Jsou dány funkce f(x) a g(x). a. Pomocı vzorce pro soucet

Jsou dány funkce f (x) a g(x).
a. Pomocı́ vzorce pro součet geometrické řady určete prvnı́ 3 nenulové členy rozvoje funkce
f (x) do Taylorovy řady se středem v bodě x0 = 0. Určete interval konvergence této řady.
b. Užitı́m operace násobenı́ mocninných řad určete prvnı́ 3 nenulové členy Taylorovy řady se
středem v bodě x0 = 0 funkce f (x) · g(x) .
1.
f (x) =
1
1
1/2
=
=
2+x
2(1 + x2 )
1 + x2
x2
+ ...
2
g(x) = cos x = 1 −
x ∈ (−∞, ∞)
k
1 −x
f (x) = ∞
= 12 1 − x2 +
k=0 2
2
interval konvergence:
x ∈ (−2, 2)
f (x) · g(x) = 12 1 − 12 x − 14 x2 . . .
P
2.
f (x) =
x3
+ ...
6
k
x2
4
x2
+ ...
2
+ ...
P
k
f (x) =
x2
9
k
x2
+...
2
2
P
k
x ∈ (−∞, ∞)
f (x) =
f (x) =
1
1
1/2
=
=
3
2 − 3x
2(1 − 2 x)
1 − 32 x
g(x) = arctgx = x −
7.
k
x3
+ ...
3
x ∈< −1, 1 >
1
1/3
1
=
=
2
3 + 2x
3(1 + 3 x)
1 + 23 x
g(x) = sin x = x −
x3
+ ...
6
k
x ∈ (−∞, ∞)
1
2
f (x) = ∞
= 13 1 − 23 x + 49 x2 + . . .
k=0 3 − 3 x
interval konvergence:
x ∈ (−3/2, 3/2)
1
2 2
5 3
f (x) · g(x) = 3 x − 3 x + 18
x ...
P
8.
x ∈ (−∞, ∞)
1 x
= 31 1 + x3 + x9 + . . .
f (x) = ∞
k=0 3 3
interval konvergence:
x ∈ (−3, 3) 1
4
2
f (x) · g(x) = 3 1 + 3 x + 17
18 x . . .
x2
+ ...
2
1 −3x
f (x) = ∞
= 12 1 − 32 x + 49 x2 + . . .
k=0 2
2
interval konvergence:
x ∈ (−2/3,2/3)
1
1
f (x) · g(x) = 2 1 − 2 x − 54 x2 . . .
P
+ ...
1
1
1/2
=
=
3
2 + 3x
2(1 + 2 x)
1 + 32 x
1 3
f (x) = ∞
= 12 1 + 32 x + 94 x2 + . . .
k=0 2 2 x
interval konvergence:
x ∈ (−2/3, 2/3)
1
3 2
31 3
f (x) · g(x) = 2 x − 2 x − 12
x ...
1
1
1/3
=
x =
3−x
3(1 − 3 )
1 − x3
g(x) = ex = 1 + x +
f (x) =
g(x) = ex = 1 + x +
6.
x ∈ (−∞, ∞)
1 −x
f (x) = ∞
= 13 1 − x3 +
k=0 3
3
interval konvergence:
x ∈ (−3, 3)
2
1
x
.
.
.
f (x) · g(x) = 3 1 − 3 − 7x
18
P
1/3
1
1
=
x =
3+x
3(1 + 3 )
1 + x3
g(x) = cos x = 1 −
4.
+ ...
x ∈ (−∞, ∞)
1 x
= 21 1 + x2 +
f (x) = ∞
k=0 2 2
interval konvergence:
x ∈ (−2,2)
3
1
x2
f (x) · g(x) = 2 x + 2 + x12 . . .
3.f (x) =
x2
4
1
1
1/2
=
x =
2−x
2(1 − 2 )
1 − x2
g(x) = sin x = x −
P
5.
f (x) =
1
1
1/3
=
=
2
3 − 2x
3(1 − 3 x)
1 − 23 x
g(x) = cos x = 1 −
k
x2
+ ...
2
x ∈ (−∞, ∞)
1 2
f (x) = ∞
= 13 1 + 23 x + 49 x2 + . . .
k=0 3 3 x
interval konvergence:
x ∈ (−3/2, 3/2)
1 3
x ...
f (x) · g(x) = 13 1 + 23 x − 18
P
9.
f (x) =
1
1/4
1
=
x =
4−x
4(1 − 4 )
1 − x4
g(x) = sin x = x −
x3
+ ...
6
x ∈ (−∞, ∞)
k
2
1 x
= 41 1 + x4 + x16 + . . .
f (x) = ∞
k=0 4 4
interval konvergence:
x ∈ (−4, 4) 1 2
5 3
1
x ...
f (x) · g(x) = 4 x + 4 x − 48
P
10.
f (x) =
13.
x3
+...
3
P
11.
f (x) =
2
x2
+...
2
k
f (x) =
x2
25
k
x3
+...
3
P
2
9 2
16 x
+ ...
f (x) =
k
x3
+ ...
3
x ∈< −1, 1 >
f (x) =
x3
+ ...
6
k
x ∈ (−∞, ∞)
16 2
25 x
+ ...
1
1
1/5
=
=
4
5 − 4x
5(1 − 5 x)
1 − 54 x
g(x) = cos x = 1 −
k
x2
+ ...
2
x ∈ (−∞, ∞)
1 4
2
f (x) = ∞
= 15 1 + 45 x + 16
k=0 5 5 x
25 x + . . .
interval konvergence:
x ∈ (−5/4, 5/4)
1
4
7 2
f (x) · g(x) = 5 1 + 5 x + 50
x ...
P
1
1
1/5
=
=
4
5 + 4x
5(1 + 5 x)
1 + 54 x
1 −4
= 15 1 − 45 x +
f (x) = ∞
k=0 5
5 x
interval konvergence:
x ∈ (−5/4, 5/4)
1
4 2
71 2
f (x) · g(x) = 5 x − 5 x + 150
x ...
P
1/4
1
1
=
=
3
4 − 3x
4(1 − 4 x)
1 − 43 x
g(x) = sin x = x −
16.
x ∈< −1, 1 >
1 x
f (x) = ∞
= 51 1 + x5 + x25 + . . .
k=0 5 5
interval konvergence:
x ∈ (−5, 5) 2
f (x) · g(x) = 15 x + 15 x2 − 22
75 x . . .
x ∈ (−∞, ∞)
1 3
9 2
f (x) = ∞
= 14 1 + 34 x + 16
x + ...
k=0 4 4 x
interval konvergence:
x ∈ (−4/3, 4/3)
1
11 2
3 2
x ...
f (x) · g(x) = 4 x + 4 x + 48
1
1
1/5
=
=
5−x
5(1 − x5 )
1 − x5
g(x) = arctgx = x −
k
g(x) = arctgx = x −
15.
+ ...
f (x) =
P
x ∈ (−∞, ∞)
1 −x
f (x) = ∞
= 15 1 − x5 +
k=0 5
5
interval konvergence:
x ∈ (−5, 5) 1
1
2
f (x) · g(x) = 5 1 − 5 x + 17
50 x . . .
12.
1/5
1
1
=
x =
5+x
5(1 + 5 )
1 + x5
g(x) = ex = 1 + x +
P
P
x ∈< −1, 1 >
k
x2
+ ...
2
g(x) = ex = 1 + x +
14.
1 −x
= 14 1 − x4 + x16 + . . .
f (x) = ∞
k=0 4
4
interval konvergence:
x ∈ (−4, 4) 1
1 2
3
f (x) · g(x) = 4 x − 4 x − 13
48 x . . .
1
1
1/4
=
=
3
4 + 3x
4(1 + 4 x)
1 + 43 x
3
1
= 14 1 − 34 x +
f (x) = ∞
k=0 4 − 4 x
interval konvergence:
x ∈ (−4/3, 4/3)
1
1
5 2
f (x) · g(x) = 4 1 + 4 x + 16
x ...
1
1
1/4
=
x =
4+x
4(1 + 4 )
1 + x4
g(x) = arctgx = x −
f (x) =