The exponential function is a general description of a random process, where the probability of a certain event is independent of time C 0 / 2 C 0 / 2 C 1 / 2 C 1 / 2 C 2 / 2 C 2 / 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0 5 10 15 x C x Lecture 2 Some basic functions and their application in biology 0 0 ln(2) 0.693 0 1 2 x x x x C C Ce Ce 0 kx x C Ce 1 0 (1 ) kx x C C e 2 / 1 2 / 1 ) 2 ln( ) 2 ln( x k k x
Lecture 2 Some basic functions and their application in biology. C 0 /2. C 0 /2. C 1 /2. C 1 /2. C 2 /2. C 2 /2. The exponential function is a general description of a random process , where the probability of a certain event is independent of time. - PowerPoint PPT Presentation
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0 0
ln(2) 0.6930
1
2
xx x
xC C C e C e
0kx
xC C e
1 0 (1 )kxxC C e
The exponential function is a general description of a random process, where the probability of a certain event is independent of time
C0/2 C0/2
C1/2 C1/2
C2/2 C2/2
0
0.1
0.2
0.3
0.4
0.5
0.6
0 5 10 15
x
Cx
Lecture 2Some basic functions and their application in biology
2/1
2/1
)2ln(
)2ln(
xk
kx
C14 has a half time of 5568 years.
How long would it take until of one mol C14 only one atom remains?
One mol contains 6.0210*1023 atoms
ktt eNN 0
kte 23100210.615568
)2ln()2ln(
2/1
t
k
5568/)2ln(23100210.61 te
439841)2ln(/)100210.6ln(5568
5568/)2ln()100210.6ln(23
23
t
t
The age of fossilized organic matter can be determined by the C14-method of radioactive decay. The half-live of C14 is 5568 years. The equilibrium content of C14
in living plants is about 10-6 ppm (parts per million). How old is a fossilized plant with a C14 content of 1.5*10-7 ppm?
1. Start with a triangle, 2. shrink to 1/2 size, 3. make three copies 4. arrange the three copies in quadrants 2,3, and
4 5. goto (2).
Self similar objects
"All the branches of a tree at every stage of its height when put together are equal in thickness to the trunk„ (Leonardo da Vinci, Notebooks,
1510)
zz
x
x
x
x
a
a
y
y
2
1
2
1
2
1zy ax
Self similarity
How life makes a complex pattern
1000 km500 km
y = 157000x0.30
1000
10000
100000
1000000
0.001 0.01 0.1 1
Scaling factor
Leng
th o
f Eur
ope
[km
]Scaling factor = 1 / unit of measurement
1000 km500 km
Scaling factor = 1 / unit of measurement
= magnification
dddd
assll
L
1
a is called the normalization constant
dddd
ssasll
A
111
1
)(1
Ruler length l
sbblLEuclid
1
What is the value of a?
If our object is classical Euclidean d = 0
11)( ddd bss
sbssaL
Both equations match if b = cs1)1(11 dd csscsLNow we consider an area. The area scales to the square of the ruler length
22 1
sbblAEuclid
112
1 1)( ddd bss
sbssaA Both equations match
if b = cs2
1)2(12 dd csscsA
1)3(13 dd csscsV
dddd
ssasll
L )(1
33 1
sbblVEuclid
Euclidean dimension E=1
Euclidean dimension E=2
Euclidean dimension E=3
. Euclidean dimension E=0
33 dscsV
22 dscsA
11 dscsL 1)( dEcsY
E+d takes always values between the actual Euclidean dimension and the next higher one. It is commonly termed the fractal dimension of an object.
0 < d < 1
An important class of fractal objects are self similar objects. We describe them by power functions.
rK2LK
rK
LKrK2LK
rK
LK
b
n n
K K
A r
A r
x
n n
K K
M V
M V
X = 0.75
a
n n
K K
V r
V r
What are the relation between radius, volumen and surface in such a branching pattern?
M A
B V0.75M B
A branching pattern
Calculate the total leaf area of this fern
We have to measure two leafs at different scale to get the scaling exponent of the area - length relationship
zlAA 0z
ll
AA
2
1
2
1
Let the average length of the smallest leaflets be 1 cm and its area 3 cm2. At the next higher scale leaflet length might be 10 cm and the respective area 35 cm2. The whole fern is 1 m long.
07.1)10ln()1ln(
)35ln()3ln(
10
1
35
3
z
cm
cm
cm
cmz
207.1
2
10
10010100 408
10100
35 cmcmcm
cmll
AAz
(2 ) 1( ) 2.28dA s bs D
1)()( dDbssL
How should population density scale to body weight?0 5 10 15 20 25
020406080
100120140160
f(x) = 3 x^1.28
Magnification
Da
rk a
rea
y = 4.36x-0.89
R2 = 0.81
0.001
0.1
10
1000
100000
10000000
1E-06 0.0001 0.01 1 100 10000
Mean weight [g]
Me
an
de
nsi
ty /
m2
How should population density scale to body weight?
sL /1 VN 2/3AV 28.1sA 4/3WM
64.092.13/192.12/328.1 )()( WWssV
33 slW MN /1
39.175.064.0 WWWN
Population density is proportional to available space and to available energy.
The mean number of bee species per km2 in Poland [312685 km2] is 110, the total number of Polish bees is 463. Estimate the number of bees in the district of Kujaw-Pommern [17970 km2].
The mean number of bird species in Poland is about 430, the total European [10500000 km2] species number is about 800. How many species do you expect for France [543965 km2]?
Would it make sense to estimate the species number of Luxembourg [2586 km2]? What about Kujaw-Pommern [17970 km2]?