Download i + 2 - UBC Math

(1) Compute the following sums:
P150
(a)
i=1 2
P50
(b)
n=1 n
P60
(c)
n=10 n
P25
2
(d)
n=1 3n
P55
(e)
i=1 (i + 2)
P500
(f)
k=100 k
P50
2
(g)
k=2 (k − 2k + 1)
P20
3
(h)
m=10 m
Solution: For the solutions to these, we will use the following results, which have been
established by Gauss’s formula, and the notation shown below for convenience:
S0 (n) =
n
X
1=n
S1 (n) =
i=1
n
X
i=1
i=
n(n + 1)
2
2
n
X
n(n + 1)
n(n + 1)(2n + 1)
3
S3 (n) =
i =
i =
S2 (n) =
6
2
i=1
i=1
P150
P150
(a)
i=1 2 = 2
i=1 1 = 2(150) = 300
P50
(50)(50+1)
= 1275
(b)
n=1 n =
2
P60
P60
P9
(60)(61)
(c)
− (9)(10)
= 1785
n=10 n =
n=1 n −
n=1 n =
2
2
P25
P25
(25)(26)(51)
2
= 5525
(d)
n=1 3n = 3
n=1 = 3
6
P55
P55
P55
(55)(56)
(e)
+ 2(55) = 1650.
i=1 (i + 2) =
i=1 i +
i=1 2 =
2
P500
P99
P500
(500)(501)
− (99)(100)
= 120300
(f)
k=100 k =
k=1 k −
k=1 k =
2
2
P
P
P
P50
50
50
50
2
2
(g)
k=2 (k − 2k + 1) =
k=2 k − 2
k=2 k +
k=2 1
P50
P
P
50
50
= ( k=1 k 2 − 1) − 2( k=1 k − 1) + ( k=1 1 − 1)
n
X
2
− 1) − 2( (50)(51)
− 1) + (1(55) − 1) = 40430
= ( (50)(51)(101)
6
2
P20
P
P
20
9
(20)(21) 2
3
3
3
(h)
] − [ (9)(10)
]2 = 46125
m=10 m =
m=1 m −
m=1 m = [
2
2
(2) A clock at London’s Heathrow airport chimes every half hour. At the beginning of the
n’th hour, the clock chimes n times. (For example, at 8:00 am the clock chimes 8 times,
at 2:00 pm the clock chimes fourteen times, and at midnight the clock chimes 24 times.)
The clock also chimes once at half-past every hour. Determine how many times in total
the clock chimes in one full day. Use sigma notation to write the form of the series before
finding its sum.
Solution: ¡¡ There is no solution for this problem. ¿¿
1
(3) Use the sigma summation notation to set up the following problems, and then apply
known formulae to compute the sums.
(a) Find the sum of the first 100 even numbers.
(b) Find the sum of the first 100 odd numbers.
(c) The sum of the first 100 integers.
Solution:
(a) An even number can be written in the form 2n. Thus the sum of the first fifty such
P50
P50
numbers is n=1 (2n) = 2 n=1 n = 2S1 (50) = 2550.
(b) An odd number can be written in the form 2n−1 where n = 1, 2... Thus the sum of the
P50
P50
P50
first fifty such numbers is n=1 (2n −1) = 2 n=1 n − n=1 1 = 2S1 (50) −S0 (50) =
2550 − 50 = 2500
(c) The sum of the first 100 integers is
P100
n=1
n = S1 (100) = 5050 which is the same as
the total of the sums in parts (a) and (b)
(4) Consider all the integers that are of the form n(n − 1) where n = 1, 2, 3... Find the
sum of the first 50 such numbers.
Solution: ¡¡ There is no solution for this problem. ¿¿
(5) A set of Japanese lacquer boxes have been made to fit perfectly one inside the other.
(When objects fit inside one another perfectly, we say that they form a ”nested” set.) All
the boxes are cubical, and they have sides of lengths 1,2,3..15 inches. Find the total volume
enclosed by all the boxes combined. (You may assume that the walls are negligeably thin).
Solution: ¡¡ There is no solution for this exercise. ¿¿
(6) A framing shop uses a square piece of matt cardboard to create a set of square frames,
one cut out from the other, with as little wasted as possible. The original piece of cardboard
is 50 cm by 50 cm. Each of the ”nested” square frames (see Prob 5 for defn of nested)
has a border 2 cm thick. How many frames in all can be made from this original piece of
cardboard? What is the total area that can be enclosed by all these frames together?
Solution: ¡¡ There is no solution for this exercise. ¿¿
(7) The geometric series: Refresh your memory on the sum of a (finite) geometric series
before attempting this problem.
(a) Find the sum of the first 10 numbers of the form 1.1k ,
k = 1, 2, ... Now find the sum
of the first 20 such numbers, the first 30 such numbers, the first 40 such numbers, and
2
the first 50 such numbers.
(b) Repeat the process but now find sums of the numbers 0.9k ,
k = 1, 2...
(c) What do you notice about the pattern of results in part (a) and in part (b) ? Can
you explain what is happening in each of these cases and why they are different?
(d) Now consider the general problem of finding a value for the sum
N
X
rk
k=0
when the number N gets larger and larger. Suggest under what circumstances this
sum will stay finite, and what value that finite sum will approach. To do this, you
should think about the formula for the finite geometric sum and determine how it
behaves for various values of r as N gets very large. This idea will be very important
when we discuss infinite series.
Solution:
20+1
30+1
1−1.110+1
= 18.53, First 20: 1−1.1
= 64, First 30: 1−1.1
1−1.1
1−1.1
1−1.1
40+1
50+1
First 40: 1−1.1
= 487.8, First 50: 1−1.1
= 1281.3.
1−1.1
1−1.1
10+1
20+1
30+1
= 6.86. First 20: 1−0.9
= 8.9 First 30: 1−0.9
=
First 10: 1−0.9
1−0.9
1−0.9
1−0.9
40+1
50+1
= 9.86 First 50: 1−0.9
= 9.95.
40: 1−0.9
1−0.9
1−0.9
(a) First 10:
= 181.9,
(b)
9.6 First
¡¡ Solutions (c) and (d) are missing. ¿¿
(8) Find the area between the x axis and the graph of the function y = f (x) = 2 − x
between x = 0 and x = 1, (a) By using your knowlege about the area of a triangular
region, and (b) By setting up the problem as a sum of the areas of n rectangular strips,
using the appropriate summation formula, and letting the number of strips (n) get larger
and larger to arrive at the result. Show that your answer in (b) is then identical to the
answer in (a).
Solution: The intercepts of the line y = f (x) = 2−x on the x and y axes are both 2 so that
the height and the base of the triangle have lengths 2. Thus its area is A = (1/2)2 × 2 = 2.
We now subdivide the iterval [0, 2] on the x axis into n equal pieces (each has width
h = 2/n) and consider the strips that are so defined. The coordinate of the right endpoint of
the k’th interval is xk = 2(k/n). The y coordinate on the straight line corresponding to xk
(which is also the height of the k’th strip is yk = f (xk ) = 2 −xk = 2 −2(k/n) = 2(1 −k/n).
Thus the area of the (approximately rectangular) k’th strip is Ak = width × height =
Pn
Pn
2/n×2(1−k/n). The total area of n strips is A(n) = k=1 (4/n)(1−k/n) = (4/n)[ k=1 1−
3
(1/n)
Pn
k=1
k] = (4/n)n − (4/n2 )(n(n + 1)/2) = 4 − 2(n + 1)/n. If we now increase the
number of strips, n → ∞, we find that this area approaches the true area of the triangle,
A = 4 − 2 = 2. (This uses the fact that the ratio (n + 1)/n approaches 1 as n gets large.
(9) Find the area between the graphs of the following two functions: y = f (x) = 2x and
y = g(x) = 1 + x2 , and the y-axis. (Hint: what do we mean by ”between” ? Where
does this region begin and where does it end?) This problem should be set up in the form
of a sum of areas of rectangular strips. You should NOT use previous familiarity with
integration techniques to solve it.
Solution: The region is below the graph of y = g(x) = 1 + x2 , above y = f (x) = 2x and
to the right of the y axis. The two functions have graphs that intersect when 2x = 1 + x2 ,
i.e. when x2 − 2x + 1 = 0 which implies that x = 1, (the graphs are actually tangent
at that point) so we want to find the area between the two graphs for 0 < x < 1. We
subdivide this interval into N equal pieces, each of width 1/N . The coordinates are
x0 = 0, x1 = 1/N,
xk = k/N, ...,
xN = 1. The area of the k’th rectangle is ak =
height × width where the height is g(xk ) − f (xk ) = 1 + x2k − 2xk = 1 + (k/N )2 − 2(k/N )
PN
The total area of the region is: A = (1/N ) k=1 (1 + (k/N )2 − 2(k/N )). This expression
PN
PN
can be simplified to A = (1/N )N + (1/N 3 ) k=1 k 2 − 2(1/N 2 ) k=1 k. We find that
A = 1 + N (N + 1)(2N + 1)/(6N 3 ) − 2N (N + 1)/(2N 2 ). We take a limit as the number of
pieces, N gets very large, and find that A = 1 + (1/3) − 1 = 1/3.
(10) Use methods similar to those developed in class and previous homework to find the
area under the graph of the function y = f (x) = e2x over the interval between x = 0 and
x = 2. (Hint: you will need to sum a geometric series to do this.)
Solution: As before, we subdivide the interval into N pieces, each of width 2/N . Then
xk = 2k/N and yk = f (xk ) = e2xk = e2k(2/N) = e4k/N . The area of the k’th rectangle
PN
is ak = (2/N )e4k/N and the total area of all the rectangles is A = (2/N ) k=1 e4k/N =
PN
(2/N ) k=1 r k where r = e4/N . This is a finite geometric series. Because the series starts
with k = 1 and not with k = 0, the sum is A = (2/N )[(1 − r N+1 )/(1 − r) − 1]. After some
simplification we find that A = (2/N )e4/N (1 − e4 )/(1 − e4/N ) = 2(1 − e4 )/[N (e−4/N −
1)]. We need to determine what happens when N gets very large. We can use a linear
approximation for e−4/N = 1 − 4/N.. to evaluate the limit of the term in square brackets,
and we find that A = 2(1 − e4 )/[−N (4/N )] = (e4 − 1)/2.
4
(11) Prove the following summation formula:
k
X
n2 =
n=1
1
k(k + 1)(2k + 1).
6
Hint: Use the fact that (n + 1)3 = n3 + 3n2 + 3n + 1 and follow steps similar to those used
to establish Gauss’s formula.
Solution: ¡¡ There is no solution to this exercise. ¿¿
(12) Write the sum
1 + 16 + · · · + k 4
in
P
notation. Find a formula for it.
Solution:
k
X
i4
i=1
k(k − 1)(k − 2)
k(k − 1)
+ 50 ·
2
3!
k(k − 1)(k − 2)(k − 3)
k(k − 1)(k − 2)(k − 3)(k − 4)
+ 60 ·
+ 24 ·
4!
5!
or the equivalent polynomial in powers of k.
P59
(13) Find formulas for (a) k=0 k;
P120
(b) k=20 k;
Pn
(c) k=0 k;
Pn−1
(d) k=0 (k 2 + 1);
Pn
(e) k=1 (k 2 + 1);
Pn
(f) k=0 (k 3 + k 2 − k);
Pk−1
(g) n=0 n2 .
0 + 1 · k + 15 ·
Solution: Not in quite the same order:
(a) 60(59)/2 = 1770.
(b) One way to do this is write
120
X
k=20
k=
120
X
k−
k=1
Another is to find a formula for
19
X
k = 121(120) − 20(19) /2 = 7070
k=1
20+n
X
k=20+0
5
k
and then set n = 100. The second probably works in more general circumstances. Note
that the way to do all sums of this kind is actually not at all obvious—the first step is to
solve an apparently more difficult problem by introducing a variable, which you then set
at the end to a specific value.
(c) n(n + 1)/2.
3
2
Pn
Pn
+ n + n.
(e) k=1 k 2 + k=1 1 = 2n + 3n
6
2(n − 1)3 + 3(n − 1)2 + (n − 1)
(d)
+n
6
3
2
2(k − 1) + 3(k − 1) + (k − 1)
(g)
6
(f)
n+
(14) Use the
P
9n(n − 1) 7n(n − 1)(n − 2) n(n − 1)(n − 2)(n − 3)
+
+
2
3
4
summation notation to set up the following problems, and then compute.
(a) Find the sum of the first 50 even numbers.
(b) Find the sum of the first 50 odd numbers.
(c) Find the sum of the first 50 integers of the form n(n + 1) (where n = 1, 2, 3, . . . )
P50
P50
Solution: (a) i=1 2i = 2 1 i = 50(51) (Sum starting with 0 also OK.)
P49
P49
(b) i=0 2i + 1 = 50 + 2 0 i = 2, 500
3
2
P50
P50
P50
+ n + n(n + 1)
(c) i=1 i(i + 1) = i=1 i2 + i=1 i = 2n + 3n
= 44, 200
6
2
[n=50]
(15) The Great Pyramid of Giza, Egypt, built around 2,720-2,560 BC by Khufu (anglicized
to Cheops) has a square base. We will assume that the base has side length 200 meters.
(Actual measurement: 230 m) The pyramid is made out of blocks of stone whose size is
roughly 1 · 1 · 0.73m3 (Actual: 1.17 · 1.17 · 0.73m3 ). There are 200 layers of blocks, so that
the height of the pyramid is 200 · 0.73 = 146 m. Assume that the size of the pyramid steps
(i.e. the horizontal distance between the end of one step and the beginning of another) is
0.5 m. We will also assume that the pyramid is solid, i.e. we will neglect the (relatively
small) spaces that make up passages and burial chambers inside the structure.
(a) How many blocks are there in the layer that makes up the base of the pyramid?
How many blocks in the second layer ?
(b) How many blocks are there at the very top of the pyramid?
(c) Write down a summation formula for the total number of blocks in the pyramid
and compute the total. (Hint: you may find it easiest to start the sum from the top layer
and work your way down.)
6
Solution: (a) There are 200 · 200 = 2002 = 40, 000 stones in the base and the blocks are
indented 0.50.5 meters in the next layer up, so that the side lengths are 199 meters and
thus there are 199 ∗ 199 = 39601 blocks.
(b) Each level reduces the side length by 1 meter, so we are left with one block at the
top.
(c) We want to sum the number of blocks in all the layers, so we get
P200
n=1
n2 . We
use the summation formula
N
X
2N 3 + 3N 2 + N
n =
.
6
n=1
2
Plugging in N = 200 we get 2, 686, 700 blocks in the pyramid.
You might be amused by the web page
http://www.geocities.com/Athens/Oracle/2451/pyramid/desk2.htm
(16) Your local produce store has a special on oranges. Their display of fruit is a triangular
pyramid with 100 layers, topped with a single orange (i.e. top layer: 1). The layer
second from the top has three (3 = 1 + 2) oranges, and the one directly under it has six
(6 = 3 + 2 + 1). The same pattern continues for all 100 layers. (This results in efficient
hexagonal packing, with each orange sitting in a little depression created by three neighbors
right under it.)
(a) How many oranges are there in the fourth and fifth layers from the top ? How
many in the N -th layer from the top ? Draw a picture of the second, third, and fourth
layers.
(b) If the pyramid of oranges only has 3 layers, how many oranges are used in total?
What if the pyramid has 4, or 5 layers?
(c) Write down a formula for the sum of the total number of oranges that would be
needed to make a pyramid with N layers. Simplify your result so that you can use the
P
P 2
summation formulae for
n and for
n to determine the total number of oranges in
such a pyramid.
(d) Determine how many oranges are needed for the pyramid with 100 layers.
Solution: (a) The layers contain 1, 3, 6, 10, 15 oranges starting at the top. The N -th
PN
layer from the top contains n=1 n = N (N + 1)/2
(b) A pyramid with 1, 2, 3, 4, 5 layers would have 1, 4, 10, 20, 35 oranges, respectively.
7
(c ) If the pyramid has N layers, there will be
N
X
k(k + 1)/2 = N + N (N − 1) +
k=1
N (N − 1)(N − 2)
6
(d) For 100 layers, we would need
100 + 100(99) + 100 · 99 · 98/6 = 171, 700
oranges.
(17) (Geometric Seri) A (finite) geometric series with k terms is a series of the form
1+r +r +r +···+r
2
3
k−1
=
k−1
X
rn .
n=0
The sum of this series is S = (1 − r k )/(1 − r) as long as r 6= 1. (a) If r = 1, what is the
sum? (b) Find the sum of the series 1 + 2 + 22 + 23 + · · · + 210 . (c) Find the sum of the
series 1 + (0.5) + (0.5)2 + (0.5)3 + · · · + (0.5)10 .
Solution: (a) 2047; (b) 2 − 1/1024 = 2047/1024.
(18) A branching colony of fungus starts as a single spore. The spore puts out a single
branch which grows for 0.1 mm, producing a segment of that length. The tip splits into
two new tips, and each grows for 0.1 mm, producing two new segments. (The total length
at this point would be 0.1 + 0.2 = 0.3 mm. Each tip splits again, and then the four tips
grow for 0.1 mm. Suppose that this pattern of growth has continued for 20 generations
(a generation is the time between branching events). How many tips will there be? What
will be the total length of all the filaments in the fungus?
Solution: In generation 1, 1 tip is produced; in generation 2, 2; in generation n 2n−1 tips
are produced and present. (Tips are growing ends.) But the total length is 1 + 2 + · · · +
2n−1 = 2n − 1.
(19) (a) Find a formula for the area in N uniform strips of width 1/N both under and
Pn
over the curve y = x4 , using the formula for the sum 1 k 4 . The strips cover the interval
[0, 1]. (b) Draw carefully the region 0 ≤ x ≤ 1, 0 ≤ y ≤ x4 , and then find its area.
8
Solution: Inside rectangle are is
N−1
X
4
1
i
N
N
i=0
1 X 4
= 5
i
N i=0
=
N (N − 1)(N − 2)
1
N (N − 1)
+ 50 ·
1 · N + 15 ·
5
N
2
3!
N (N − 1)(N − 2)(N − 3)
N (N − 1)(N − 2)(N − 3)(N − 4)
+ 60 ·
+ 24 ·
4!
5!
=
!
1
(1 − 1/N )
(1 − 1/N )(1 − 2/N )
+ 15 ·
+ 25 ·
4
3
N
2N
3N 2
(1 − 1/N )(1 − 2/N )(1 − 3/N ) (1 − 1/N )(1 − 2/N )(1 − 3/N )(1 − 4/N )
+5·
+
2N
5
!
and as N → ∞ this goes to 1/5.
(20) A right circular cone is the shape of an upside-down ice cream cone, circular at the
bottom, say of radius r, and with the bottom perpendicular to the axis along the centre
of the cone, say of height h.
epsfxsize=3.0in
9
10
In this exercise you will find a formula for the volume of such a cone. (1) Make N uniform
slices of the cone, each one parallel to the bottom and of height (or thickness) h/N . Inside
each slice put a cylinder of the same height, as wide as possible. The radii of the circular
slices varies from 0 at the top to a bit less than r at the bottom. At the left is a sideways
view, with N = 10. Use similar triangles to answer these questions: (a) What is the radius
of the smallest cylinder other than the one of radius 0? (b) Of the bottom cylinder? (2)
Express the total volume of the N cylinders as a sum, and then find a simple formula for
this sum. (3) As N gets larger and larger, the volume of the cylinders has as limit the
volume of the cone. Explain what happens to the formula in (2) as N gets large, and tell
what the volume of the cone is.
Solution: The radius of the i-th inside disk from the top is ir/N for i = 0, . . . N − 1. The
volume od the inside disks is therefore
N
X
i=1
N−1
πr 2 h X 2
πr 2 h 2(N − 1)3 + 3(N − 1)2 + 2(N − 1)
π(ir/N ) (h/N ) =
i
=
N3
N3
6
2
i=0
As N → ∞ this goes to πr 2 h/3.
(21) (a) Use an integral to estimate the sums
PN
k=1
√
k. (b) For N = 4 draw a sketch in
which the value computed by summing the four terms is compared to the value found by
the integration. (Your sketch should show the graph of the appropriate function and a set
of steps that represent the above sum.)
√
Solution: We will take f (x) = x. Let’s look at a couple of pictures, with N = 4 and 16.
11
Let S be the sum
S=
We can see that the integral of
√
1+
√
2+···+
√
N.
√
x is sandwiched in between the areas in the lower and
upper sets of rectangles, so
Z
√
√
√
√
0+ 1+···+ N − 1 = S − N <
N
√
x dx = 2N 3/2 /3 <
√
√
√
1+ 2+···+ N = S .
0
The integral is a good approximation to either sum, but it is best estimated by the average
of the two rectangular areas, which amounts to using the trapezoid rule. This gives us
2N
3/2
/3 ∼
S−
√
N +S
N
=S−
,
2
2
√
12
√
2N 3/2
N
S∼
−
.
3
2