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2.1 Geometric Progressions

For Fred Greenleaf’s QR Textbook

Compiled by Sam Marateck © 2009

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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

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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 23.15If r is the rate of growth, what is the general formula for growth?

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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.

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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 givesAs 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.

.

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Napier used as the base of his logarithms1- 10-7 . Henry Briggs (1561 – 1631), aBritish mathematician, proposed to Napiermaking the base 10. Thus log of 1 would be0. The tables we use today are based onBrigg’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 itselfmany times and assign it to y, y= xxxx.. thenlog(y) = log(x)+ log(x)+ log(x)+ log(x)+…So log(xn) = n log(x).

Look at y = 103. Take the log of both sideslog10 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.

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Earth quake magnitude measured in Richter Scale

• Magnitude Notes • -3.0 1.5 foot-pounds (18 inch-pounds) • -2.0 47 foot-pounds • -1.0 1,500 foot-pounds • 2.0 Felt only nearby, if at all • 4.0 Often felt up to 10's of miles away • 6.9 1995 Kobe, Japan, Earthquake • 7.3 1933 Salcha Earthquake• 9.0 2011 Honshu Japan Earhquake • 9.2 1964 Alaska Earthquake • 9.5 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 0.212E-05 0.477E-09 • 2.0 0.671E-04 0.151E-07 • 3.0 0.212E-02 0.477E-06 • 4.0 0.671E-01 0.151E-04 • 5.0 2.12 0.477E-03 • 6.0 67.1 0.151E-01 • 7.0 212E+01 0.477E+00 • 8.0 671E+02 0.151E+02 • 9.0 212E+04 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

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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.

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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

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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

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Let A(t)/A0 = (1/2)1.008 = .497 so A0 is about

twice as much as there is now.

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The meltdown at Chernobyl Russia in 1986released radio-active iodine into the atmos-phere. It contaminated many square kilo-meters. The original contamination was5 x 10-6 grams per meter2. A safe level is1 X 10-9 grams per meter2 . The half-life ofradio-active (131I) is 8.05 days. The thyroid absorbs iodine and if it absorbsradio-active iodine, cancer may develop. After how many days after the meltdown, were the levelssafe?

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We use the formula A(t) = A0(1/2)t/T whereT is 8.05 days, A0 is 5 x 10-6 grms/m2 andA(t) is 1 x 10-9 grams/m2. Lets call A0 = 5000 x 10-9 . Since the levels use one signif-Icant 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.

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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

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log(1/5 x 10-3 ) = log( (½)t/8 ) becomeslog(1/5 x 10-3 ) = t/8 log( (½) ). Since1/5 is 5-1 and ½ is 2-1 our equation becomeslog(5-1 x 10-3) = t/8 log(2-1). Sincelog(AB) = log(A) + log(B), we getlog(5-1 ) + log(10-3) = t/8 log(2-1). Usinglog(An) = nlog(A), we get:-log(5) - 3log(10) = - t/8 log(2).

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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.

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2.2 Compound Interest

Compound interest is an example of a geo-

metric 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 quart-erly?

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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= $110.38This is $0.38 greater than the non-compoundedcase. For t years, we get, A =(1+0.025)4t100.What if the interest is compounded n timesa year?

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we get A =(1+r/n)ntP.

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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?

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The answer: 1.01383*4 24 =$100,128,077.29

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What happens if r is 100% and you com-pound every second for 1 year? There are31536000 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…

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If we compound continuously, the principal

after t years at rate r is:

A = Pert

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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)

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At what annual interest rate, r, would yourmoney 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)

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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%

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There is a fast way of solving doubling timeproblems for continuous growth problems. We have 2 = e20r. Solving for r we getln(2) = 20r. Now ln(2) = .69 so

.69 = 20r.

If we express r in %, e.g., .05 becomes 5%. Thismeans we should multiply both sides by 100getting 69 = 20 r (expressed in %). Solving for r we getr = 3.5%

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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

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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 RadiationRegion Wavlen Wavlen Freq Energy (Angstr) (cm) (Hz) (ev)Radio >109 >10 <3x109 <10-5

Microwave 109 – 106 10 - 0.01 3x109-3x1012 10-5-0.01Infrared 106 – 7000 0.01 - 7x10-5 3x1012-4.3x1014 0.01–2Visible 7000 – 4000 7x10-5-4x10-5 4.3x1014-7.5x1014 2 – 3Ultraviol 4000 – 10 4x10-5- 10-7 7.5x1014-3x1017 3 – 103

X-Rays 10 - 0.1 10-7 - 10-9 3x1017- 3x1019 103 – 105

Gamma < 0.1 < 10-9 > 3x1019 > 105

1 Angstrom = 10-8 cm

From University of Tennessee Knoxville, Astronomy 161