2.1 Geometric Progressions
For Fred Greenleaf’s QR Textbook
Compiled by Sam Marateck © 2009
1
Examples of geometric progressions
• A geometric progression is a sequence of
numbers in which a given number is a constant
ratio, let’s say a, times the previous number.
Examples:
• 1, 2, 4, 8, 16, 32, i.e., An = aAn-1, A0 = 1, a =2
• 80, 40, 20, 10, 5, 2.5, i.e., A0 = 80, a = 1/2
2
Given: a population increases by 5% each
year. Assume that A0 =20, i.e., the
population at year zero is 20. Then at year
one,
A1 =(1+0.05)20 or 1.05*20 or 21, at year two
A2 =(1+0.05)21 or (1+0.5)220 or 22.05,
A3 = (1+0.05)320 or 23.15
If r is the rate of growth, what is the general
formula for growth?
3
In general, An = (1 + r)nA0
For our preceding population growth problem,
what is the value of n when An = 2A0 , i.e.,
what is the doubling time?
In our population growth problem, since A0 =20, we
set A to 2*20 or 40 and we solve the
problem 40 = 1.05n 20 or 2 = 1.05n for n. The
easiest way to do this is to use logarithms.
4
In y = 103, we say that 3 is the logarithm and
10 is the base, or the logarithm of y to the
base 10 is 3. We write this as log10 y = 3.
Let’s multiply two numbers with the same
base but different exponents (logarithms), e.g.,
anam. The result is an+m, i.e., we add the
exponents
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The use of this book is quite large, my dear friend,
No matter how modest it looks,
You study it carefully and find that it gives
As much as a thousand big books.
John Napier, Baron Merchiston (1550-1617)
inventor of logarithms on his book,
Mirifici Logarithmorum Canonis Descriptio
(Description of the wonderful law of logarithms)
and popularizer of the use of the decimal point.
.
6
Napier used as the base of his logarithms
1- 10-7 . Henry Briggs (1561 – 1631), a
British mathematician, proposed to Napier
making the base 10. Thus log of 1 would be
0. The tables we use today are based on
Brigg’s generation of a log table.
Let’s review logarithms
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When we divide two numbers with the same
base but different exponents e.g.,
an / am, the result is an-m, i.e., we subtract the
exponent in the denominator from the one in the
numerator. So if x = 105 and y = 103, then
xy = 108. So log10 (xy) = log10 (x) + log10(y).
Similarly x/y = 105-3 or 102. So log10(x/y) =
log(x) – log(y).
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If we multiply the same number x by itself
many times and assign it to y, y= xxxx.. then
log(y) = log(x)+ log(x)+ log(x)+ log(x)+…
So log(xn) = n log(x).
Look at y = 103. Take the log of both sides
log10 y = log10(103) = 3 log10(10). But log10(10) = 1 since 10=101.
So we get log10 y = 3 which conforms with the definition.
Let’s now solve
2 = 1.05n
for n. Take the log of both sides:
log(2) = log(1.05n ) = n log(1.05)
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We solve log(2) = n log(1.05) for n.
n = log(2)/log(1.05) . From the log table we
See that log(2) is 0.301 and log(1.05) is
0.0211. Thus the doubling time, n, is 14.3
years.
10
Earth quake magnitude measured
in Richter Scale
•
•
•
•
•
•
•
•
•
•
•
Magnitude
-3.0
-2.0
-1.0
2.0
4.0
6.9
7.3
9.0
9.2
9.5
Notes
1.5 foot-pounds (18 inch-pounds)
47 foot-pounds
1,500 foot-pounds
Felt only nearby, if at all
Often felt up to 10's of miles away
1995 Kobe, Japan, Earthquake
1933 Salcha Earthquake
2011 Honshu Japan Earhquake
1964 Alaska Earthquake
1960 Chile - Largest Recorded Earthquake
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Comparison of earthquake and
Hiroshima A-Bomb
• Magnitude
Hiroshima Bombs TNT
(Richter scale)
(megaton)
•
•
•
•
•
•
•
•
•
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
0.212E-05
0.671E-04
0.212E-02
0.671E-01
2.12
67.1
212E+01
671E+02
212E+04
0.477E-09
0.151E-07
0.477E-06
0.151E-04
0.477E-03
0.151E-01
0.477E+00
0.151E+02
0.477E+03
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If we represent the doubling time by T,then
our equation for An = (1 + r)nA0 for any time
t becomes
At =2(t/T)A0
13
Radioactivity
The time T at which half of a radioactive
substance decays is called the half-life of
the substance.
• At time T, ½ of the substance remains.
• At time 2T, ¼ of the substance remains.
• At time 3T, 1/8 of the substance remains.
• At time nT, (½)n of the substance remains.
14
In general, in t years, the amount remaining
of the original substance A0 is
A(t) = A0(½)t/T.
Note that t is a multiple of T, the half-life
15
A problem involving U238
The half-life of U238 is 4.51 x 109 years. How
much U238 was there when the solar system
first formed? The age of the solar system is
4.55 x 109 years.
We know that A(t) = A0(1/2)t/T, so
A(t)/A0 = (1/2)4.55x 10^9/4.51x10^9 =(1/2) 1.008
16
Let A(t)/A0 = (1/2)1.008 = .497 so A0 is about
twice as much as there is now.
17
The meltdown at Chernobyl Russia in 1986
released radio-active iodine into the atmosphere. It contaminated many square kilometers. The original contamination was
5 x 10-6 grams per meter2. A safe level is
1 X 10-9 grams per meter2 . The half-life of
radio-active (131I) is 8.05 days.
The thyroid absorbs iodine and if it absorbs
radio-active iodine, cancer may develop.
After how many days after the meltdown, were the levels
safe?
18
We use the formula A(t) = A0(1/2)t/T where
T is 8.05 days, A0 is 5 x 10-6 grms/m2 and
A(t) is 1 x 10-9 grams/m2. Lets call
A0 = 5000 x 10-9 . Since the levels use one signifIcant digit, let’s make 8.05 the digit 8.
We must solve:
1 x 10-9 = (½)t/8 5000 x 10-9
for t.
19
In 1 x 10-9 = (½)t/8 5000 x 10-9 ,we cancel
the 10-9 on both sides. So we have to solve
1 = (½)t/8 5000 for t. Divide both sides by
5000, getting 1/ 5000 = (½)t/8 . So
1/5 x 10-3 = (½)t/8 . Lets take the log of
both sides: log(1/5 x 10-3 ) = log( (½)t/8 )
But log(xn) = n log(x); our equation
20
log(1/5 x 10-3 ) = log( (½)t/8 ) becomes
log(1/5 x 10-3 ) = t/8 log( (½) ). Since
1/5 is 5-1 and ½ is 2-1 our equation becomes
log(5-1 x 10-3) = t/8 log(2-1). Since
log(AB) = log(A) + log(B), we get
log(5-1 ) + log(10-3) = t/8 log(2-1). Using
log(An) = nlog(A), we get:
-log(5) - 3log(10) = - t/8 log(2).
21
We know that log10(10) = 1, since 10=101 ,
so -log(5) - 3log(10) = - t/8 log(2) is
-log(5) - 3 = - t/8 log(2). Multiplying by
-1 we get: log(5) + 3 = (t/ 8) log(2). Then
t =8 (log(5) + 3 )/ log(2). So t =98.3 days.
22
2.2 Compound Interest
Compound interest is an example of a geometric progression. Let’s say a bank gives
10% interest annually. What will the balance
be after 4 years for an initial balance of
$100?
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•
•
•
•
•
•
After one year, A = (1+0.1)100=$110
After two years, A =(1.1)2 100 or $121
After three years,A= (1.1)3 100 or $133
After t years, A= (1.1)t 100
In general, A= (1+r)t P where P is the
original balance and r is the interest rate.
What if the interest is compounded quarterly?
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• After the 1st quarter, A=(1+0.025)100= $102
• After the 2nd quarter, A=(1+0.025)2100 = $105
• After the 3rd quarter, A =(1+0.025)3100= $107.68
• After the 4th quarter, A =(1+0.025)4100= $110.38
This is $0.38 greater than the non-compounded
case.
For t years, we get, A =(1+0.025)4t100.
What if the interest is compounded n times
a year?
25
we get A =(1+r/n)ntP.
26
What is the principal for $24 compounded
annually since 1626 for an interest rate of
4%. Now 2009-1626 = 383 years. So
A =1.04 383 24 = $80,164,150.21
What is the answer if the interest is
compounded
quarterly?
27
The answer: 1.01383*4 24 =$100,128,077.29
28
What happens if r is 100% and you compound every second for 1 year? There are
31536000 seconds in a year. So
A =(1 +1/ 31536000) 31536000 =
2.7182817784689974.
It can be shown in calculus that
lim n→ ∞ (1+1/n)n = a constant called e, where e
is 2.71828…
29
If we compound continuously, the principal
after t years at rate r is:
A = Pert
30
In order to facilitate calculations with P=Aert,
we introduce loge, i.e., a logarithm to the
base e. We refer to this as ln. It has the
same properties as log10. So:
•
ln(AB) = ln(A) + ln(B)
•
ln(A/B) = ln(A) – ln(B)
•
ln(An) = nln(A)
31
At what annual interest rate, r, would your
money double in twenty years?
We use A = Pert where A= 2P and t = 20.
2 = e20r
ln(2) = ln(e20r)
ln(2) = 20r lne(e)
32
ln(2) = 20r lne(e)
But lne(e) = 1 so
20r = ln(2)
r = ln(2) /20
So r = .6931/20 = 0.035 or 3.5%
33
There is a fast way of solving doubling time
problems for continuous growth problems.
We have 2 = e20r. Solving for r we get
ln(2) = 20r. Now ln(2) = .69 so
.69 = 20r.
If we express r in %, e.g., .05 becomes 5%. This
means we should multiply both sides by 100
getting 69 = 20 r (expressed in %). Solving for r we get
r = 3.5%
34
The initial amount of 60Co (cobalt 60) is
57.30 gms. After 6 months 53.64 remain.
What is 60Co half-life?
We use Nt = N0(1/2)t/T
35
Using Nt = N0(1/2)t/T and setting
R=53.64/57.30 we have log(R) = (6/T)log(.5)
So 6/T = log(R)/log(0.5) or
6/[log(R)/log(0.5)] = T or
T = 63 months or 63/12 years.
T = 5.25 years
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Spectrum of Electromagnetic Radiation
Region Wavlen Wavlen
Freq Energy
(Angstr)
(cm)
(Hz)
(ev)
Radio
Microwave
Infrared
Visible
Ultraviol
X-Rays
Gamma
>109
109 – 106
106 – 7000
7000 – 4000
4000 – 10
10 - 0.1
< 0.1
>10
<3x109
<10-5
10 - 0.01
3x109-3x1012
10-5-0.01
0.01 - 7x10-5 3x1012-4.3x1014 0.01–2
7x10-5-4x10-5 4.3x1014-7.5x1014 2 – 3
4x10-5- 10-7
7.5x1014-3x1017 3 – 103
10-7 - 10-9
3x1017- 3x1019 103 – 105
< 10-9
> 3x1019
> 105
1 Angstrom = 10-8 cm
From University of Tennessee Knoxville, Astronomy 161
37