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171
y
434
1
ppp
-
3
1
(
)
1
p
1
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2312312
ì
ï
x
ï
y
22
yxzr +==
and
arg
ï
==
z
í
ï
ï
ï
î
pj
x
y
p
x
>
,0 w,arctg
xhen
><+
yxwhen
<<-
yxhen
We get:
( )
11
22
22
211
=+-== zr
2
2
231
=+== zr
,
arg
== arctgz
11
,
arg
j
22
1
pj
3
p
=== arctgz
.
To go from the algebraic form of writing complex numbers to trigonometric and exponential use this formula:
j
i
jj
sincos irz +=
и
rez =
.
Using the results we got earlier, we get:
+=
sin
3
pp
ö
sin
iz
÷
,
4
ø
pp
ö
÷
,
3
ø
3
æ
cos2
1
ç
4
è
3
i
,
4
2
ez =
æ
ç
2
è
2
ez =
+=
cos2
iz
3
p
i
3
.
2)
а)
2
( )
1
2
2
( )
2
æ
ç
è
2
2
ö
÷
ø
( )
2322322322322321
iiiiiii
=+++-=+-+-=-×--=
++×--=+×--=× 3321133211311
,0,0 ,arctg
.0,0 w,arctg
=+-=+
,
=-+×--=
iiiiiiizz
( ) ( )
z
2
=
b)
z
1
=
-++=
+
31
i
=
+-
1
i
;322322
i
131
( )( )
3131
i +
=
11
+-
2
--+
ii
=
--+-
ii
-
i
( )
;
----
331
iii
2
--
1
=
2
i
+---
331
ii
=
+
11
172
c) Let’s use the formula
=
=
=
-
=
()()(
)
=
2
+
+ iii
+
k
+
n
æ
ç
3
=+-=
iz
sin
2
:
2
4
p
ö
÷
ø
ç
ç
è
3
pp
4
3
p
+
sin
+
sin
æ
sin
i
ç
è
æ
nn
cos -=
= nk
rz
ç
è
1
æ
ç
cos2
ç
ç
è
æ
ç
cos2
ç
ç
è
when
æ
ç
cos2
ç
ç
è
æ
6
ç
ç
è
3
:
0
k
3
4
3
:
1
k
3
p
4
3
k
3
p
+
4
3
5
æ
cos2
-=
ç
12
è
3
when
when
+
sin
i
3
p
cos21
ö
æ
÷
ç
÷
÷
è
ø
3
p
+
4
3
3
p
+
4
4
3
5
ö
pp
ö
÷
-+
÷
÷
12
ø
ø
Exercise 2. Find out the curve type
Solution.
ì
So
í
î
Show t in each equation:
,tg5tytx
-=
.sec3
ì
t
ï
ï
í
ï
t
ï
î
Exclude t from the equations:
2
n
4
3
cos2
2
p
p
.
=
k
pjpj
ö
÷
ø
+
2
p
4
ö
÷
÷
÷
ø
ö
÷
÷
÷
ø
titgttiytxtz sec35 -=+=
.
arctg
arccos
arccos
( )
1...,,1,0,
.
3
i
sin
cos2
æ
ç
è
sin
cos2
pp
4
19
12
ö
÷
ø
p
4
666
11
12
+=
k
+
+=
66
æ
66
ç
è
ö
+
2
k
p
÷
÷
3
÷
ø
+=
12
ö
2
÷
;
÷
2
ø
pp
ö
÷
;
ø
ö
=
÷
ø
æ
2
ç
2
ç
2
è
11
+=
sin
ii
19
pp
sin
ii
12
titz sec3tg5
.
x
,
5
ö
æ
3
÷
ç
-=
.
÷
ç
y
ø
è
ö
æ
3
÷
ç
=
-
÷
ç
y
ø
è
arctg
x
.
5
173
ö
9
25
9
3
6
p
p
arg;
62
pp
ö
æ
3
÷
÷
ç
ç
è
+= x
1
=
-
÷
÷
y
ø
ø
ö
÷
=
÷
y
ø
,
25
– the equation of a hyperbole.
x
æ
arctgcos
ç
è
x
ö
÷
,
5
ø
53
22
5
+
=-
,
y
53
,
22
5
+
x
25
æ
ç
arccoscos
ç
è
æ
ç
arccoscos
-
p
ç
è
2
y
2
22
xy
=-
Exercise 3. Draw the plane domain of z, defined by these inequali-
ties:
ì
ï
í
ï
6
î
The wanted set is an intersection of rings
part of the angle
arg
£-<
iz
( )
,31
2
pp
<-£
3
arg
;
( )
2
<-£ iz :
31 £-< iz
and the inner
iz
Fig.19 – graphic representation of a conditions of an exercise 3
Try to do these yourself:
Draw the plane domain of z, defined by these inequalities:
14,
zi
ì<-£
ï
í
zi
£-<
ï
î
( )
Fig.19
174
Section 10. Basics of probability analysis
=-=
=
=
(
)
(
)
5,0,5
==
2
£
(2)(012)(0)(1)(2)
£=====++=
0,50,50,50,50,50,50,5
323232
Topic 25. Probability theory
Introduction. We will review the main aspects of the theory of probability. However, a quantity of theoretical positions from this section of
mathematics will not be discussed because not enough time. In this
workshop we will learn how to calculate probability using the classic
definition. We learn how to calculate the probability of an event a certain
number of times.
Exercise 1. Four coins are being flipped. What are the odds you will
get three “heads”?
Solution. We will consider flipping a coin as an experiment. According to the terms of the exercise, there are 4 identical experiments. The
probability of success (flipping a «heads») at each experiment is
5.01;5.0
pqp
ments
will be successful. To solve this problem let’s use the for-
3
k
mula of binomial distribution of a discrete random variable.
knkk
-
=
qpCkP
nn
Answer: 0.25.
Exercise 2. A coin is flipped 5 times. What is the probability of get-
ting no more than 2 “heads”?
Solution: In this exercise
la let’s find the probability of event
PmPm или m или mPPP
051423
012
CCC=++=++=
( )( ) ( )( ) ( )( )
555
Exercise 3. The odds of hitting the target every time you shoot is
0,75. Find the probability of hitting the target 8 times out of 10 shots.
Does the probability of success chance if you shoot 10 times as much?
. We have to find, that out of all the experi-
. Under terms of our problem
CP
44
np
-
m
43433
25.05.0*45.05.03
===
.
. Using the Bernoulli formu-
.
555
1510
.
175
Solution. The probability of hitting the target on the first shot is
0.7510.750.25;10;8
=Þ=-===
(8)0.750.2545*0.1001*0.06250.281
8!2!
Y 5 6 7 10 p 0,1 0,1 x 0,3
=
=
()()()(
)
pqnk
. Using the Bernoulli for-
mula let’s find:
P ===
10
10!
82
;
Try to do these yourself:
The odds of hitting the target every time you shoot is 0,65. Find the
probability of hitting the target 7 times out of 10 shots. Does the probability of success chance if you shoot 10 times as much?
Topic 26. Elements of mathematical statistics
Introduction. In this topic we will discuss the most important concepts of mathematical statistics and learn how to calculate the variance
of two ways.
Exercise 1. We have a number distribution of a discrete random variable Y. Find the value of x and calculate the mathematical expectation of
a discrete random variable Y.
Solution. Let’s find x from of the given terms
n
å
=
i
i
1
xxp
Knowing x, we are able to calculate the mathematical expectation.
pYYM
å
ii
1
=ni
Answer:
YMx
.6,7)(;5,0
Exercise 2. DX = 3. Using the properties of dispersion, find D(4X-2).
Solution.
Answer: 48.
5,013,01,01,0,1
=Þ=+++=
.
6,73,0*105,0*71,0*61,0*5)(
=+++==
22
.
48316424
=×==-®=+ XDXDXDABAXD
176
The mode is a value of a property, which occurs most commonly in
85.9;89.1;72.3;82.5;70.6
5;
=
85.9;89.1;
72.3;82.5;70.6
(85.989.172.382.570.6)80.8,
()271.4954.29,
54.297.37.
X 2 3 4 5 Y 3 4 6 8
the distribution series. The mode is determined by various means depending on the type of the variational series. In the discrete variational
series the mode is the variant with the highest frequency in the target
area.
Example 3. According to the statistical observation, we have obtained values X = {5, 3, 1, 2, 1, 4, 1, 5, 2, 1, 4, 2, 1, 1, 6}. Find the mode.
Let’s create the variational series
X 1 1 1 1 1 1 2 2
2 3 4 4 5 5 6
The corresponding grouped variational series looks like this:
X 1 2 3 4 5 6
F 6 3 1 2 2 1
Value of the Х parameter, which has the highest frequency (6) is
equal to 1. Therefore, the mode for this variational series = 1.
Example 4. We have 5 experimental observations of a random car
velocity on one part of the highway
(km/h):
XXXXX=====
12345
. Calcu-
late the dispersion and mean-square distance.
Solution. Under the terms we know, that
XX==
n
12
XXX===
345
. Using
this data, let’s calculate:
n
11
XX
==++++=
å
i
n
i
=
11
22
SXX
=-=×=
å
Xi
n
2
S ==
X
5
1
n
1
i
=
5
Exercise 5. Random value X is the number of years employees have
worked in a trading company; Y – the number of vacations they took
while working at this company. The results of observations of random
variables X и Y: are given in this table:
177
Create the equation of linear regression of Y by X and X by Y. Find
111
4;(2345)3.5;(3468)5.25;(23344658)20.5;
444
nXYXY
==+++==+++==×+×+×+×=
(23.5)(33.5)(43.5)(53.5)1.25
4
(35.25)(45.25)(65.25)(85.25)3.69
4
1.7;0.58.
1.253.69
-×-×
5.251.7(3.5)1.70.7
3.50.58(5.25)0.580.45.
1.70.580.99
Cow
Number of
the sample linear correlation coefficient r*XY.
Solution. Under terms, let’s find:
1
22222
S
X
22222
S
Y
efficients of Y by X and X by Y.
and the sample linear correlation coefficient.
the table.
éù
=-+-+-+-=
ëû
1
éù
=-+-+-+-=
ëû
Using the suggested formulas, let’s calculate linear regression co-
r==r==
YXXY
And using the formulas, let’s create equations of linear regressions
YXYX-=-Þ=-
XX
XYXY-=-Þ=+
YY
*
r =×=
XY
Example 6.
A farmer has 9 cows, the number of calvings of which are given in
Find the median.
Calculating the median is executed by this algorithm:
20.53.55.2520.53.55.25
//
;
.
;
;
№ п/п
calvings
1 2 3 4 5 6 7 8 9
4 2 1 5 2 3 4 3 1
178
To begin solving the problem, first establish which series type it is.
85.5;89.4;73.3;82.5;70.6
Because values take form of numbers, the series is discrete.
Notice! The most common mistake is that the median is being sought
for the unranked series, i.e. the median is equal to two calvings (cow
№5), will not be the correct solution.
We have to build a ranked series for a discrete series. It will look like
this: 1, 1, 2, 2, 3, 3, 4, 4, 5
If the number of members is odd, the ranked series will have a middle, so that the number of variants would be equal at both sides. In our
case it will be variant № 5 (3 calvings), there are 4 variants both on the
right and on the left of this one.
Try to do these yourself:
We have 5 experimental observations of a random car velocity on one
part of the highway
(km/h):
XXXXX=====
12345
. Calcu-
late the dispersion and mean-square distance.
List of literature.
12. Calculus: Early Transcendentals, James Stewart Cengage Learn-
ing, 2010.
13. Introduction to Analysis Edward D. Gaughan, American Mathe-
matical Society, 2009.
14. Calculus. Gilbert Strang by Wellesley-Cambridge Press, 1991
15. A First Course in Linear Algebra, Robert A. Beezer, Tacoma,
Washington, http://buzzard.ups.edu/ , Version 3.20 (Created: 2014-0224T20:51:53-08:00) 2013
16. Principles of Mathematical Modeling By Clive Dym, Harvey
Mudd College, Claremont, California, U.S.A. 2004. (ISBN: 978-0-12226551-8)
17. Mathematics, Statistics & Mathematical Education Abstract Book
18. From the 5th Annual International Conference on Mathematics,
Statistics & Mathematical Education, 13-16 June 2011, Athens, Greece.
Edited by Gregory T. Papanikos
19. Mathematical modeling . Ckassroom Notes in Applied mathemat-
ics. Edited Murray S, Klamkin. Philadelphia 1987.
20. http://www.collegeopentextbooks.org/
21. http://ocw.mit.edu/
22. http://en.wikibooks.org/
179
CHAPTER 3
Section 1. Linear algebra
and analytical geometry
Topic 1. Matrix. Matrix Operations
The exercises: Add matrices
121
ö
÷
and
÷
013
ø
121
ö
÷
and
÷
0113
ø
1221
ö
÷
and
÷
013
ø
121
ö
÷
and
÷
013
ø
121
ö
÷
and
÷
0313
ø
121
ö
÷
and
÷
013
ø
1321
ö
÷
and
÷
1013
ø
121
ö
÷
and
÷
013
ø
1241
ö
÷
and
÷
013
ø
121
0133
ö
÷
÷
ø
B
B
B
and
æ
ç
=
ç
è
=
B
=
B
æ
ç
=
ç
è
=
B
=
B
=
B
æ
ç
=
ç
è
=
B
B
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
æ
ç
=
A
ç
è
æ
ç
=
A
ç
è
æ
ç
=
A
ç
è
æ
ç
=
A
ç
è
æ
ç
=
A
ç
è
æ
ç
=
A
ç
è
æ
ç
=
A
ç
è
æ
ç
=
A
ç
è
æ
ç
=
A
ç
è
æ
ç
=
A
ç
è
31
ö
÷
÷
21
ø
31
æ
ç
ç
è
æ
ç
ç
è
æ
ç
ç
è
æ
ç
ç
è
æ
ç
ç
è
ö
÷
÷
-
21
ø
31
ö
÷
÷
21
ø
331
ö
÷
÷
21
ø
31
ö
÷
÷
21
ø
331
ö
÷
÷
21
ø
31
ö
æ
÷
ç
÷
ç
21
ø
è
31
ö
÷
÷
211
ø
31
ö
÷
÷
214
ø
31
ö
æ
÷
ç
=
÷
ç
21
ø
è
180
Topic 2. Matrix Multiplication
Example. Solve step by step:
æ
ç
ç
ç
è
Run the solution in action:
321
æ
ç
311
ç
ç
542
è
æ
ç
ç
=
ç
è
æ
ç
×
3
ç
ç
è
æ
ç
ç
ç
è
321
311
542
Answer:
The exercises: Multiply matrices
æ
ç
ç
ç
è
1)
æ
ç
ç
ç
è
2)
2
321
311
542
æ
ö
ç
÷
*
ç
÷
ç
÷
è
ø
24169
ö
÷
21158
÷
÷
432816
ø
123
ö
÷
=
311
÷
÷
522
ø
2
ö
÷
-
3
÷
÷
ø
æ
ö
ç
÷
-
3
ç
÷
ç
÷
è
ø
321
ö
÷
=
311
÷
÷
542
ø
.
æ
ç
ç
ç
è
æ
ç
ç
ç
è
æ
ç
ç
ç
è
369
933
1556
123
311
522
21100
12125
21100
ö
÷
12125
÷
÷
282310
ø
and
21110
ö
÷
121215
÷
÷
282310
ø
and
123
ö
÷
311
÷
÷
522
ø
æ
ç
ç
ç
è
ö
÷
÷
.
÷
ø
æ
ö
ç
÷
=
ç
÷
ç
÷
è
ø
ö
÷
÷
.
÷
282310
ø
æ
ç
ç
ç
è
æ
ç
ç
ç
è
++++++
5*33*23*14*31*22*12*31*21*1
ö
÷
++++++
5*33*13*14*31*12*12*31*11*1
÷
=
÷
++++++
5*53*43*24*51*42*22*51*41*2
ø
24169
ö
÷
-
21158
÷
÷
432816
ø
211076
ö
÷
121275
÷
÷
282310
ø
;
21100
ö
÷
12125
÷
÷
282310
ø
;
369
æ
æ
ç
ç
ç
è
ö
ç
÷
=
933
ç
÷
ç
÷
1556
è
ø
21100
ö
÷
12125
÷
.
÷
282310
ø
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