modele dyskretne, studia, 4 semestr, modele dyskretne + projekt
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k
R
k
.
x ∈
t = 0
R
k
→
R
k
,
: [0,∞) ×
(t,x)
t,
x
0.
(0,x) = x
x
R
k
(t,(s,x)) = (t + s,x)
t,s ≥ 0
x ∈
0
.
,
.
t
[0,∞)
N
0).
f(x) := (1,x),
(t,x) = f
t
(x)
t = 1,2,...
R
k
f : R
k
R
k
.
R
k
→
f
X ⊂
R
k
f
f
−1
0
f
n
: X → X
n ∈
N,
n ∈
Z.
f : X → X
x
0
∈ X.
R
k
.
X
x
(x) = {x
m
: m ∈
Z
}
x
0
= x
f
n
(x
m
) = x
n+m
,
n ∈
N, m ∈
Z.
x
+
(x) = {x
m
: m ≥ 0}
x
m
= f
m
(x),
{x
m
: m ≤
0}
x
(x)
m < 0
x
m
= (f
−1
)
−m
(x).
f : X → X
x
0
∈ f(X)
(x) = {x}.
f(x) = x.
(x) = {x
m
: m ∈
Z
},
x
m
= z
m ≥ m
0
> 0.
f(z) = z,
z
R, f(x) = x(x
2
− 1),
x
0
= −1 : x
−1
=
f
−1
(−1), x
−2
= f
−1
(x
−1
), ..., x
−m−1
= f
−1
(x
−m
),
f(x) = y
f : R
→
√
y < −2
3/9
x
1
= 0 = x
m
m > 1
{x
m
: m ∈
Z
},
n
0
> 1
x
m+n
0
= x
m
m.
m
0
x
m+n
0
= x
m
m ≥ m
0
> 0.
x
m
0
f : R
→
R,
8
<
:
x+3
x < −1,
2
−x
|x| ≤ 1,
f(x) =
x−3
2
x > +1.
x
0
= 5, x
m
= (−1)
m+1
m ≥ 1
x = 5
x
−m−1
= 2x
−m
+ 3
m ≥ 0
±1
2.
x
m
m
f : R
→
R, f(x) = x/2.
1
x
m
= 2
−m
m ∈
Z
!−
x ∈ X
f : X →
f
n
(x), n ∈
X
N
!(x).
!(x)
X
x.
−
(x) :
x
−m
k
x.
(x)
f
−n
(x), n ∈
N.
x
!(x) = {x}.
(x) =
{x},
!
(5) = ∅.
5
f : R
→
R
f
y = x.
x
0
∈
R.
(x
0
,0)
x = x
0
(x
0
,f(x
0
)).
y = f(x
0
)
y = x.
x
1
= f(x
0
).
y = x.
x
2
= f(x
1
)
f
y = x).
y = x,
2.
f
[a,b]
x0
x[N].
f : [0,1] →
[0,1], f(x) = µx(1 − x),
µ = 2,7
0,3.
1
0.8
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8
1
µ = 3,4
1
0.8
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8
1
µ = 4
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