11.1.1
An Introduction to Sequences and Series
The spectrum of the element
hydrogen is shown to the left. The dark
‘spectral’ lines that make-up the fingerprint
of hydrogen only exist at specific
wavelengths. This allows astronomers to
identify this gas in many different bodies in
the universe.
The energy corresponding to each
line follows a simple mathematical series
because at the atomic-scale, energy
comes in the form of specific packets of
light energy called quanta.
The Lyman Series of hydrogen lines is determined by the term relation:
2
1
1
13.7
n
E
n
⎛
−
⎜
⎝⎠
=
⎞
⎟
electron Volts
where n is the energy level, which is a positive integer from 1 to infinity, and E
n
is the
energy in electron Volts (eV) between level n and then lowest ‘ground state’ level
n=1. E
n
determines the energy of the light emitted by the hydrogen atom when an
electron loses energy by making a jump from level n to the ground state level.
Problem 1 – Compute the energy in eV of the first six spectral lines for the hydrogen
atom using E
n
.
Problem 2 – Suppose an electron jumped from an energy level of n=7 to a lower
level where n = 3. What is the absolute magnitude of the energy difference between
level n = 3 and level n = 7?
Space Math http://spacemath.gsfc.nasa.gov
Answer Key
11.1.1
Problem 1 – Compute the energy in eV of the first six spectral lines for the hydrogen
atom using the formula for E
n
.
Answer: Example: E
2
= 13.7 (1-1/4) = 13.7 x ¾ = 10.3 eV.
E
2
= 10.3 eV
E
3
= 12.2 eV
E
4
= 12.8 eV
E
5
= 13.2 eV
E
6
= 13.3 eV
E
7
= 13.4 eV
Problem 2 – Suppose an electron jumped from an energy level of n=7 to a lower level
where n = 3. What is the absolute magnitude of the energy difference between level n
= 3 and level n = 7?
Answer: E
3
= 12.2 eV and E
7
= 13.4 eV so E
7
-E
3
= 1.2 eV
Space Math http://spacemath.gsfc.nasa.gov
11.1.2
An Introduction to Sequences and Series
Long before the planet Uranus was
discovered in 1781, it was thought that
their distances from the sun might have to
do with some mathematical relationship.
Many proposed distance laws were
popular as early as 1715.
Among the many proposals was one
developed by Johann Titius in 1766 and
Johann Bode 1772 who independently
found a simple series progression that
matched up with the planetary distances
rather remarkably.
Problem 1 – Compute the first eight terms, n=0 through n=7, in the Titius-Bode Law
whose terms are defied by Dn = 0.4 + 0.3*2
n
where n is the planet number beginning
with Venus (n=0). For example, for Neptune, N = 7 so Dn = 0.4 + 0.3 (128) = 38.8
AU.
Problem 2 – A similar series can be determined for the satellites of Jupiter, called
Dermott’s Law, for which each term is defined by Tn = 0.44 (2.03)
n
and gives the
orbit period of the satellite in days What are the orbital periods for the first six
satellites of Jupiter?
Space Math http://spacemath.gsfc.nasa.gov
Answer Key
11.1.2
Problem 1 – Compute the first eight terms in the Titius-Bode Law whose terms are
defied by Dn = 0.4 + 0.3*2
n
where n is the planet number beginning with Venus (n=0).
For example, for Neptune, N = 7 so Dn = 0.4 + 0.3 (128) = 38.8 AU.
Answer: For n = 0, 1, 2, 3, 4, 5, 6 and 7 the distances are
Venus: d0 = 0.7 actual planet distance = 0.69
Earth: d1 = 1.0 actual planet distance = 1.0
Mars: d2 = 1.6 actual planet distance = 1.52
Ceres: d3 = 2.8 actual planet distance = 2.77
Jupiter: d4 = 5.2 actual planet distance = 5.2
Saturn: d5 = 10.0 actual planet distance = 9.54
Uranus: d6 = 19.6 actual planet distance = 19.2
Neptune: d7 = 38.8 actual planet distance = 30.06
Note: Ceres is a large asteroid not a planet.
Problem 2 – A similar series can be determined for the satellites of Jupiter, called
Dermott’s Law, for which each term is defined by Tn = 0.44 (2.03)
n
and gives the orbit
period of the satellite in days What are the orbital periods for the first six satellites of
Jupiter?
Answer: T0 = 0.44 days
T1 = 0.89 days
T2 = 1.81 days
T3 = 3.68 days
T4 = 7.47 days
T5 = 15.17 days
Space Math http://spacemath.gsfc.nasa.gov
11.2.1
Arithmetic Sequences and Series
Arithmetic series
appear in many different
ways in astronomy and space
science. The most common is
in determining the areas
under curves.
For example, an
arithmetic series is formed
from the addition of the
rectangular areas a
n
in the
figure to the left.
Imagine a car traveling at a speed of 11 meters/sec and wants to accelerate
smoothly to 22 meters/sec to enter a freeway. As it accelerates, its speed changes
from 11 m/sec at the first second, to 12 m/sec after the second second and 13 m/sec
after the third second and so on.
Problem 1 – What is the general formula for the Nth term in this series for V
n
where
the first term in the series, V
1
= 11 m/sec?
Problem 2 – What is the value of the term V
8
in meters/sec?
Problem 3 – For what value of N will V
n
= 22 meters/sec?
Problem 4 – What is the sum, S
12
, of the first 12 terms in the series?
Problem 5 – If the distance traveled is given by D = S
12
x T where T is the time
interval between each term in the series, how far did the car travel in order to reach
22 meters/sec?
Space Math http://spacemath.gsfc.nasa.gov
Answer Key
11.2.1
Problem 1 – What is the general formula for the Nth term in this series for V
n
where
V
1
= 11 m/sec?
Answer: V
n
= 10 + 1.0n
Problem 2 – What is the value of the term V
8
in meters/sec?
Answer: V
8
= 10 + 1.0 *(8) so V
8
= 18.0 m/sec
Problem 3 – For what value of N will V
n
= 22 meters/sec?
Answer: 22 = 10 + 1.0 N so N = 12
Problem 4 – What is the sum, S
12
, of the first 12 terms in the series?
Answer:
The first 12 terms in the series are:
11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22
The sum of an arithmetic series is given by S
n
= n (a
1
+ a
n
)/2
So for n = 12, a
1
= 11 and a
12
= 22 we have S
12
= 211 m/sec.
Problem 5 – If the distance traveled is given by D = S
12
x T where T is the time
interval between each term in the series, how far did the car travel in order to reach 22
meters/sec?
Answer: In the series for V
n
, the time interval between terms is 1.0 seconds. Since S
12
= 211 m/sec we have D = 211 m/sec x 1.0 sec so D = 211 meters.
Space Math http://spacemath.gsfc.nasa.gov
11.2.2
Arithmetic Sequences and Series
The $150 million Deep Space 1
spacecraft launched on October 22, 1998
used an ion engine to travel from Earth to
the Comet Borrelly. It arrived on
September 22, 2001.
By ejecting a constant stream of
xenon atoms into space, at speeds of
thousands of kilometers per second, the
new ion engine could run continuously for
months. This allowed the spacecraft to
accelerate to speeds that eventually could
exceed the fastest rocket-powered
spacecraft.
Problem 1 – The Deep Space 1 ion engine produced a constant acceleration,
starting from a speed of 44,000 km/hr, reaching a speed of 56,060 km/hr as it passed
the comet 36 months later. The series representing the monthly average speed of
the spacecraft can be approximated by a series based upon its first 7 months of
operation given by:
n 1 2 3 4 5 6 7
V
n
44,000 44,335 44,670 45,005 45,340 45,675 46,010
What is the general formula for Vn?
Problem 2 – Suppose the Deep Space I ion engine could be left on for 30 years!
What would be the speed of the spacecraft at that time?
Problem 3 – The sum of an arithmetic series is given by S
n
= n (a
1
+ a
n
)/2. What is
the sum, S
36
, of the first 36 terms of this series?
Problem 4 – The total distance traveled is given by D = S
36
x T where T is the time
between series terms in hours. How far did the Deep Space 1 spacecraft travel in
reaching Comet Borrelly?
Space Math http://spacemath.gsfc.nasa.gov
Answer Key
11.2.2
Problem 1 – The Deep Space 1 ion engine produced a constant acceleration, starting
from a speed of 44,000 km/hr, reaching a speed of 56,060 km/hr as it passed the
comet 36 months later. The series representing the monthly average speed of the
spacecraft can be approximated by a series based upon its first 7 months of operation
given by:
n 1 2 3 4 5 6 7
V
n
44,000 44,335 44,670 45,005 45,340 45,675 46,010
What is the general formula for Vn?
Answer: Vn = 44,000 + 335(n-1)
Problem 2 – Suppose the Deep Space I ion engine could be left on for 30 years!
What would be the speed of the spacecraft at that time?
Answer: 30 years = 30 x 12 = 360 months so the relevant term in the series is V
360
which has a value of V360 = 44,000 + 335x(360-1) so V
360
= 164,265
kilometers/hour.
Problem 3 – What is the sum, S
36
, of the first 36 terms of this series?
Answer: V
36
= 44,000 + 11,725 = 55,725 km/hour. Then S
36
= 36 (44000 +
55,725)/2 so S
36
= 1,795,050 kilometers/hour.
Problem 4 – The total distance traveled is given by D = S
36
x T where T is the time
between series terms in hours. How far did the Deep Space 1 spacecraft travel in
reaching Comet Borrelly if there are 30 days in a month?
Answer: The time between each series term is 1 month which equals 30 days x
24hours/day = 720 hours. The total distance traveled is then
D = 1,795,050 km/hr x 720 hours
D = 1,292,436,000 kilometers.
Note, this path was a spiral curve between the orbit of Earth and the comet. During this
time, it traveled a distance equal to 8.7 times the distance from the Sun to Earth!
Space Math http://spacemath.gsfc.nasa.gov
11.3.1
Geometric Sequences and Series
When light passes through a dust
cloud, it decreases in intensity. This
decrease can be modeled by a geometric
series where each term represents the
amount of light lost from the original beam
of light entering the cloud.
The image to the left shows the dark
cloud called Barnard 68 photographed by
astronomers at the ESO, Very Large
Telescope observatory. The dust cloud is
about 500 light years from Earth and about
1 light year across.
Problem 1 – A dust cloud causes starlight to be diminished by 1% in intensity for
each 100 billion kilometers that it travels through the cloud. If the initial starlight has a
brightness of B
1
= 350 lumens, what is the geometric series that defines its
brightness?
Problem 2 – What are the first 8 terms in this series for the brightness of the light?
Problem 3 - How far would the light have to penetrate the cloud before it loses 50%
of its original intensity?
Space Math http://spacemath.gsfc.nasa.gov
Answer Key
11.3.1
Problem 1 – A dust cloud causes starlight to be diminished by 1% in intensity for each
100 billion kilometers that it travels through the cloud. If the initial starlight has a
brightness of B
1
= 350 lumens, what is the geometric series that defines its
brightness?
Answer: B
1
= 350 and r = 0.01 so if each term represents a step of 100 billion km in
distance, the series is B
B
n
= 350 (0.99)
n-1
Problem 2 – What are the first 8 terms in this series for the brightness of the light?
Answer: Calculate B
n
for n = 1, 2, 3, 4, 5, 6, 7, 8
N 1 2 3 4 5 6 7 8
Bn 350 346 343 340 336 333 330 326
Problem 3 - How far would the light have to penetrate the cloud before it loses 50% of
its original intensity?
Answer: Find the term number for which B
n
= 0.5*350 = 175. Then
175 = 350 (0.99)
n-1
solve for n using logarithms:
Log(175) = Log(350) + (n-1) log(0.99)
so n-1 = (log(175) – Log(350))/log(0.99)
and so n-1 = 68.96
or n = 68.
Since the distance between each term is 100 billion km, the penetration distance to
half-intensity will be 68 x 100 billion km = 6.8 trillion kilometers.
Note: 1 light year = 9.3 trillion km, the distance is just under 1 light year.
Space Math http://spacemath.gsfc.nasa.gov
11.4.1
Infinite Geometric
Series
The star field shown above was photographed by NASA’s WISE satellite
and shows thousands of stars, and represents an area of the sky about the size
of the full moon. Notice that the stars come in many different brightnesses.
Astronomers describe the distribution of stars in the sky by counting the number
in various brightness bins.
Suppose that after counting the stars in this way, an astronomer
determines that the number of the can be modeled by an infinite geometric series:
B
B
m
= 100 a where a is a scaling number between 1/3 and 1/2. The series
term index, m, is related to the apparent magnitude of the stars in the star field
and ranges from m:[1 to +infinity].
m-1
Problem 1 –What are the first 7 terms in this series for a=0.398?
Problem 2 - What is the sum of the geometric series, B
m
, for A) a=0.333? B)
a=0.398? C) a=0.5?
Space Math http://spacemath.gsfc.nasa.gov
Answer Key
11.4.1
Problem 1 – Suppose that the brightness of this field can be approximately given by
the geometric series B
m
= 100 a
m-1
where a is a number between 1/2 and 1/3. The
series term index, m, is related to the apparent magnitude of the stars in the star field
and ranges from m:[1 to +infinity]. What are the first 7 terms in this series for a=0.398?
Answer: a = 0.398 then:
m 1 2 3 4 5 6 7
Bm 100 40 16 6.3 2.5 1.0 0.4
Problem 2 - What is the sum of this geometric series for A) a=0.333? B) a=0.398? C)
a=0.5?
Answer: A) The common ratio is 0.333 and the first term has a value of B
1
= 100, so B
= 100 / (1-0.333) and so B= 150.
B) The common ratio is 0.398 and the first term has a value of B
1
= 100, so B = 100 /
(1-0.398) and so B= 166.
C) The common ratio is 0.50 and the first term has a value of B
1
= 100, so B = 100 /
(1-0.50) and so B= 200.
Note: Normally, star counts are always referred to a specific magnitude system since
the brightness of stars at different wavelengths varies.
Space Math http://spacemath.gsfc.nasa.gov
11.4.2
Infinite Geometric Series
Rockets work by throwing mass out
their ends to produce ‘thrust’, which moves
the rocket forward.
As fuel mass leaves the rocket, the
mass of the rocket decreases and so the
speed of the rocket steadily increases as
the rocket becomes lighter and lighter.
An interesting feature of all rockets
that work in this way is that the maximum
attainable speed of the rocket is
determined by the Rocket Equation, which
can be understood by using an infinite
geometric series.
Problem 1 – The Rocket Equation can be approximated by the series V = V
1
a
n-
1
,
where a is a quantity that varies with the mass ratio of the surviving payload mass,
m, to the initial rocket mass, M. For a rocket in which the payload mass is 10% of the
total fueled rocket mass, a = 0.56. What are the first 5 terms in the equation for the
rocket speed, V, if the exhaust speed is V
1
= 2,500 meters/sec?
Problem 2 – The partial sums of the series, S
1
, S
2
, S
3
, …, reflect the fact that, as
the rocket burns fuel, the mass of the rocket decreases, so the speed will increase.
For example, after two seconds, the second time interval, S
2
= V
1
+ V
2
= 2,500 +
1,400 = 3,900 m/sec. After three seconds, the speed is S
3
= 2,500 + 1,400 + 784 =
4,684 m/sec etc. What is the speed of the rocket after A) 15 seconds? B) 35
seconds?
Problem 3 – The maximum speed of the payload is given by the limit to the sum of
the series for V. What is the sum of this infinite series for V in meters/sec?
Space Math http://spacemath.gsfc.nasa.gov
Answer Key
11.4.2
Problem 1 – The Rocket Equation can be approximated by the series V = V
0
a
n
,
where a is a quantity that varies with the mass ratio of the surviving payload mass, m,
to the initial rocket mass, M. For a rocket in which the payload mass is 10% of the total
fueled rocket mass, a = 0.56. What are the first 5 terms in the equation for the rocket
speed, V, if the exhaust speed is 2,500 meters/sec?
Answer: V
1
= 2500 (0.56)
0
= 2,500 m/sec
V
2
= 2500 (0.56)
1
= 1,400 m/sec
V
3
= 2500 (0.56)
2
= 784 m/sec
V
4
= 2500 (0.56)
3
= 439 m/sec
V
5
= 2500 (0.56)
4
= 246 m/sec
So the sequence is V=2,500 + 1,400 + 784 + 439 + 246 + …
Problem 2 – The partial sums of the series, S
1
, S
2
, S
3
, …, reflect the fact that, as the
rocket burns fuel, the mass of the rocket decreases, so the speed will increase. For
example, after two seconds, the second time interval, S
2
= V
1
+ V
2
= 2,500 + 1,400 =
3,900 m/sec. After three seconds, the speed is S
3
= 2,500 + 1,400 + 784 = 4,684
m/sec etc. What is the speed of the rocket after A) 15 seconds? B) 35 seconds?
Answer: A) Recall that the sum of a geometric series is given by S
n
= a(1-r
n
)/(1 – r)
So A) r = 0.56, a = 2500, n = 15 and so
S
15
= 2500 (1-(0.56)
15
)/(1 – 0.56)
S = 5,681 meters/sec.
B) n = 35 so S
35
= 2500 (1-(0.56)
35
)/(1-0.56)
S = 5,682 meters/sec.
Problem 3 – The maximum speed of the payload is given by the limit to the sum of the
series for V. What is the sum of this infinite series for V in meters/sec?
Answer: S = a/(1-r) = 2500/(1-0.56) = 5,682 meters/sec.
Note: This speed is equal to 20,500 km/hour.
Space Math http://spacemath.gsfc.nasa.gov
11.5.1
Recursive Rules for Sequences
This spherical propellant tank is
an important component of testing for
the Altair lunar lander, an integral part of
NASA's Constellation Program. It will be
filled with liquid methane and extensively
tested in a simulated lunar thermal
environment to determine how liquid
methane would react to being stored on
the moon.
The volume of a sphere is a
mathematical quantity that can be
extended to spaces with different
numbers of dimensions.
The mathematical formula for the
volume of a sphere in a space of N
dimensions is given by the recursion
relation
2
2
() ( 2)
R
VN VN
N
π
=
−
For example, for 3-dimensional space, N
= 3 and since from the table to the left,
V(N-2) = V(1) = 2R, we have the usual
formula
3
4
(3)
3
VR
Dimension Formula Volume
0 1 1.00
1 2R 2.00
2
π
2
R
3.14
3
4
π
R
3
3
4.19
4
5
6
7
8
9
10
π
=
Problem 1 - Calculate the volume
formula for 'hyper-spheres' of dimension
4 through 10 and fill-in the second
column in the table.
Problem 2 - Evaluate each formula for
the volume of a sphere with a radius of
R=1.00 and enter the answer in column
3.
Problem 3 - Create a graph that shows
V(N) versus N. For what dimension of
space, N, is the volume of a
hypersphere its maximum possible
value?
Problem 4 - As N increases without
limit, what is the end behavior of the
volume of an N-dimensional
hypersphere?
Space Math http://spacemath.gsfc.nasa.gov
11.5.1
Answer Key
Dimension Formula Volume
0 1 1.00
1 2R 2.00
2
πR
2
3.14
3
3
4
3
R
π
4.19
4
24
2
R
π
4.93
5
25
8
15
R
π
5.26
6
36
6
R
π
5.16
7
37
16
105
R
π
4.72
8
48
24
R
π
4.06
9
49
32
945
R
π
3.30
10
510
120
R
π
2.55
Problem 1 - Answer for N=4:
2
π
R
2
V
(4) = (4 − 2)
4
2
π
R
2
V
(4) = (2)
4
2
π
R
2
V
(4) = (
π
2
)
4
π
24
R
V
(4) =
2
V
V
R
Problem 2 - Answer for N=4:
V(4) = (0.5)(3.141)
2
= 4.93.
Problem 3 - The graph to the left
shows that the maximum hypersphere
volume occurs for spheres in the fifth
dimension (N=5). Additional points
have been calculated for N=11-20 to
better illustrate the trend.
Problem 4 - In the limit for spaces with
very large dimensions, the hypersphere
volume a
pp
roaches zero!
Hypersphere volume
6
5
4
)NV(
3
2
1
0
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
N
Space Math http://spacemath.gsfc.nasa.gov
