.
The time it takes to double the principal P so that A = 2P.
is called the doubling time.
Solving for the doubling time we have
Perx = 2P or that
erx= 2 so from the result of the next section,
rx = In(2) and that the doubling time is
x = In(2)/r ≈ 0.72/r
The last formula for the doubling time x is the 72-Rule, i.e.
x ≈ 0.72/r
This 72- Rule gives an easy estimation for the doubling time.
Doubling Time and the 72-Rule Given that r > 0, the return A = Perx grows as time x gets larger.
The time it takes to double the principal P so that A = 2P.
is called the doubling time.
Solving for the doubling time we have
Perx = 2P or that
erx= 2 so from the result of the next section,
rx = In(2) and that the doubling time is
x = In(2)/r ≈ 0.72/r
The last formula for the doubling time x is the 72-Rule, i.e.
x ≈ 0.72/r
This 72- Rule gives an easy estimation for the doubling time.
For example, if r = 8% then the doubling time is ≈ 72/8 = 9 (yrs).
Doubling Time and the 72-Rule Given that r > 0, the return A = Perx grows as time x gets larger.
The time it takes to double the principal P so that A = 2P.
is called the doubling time.
Solving for the doubling time we have
Perx = 2P or that
erx= 2 so from the result of the next section,
rx = In(2) and that the doubling time is
x = In(2)/r ≈ 0.72/r
The last formula for the doubling time x is the 72-Rule, i.e.
x ≈ 0.72/r
This 72- Rule gives an easy estimation for the doubling time.
For example, if r = 8% then the doubling time is ≈ 72/8 = 9 (yrs).
if r = 12% then the doubling time is ≈ 72/12 = 6 (yrs).
if r = 18% then the doubling time is ≈ 72/18 = 4 (yrs).