MathII

MathII, Matematyka, Egzamin

[ Pobierz całość w formacie PDF ]
Models in Biology
Modelling Biology
Basic Applications of Mathematics and
Statistics in the Biological Sciences
Part I: Mathematics
Script B
Introductory Course for Students of
Biology, Biotechnology and Environmental Protection
Werner Ulrich
UMK Toruń
2007
 2
Models in Biology
Contents
Introduction.................................................................................................................................................. 3
1. Combinations, variations and probabilities .............................................................................................. 4
2. Periodic processes ................................................................................................................................ 14
3. Vectors ................................................................................................................................................... 23
4. Matrix algebra ........................................................................................................................................ 29
5. Panta rhei ............................................................................................................................................... 42
6.1. Partial derivatives and stationary points.............................................................................................. 52
6.2. Lagrange multipliers .......................................................................................................................... 57
Online archives and textbooks ................................................................................................................... 59
Literature.................................................................................................................................................... 60
Mathematical software............................................................................................................................... 61
Latest update: 29.09.2007
 Models in Biology
3
Introduction
The following text is the second part of a lecture in basic mathematics and statistics for biologists. It
contains what might be considered an international standard of basic knowledge although many readers will
surely miss important branches. It deals with basic probability theory, trigonometry, matrix algebra, and differ-
ential equations. However, a one year course that has to deal with mathematics modelling must be to a certain
extent eclectic. Emphasis was especially paid to basic mathematical techniques and principles of biological
modelling. Again, many examples are included that show how to program simple tasks with a spreadsheet pro-
gram and how to use advanced mathematics software. The text does not repeat school mathematics.
The following text in not a textbook. It is intended as a script to present the contents of the lecture in a
condensed form. There is no need to write a textbook again. Today, the internet took over many former tasks
textbooks had. The end of this text contains therefore a small overview over important internet pages where
students can find mathematics glossaries, textbooks, and program collections.
4
Models in Biology
1. Combinations, variations and probability
Combinatory
With this lecture we start to deal with probability theory. Both fields are of major importance for biolo-
gists. At the beginning we will learn about the mathematics of counting, about variations, combinations and
permutations. The computation of possible states of a biological systems and the comparison of this number
with theoretical expectations becomes more and more a basis of many methods of hypothesis testing and mod-
elling.
Let’s start with a simple example. How many numbers with exactly 4 digits exist? These are of course
01
9
01
9
01
9
01
9
6
5
6
5
6
5
6
5
2
3
2
3
2
3
2
3
7
7
7
7
8
4
8
4
8
4
8
4
the numbers between 1000 and 9999. There are exactly 9000 (9999 – 999). We have 10 digits (0 to 9)
to combine. There are 10 one digit numbers, 10*10 = 10
2
two digit numbers and of course 10*10*10*10 = 10
4
four digit numbers. But we have to exclude all numbers that start with a zero. There are 10*10*10 = 10
3
such
numbers. Therefore, the total number of variations of four digit numbers is 10
4
– 10
3
= 9000.
A convenient way to visualize the principle of counting is a so-called
polling box model
(Fig. 1.1). In
our case we have 4 such boxes, each containing the digits 0 to 9. To make four digit numbers we take one digit
from each box. In the first box we have 9 possibilities (1 to 9 because a zero as a starting point is not allowed).
In the next box we have 10 possibilities. The total number of variations is therefore 9*10. The total number of
variations from all four boxes is then 9*10*10*10 = 9000 variations. We obtain the same result if we consider
only 1 box and take 4 times a digit while all digits remain in the box.
In general we can write that the number of
variations
of n elements into sets of k elements is
V
=
n
n
k
k
(1.1)
if repetition is allowed. In our case, it is allowed to form numbers like 1111, 2021, 3443 and so on.
What is if repetition is not allowed? In how many ways can we take numbers from the first box? There
are 10 possibilities for the first digit. If we take one, 9 numbers remain and for the second digit these 9 possi-
bilities remain. In total, we have 10*9 so-called permutations for the first two digits. For the third digit 8 num-
bers remain and we have of course 10*9*8 permutations for the first three digits and so on. We see that the
total number of permutations of 10 digits is
P
=
10
*
9
*
8
*
7
*
6
*
5
*
4
*
3
*
2
*
1
=
10
(1.2)
In general, the number of
permutations
of n elements is
P = n!
(1.3)
Fig. 1.1
!
Models in Biology
5
What about our four digit numbers? How many permutations of four digit numbers without repetition
are possible? If we take a number out of the first box (10 possibilities) in the second box 9 possibilities remain.
Because repeating is not allowed we indeed took the first number out of all four boxes. Now we take a number
from the second box. Again we have to take this number also out of the remaining two boxes. Therefore, the
third box contains only 8 numbers. After taking one of them, the fourth box contains only seven numbers. In
total, the number of four digit numbers without repetition is therefore P = 10*9*8*7 (if a zero as first number is
allowed). We can immediately generalize this result and notify that the number of
permutations
of n elements
into sets of r elements is exactly
∏
+
n
P
n
=
n
*
(
n
−
1
*
(
n
−
2
)
*
...(
n
−
r
+
1
=
i
r
We can write this result in a slightly different but more convenient form
=
n
−
r
1
Pnn n nr
n
=−− −+
( 1)( 2)...(
1)
(
(
nrnr
=
−−− −
)(
)...1 !
n
r
nrnr
)(
)...1(
nr
)!
(1.4)
This is the form permutations are most often given. Note that the last equation is a generalization of the
above equation for simple permutations. If we consider only a set of n elements we have n! / (n-n)! = n!. By
definition 0! = 1.
A next example. How many four letter words can be formed from the word permutation? If these words
can contain the same letter two or several times we can con-
sider a letter from the box but leave it. Then, we have of
course 10
4
possibilities (permutation contains 10 letters, the t
occurs twice). If all the letters of the four letter words have to
be different there are 10*9*8*7 permutations or
Permutation
Fig. 1.2
P
=
10!
=
5040
(1 0 4 ) !
−
In our above example the ordering of elements was decisive. The four letter word
perm
is of course dif-
ferent from the (non-)word
prem.
What is if the ordering of elements is not decisive? In Duży Lotek every week 6 numbers are taken from
a total of 49 numbers. How many six number combinations exist?
Our permutation equation predicts 49! / (49-6)! However, in this case the ordering of numbers is irrele-
vant. It does not matter at which step a number falls. Our permutation equation gives therefore a prediction that
is too high. To obtain the correct result we have to consider the number of ways 6 numbers can be varied. This
number is of course 6!. So, the correct answer of our Duży Lotek example is
49
!
C
=
(
49
−
6
)!
=
49
!
=
13983816
6
6
(
49
−
6
)!
Again, we can generalize this result. The number of
combinations
of k objects out of a total of n ele-
ments, when the ordering of objects is irrelevant, is
i
− −−
11
4
[ Pobierz całość w formacie PDF ]

zanotowane.pl

doc.pisz.pl

pdf.pisz.pl

mement.xlx.pl