Apostila Fundamentos Matemática
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Transcript of Apostila Fundamentos Matemática
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RN
RN
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p
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QR
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RN
N = {0, 1, 2, 3, ....}N∗ = {1, 2, 3, 4, ....}
Z = {0, 1, −1, 2, −2,...}Z∗ =
{1,
−1, 2,
−2,...
}Q = {mn
: m, n ∈ Z, n = 0}Q = { m
n
m ∈ Z e , n ∈ Z∗}
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N Z Q
2 N
a b a b c
bc = a
b
a
b
a
b
a b| a
a
2| a
2
a
2
a
2
1742
[4, 4 ∗ 1018]
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•
•
•
•
•
•
•
A
B
A
B
✞✝ ☎ ✆ x y x + y A x y B x+y
A ⇒ B
A B A ⇒ B
A
B
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A B
A
B
A
B
B
A
A
B
A
B
A B
A
B
A B
A
B
A
B
B
A
✞✝ ☎ ✆ x x + 1
x
x + 1
x + 1
x
A
x
B
x + 1
A ⇔ B
A B A ⇔ B
x
A
x
B
B
A
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A B
A
B
A B
B
∧
A B A e B
A ⇒ not(B)
não A
¬
x
A
x
B
B
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A B A ou B
∨
A ⇒ B)
x
y
x
2|
x
2|
y
a b x = 2a y = 2b x+y = 2a+2b = 2(a+b).
c
x + y = 2c
c = a + b
2|c.
a
b
c
a| b
b| c
a| c
a| b
x
b = ax
y
c = by.
z
c = az
c = (ax)y = axy = az.
A ⇒ BA ⇐ B
x
x
x + 1
⇒
x
2| x
a
x = 2a
1
x + 1 = 2a + 1,
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⇐ x + 1 b x + 1 =2b + 1.
1
x = 2b
x
a
b
a| b
b| a
a = b
a =
−6
b = 6
−6
|6
6
| −6
−6 = 6.
a| b
x
b = ax
y
a = by.
b = (by)x
b
b,
xy = 1
x = −1
y = −1
b = −1a
a = −1b
b = −a
a = −b.
∼
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x A
x ∈ A
A = {2, 4, 6, 8, 10}
A = {x ∈ N, x 1 e x ≤ 1000}
x ∈ N
A
B
A
B
A ⊆ B
A
B
A ⊆ B sse ∀x ∈ U x ∈ A ⇒ x ∈ B
A
B
x
x A x B
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⊆
∀
∀x ∈ A,
x
•
•
•
•
x
x
x
∃
∃x ∈ A, x
•
• N x
2
¬(∃x ∈ Z, x
∀x ∈ Z, ¬(x
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¬(∀x ∈ Z, x )
∃x
∈Z,
¬(x
∀x, ∃y, x + y = 0
∃y, ∀x, x + y = 0
∀x, ∃y, x + y = 0
x
y = −x.
x + y = x − x = 0
∃y, ∀x, x + y = 0
y = 2
x = 10
A
B
A
B
A
B
A ∪ B = {x : x ∈ A ∨ x ∈ B}
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A B A B
A
B
A ∩ B = {x : x ∈ A x∧ ∈ B}
A ∪ B = B ∪ A
A ∩ B = B ∩ A
A ∪ (B ∪ C ) = (A ∪ B) ∪ C
A ∩ (B ∩ C ) = (A ∩ B) ∩ C
A ∪ (B ∩ C ) = (A ∪ B) ∩ (A ∪ C )
A ∪ ∅ = A
A ∩ ∅ = ∅
A ∪ (B ∪ C ) = (A ∪ B) ∪ C
B ∪ C
B ∪ C = {x : x ∈ B ∨ x ∈ C }
A ∪ (B ∪ C ) = {x : (x ∈ A) ∨ (x ∈ B ∨ x ∈ C )}.
{x : (x ∈ A) ∨ (x ∈ B) ∨ (x ∈ C )}
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{x : (x ∈ A ∪ B) ∨ (x ∈ C )}
(A ∪ B) ∪ C.
A − B
A
B
A − B = {x : x ∈ A ∧ x /∈ B}.
A∆B
A B B A
A∆B = (A − B) ∪ (B − A).
A B A B A × B
A
B
A × B = {(a, b) : a ∈ A, b ∈ B}
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A = {1, 2} B = {3, 4}
A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}B
×A =
{(3, 1), (3, 2), (4, 1), (4, 2)
}
(A × B = B × A)
A = {1, 1, 2, 3, 4}
|A| = 4
A
B
|A × B| = |A| |B|
|A| = 2
|B| = 2,
|A × B| = 4
{(1, 3), (1, 4), (2, 3), (2, 4)}
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R
N ⊆ Z ⊆ Q ⊆ R...
R
• + : R × R → R (x, y) → x + y
• ∗ : R × R → R (x, y) → x ∗ y
•
(x + y) + z = x + y + z x, y, z ∈ R
•
x + y = y + x x, y ∈ R
•
x + 0 = x x, 0 ∈ R
•
x + (
−x) = 0 x, 0
∈R
•
(x ∗ y) ∗ z = x ∗ y ∗ z x, y, z ∈ R
•
x ∗ y = y ∗ x x, y ∈ R
•
x ∗ 1 = x 1, 0 ∈ R
•
x ∗ x−1 = 1 x, 1 ∈ R x = 0
x ∗ (y + z ) = x ∗ y + x ∗ z x, y, z ∈ R
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P ⊆ R R
•
•
x
∈ P
x = 0
x
∈ P
−(x)
∈ P
R = 0∪P ∪ (−P ) −P
• x > 0 se x ∈ P
• x y ⇒ x − y > 0
• x < y ⇒ x − y a ∧ x < b}
• [a, b] = {x ∈ R : x ≥ a ∧ x ≤ b}
• (a, b] = {x ∈ R : x > a ∧ x ≤ b}
• [a, b) = {x ∈ R : x ≥ a ∧ x < b}
• (a, +∞) = {x ∈ R : x > a}
• [a, +∞) = {x ∈ R : x ≥ a}
• (
−∞, a) =
{x
∈ R : x < a
}
• (−∞, a] = {x ∈ R : x ≤ a}
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S ∈ R. u ∈ R S ∀s ∈ S, s ≤ u
v ∈ R S ∀s ∈ S, v ≤ s
S
0
A = [0, 1]
1
0
R
S
S
sup(S )
S
inf (S )
A = {x ∈ R : 0 ≤ x ≤ 1}
0
1
A
[1, ∞) 1
inf (A) = 0
sup(A) = 1
B = {x ∈ R : 0 ≤ x
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a ∈ R a
|a| = +a se a > 0−a se a
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f (a, b) ∈ f (a, c) ∈ f b = c.
R = {(1, 2), (3, 4), (5, 6)}
(entrada, saida)
R = {(1, 2), (1, 4), (2, 7)}
1
2
4
f
f (a) = b
f (a) = c
b = c.
f = {(x, y) : x, y ∈ Z, y = x2}
f = {(1, 1), (2, 4), (3, 9), (4, 16),...}.
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✬
✫
✩
✪
f
f
f
domf
domf = {a : ∃b, (a, b) ∈ f }
domf = {a : f (a) definido}
f = {(x, y) : x, y ∈ Z, y = x2} Z
f
f f im f
im f = {b : ∃a, (a, b) ∈ f }
im f = {b : b = f (a) para algum a ∈ dom f },
f = {(x, y) : x, y ∈ Z, y = x2} Z
f : A → B
f
A
B
f
A
B
domf = A
im f ⊆ B
B
f
A B
f = {(x, y) : x, y ∈ R, y = sen(x)} domf = R im f = [−1, 1] ⊆ R
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a1
a2
f(a1)
f(a2)
f é um-a-um
a1
a2
f(a1)
f não é um-a-um
f : A → B = {(a1, b1), (a2, b2), ....} f −1
f −1 = {(b1, a1), (b2, a2),...}
f : A → B
f −1 : B → A A = {1, 2, 3, 6, 8}
B = {2, 4, 5, 4, 7}
f : A → B= {(1, 4), (2, 5), (3, 4), (6, 2), (8, 5)}
f f −1 = {(4, 1), (5, 2), (4, 3), (2, 6), (5, 8)}
(4, 1)
(4, 3)
(5, 2)
(5, 8)
domf −1
= {4, 5, 2} = B
f
(x, b), (y, b) ∈ f ⇔ x = y
f (x) = f (y) x = y
A
B
ax + b.
ax + b = ay + b
a = 0
b
a
x = y
f =
{(x, y) : x, y
∈R, y = x2
}
a2 = b2
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f é sobre f não é sobre
a = b
∀a, b ∈ R
a = −1
b = 1
f f −1 f
domf = im f −1
im f = dom f −1.
f : A → B
B
b ∈ B
a ∈ A
f (a) = b
im f = B
∀b ∈ B, ∃a ∈ A, f (a) = b
A
B
f : A
→ B
f −1
f
f
B
f −1 : B → A
f : A → B
f
A
B
f (x) = x + 1
f
x + 1 = y + 1
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1
x = y .
B
y = 2k + 1 A
x = 2k,
f (x) = x + 1 = 2k + 1
f
f
A
B
A
B
f : A → B
•
|A| > |B|
f
•
|A| |B|
b
A
b
B
A
|A|
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g f f (a)
f
a
g
dom (g ◦ f )(a) = dom f
f
g,
f
g
im f ⊆ dom g
(g ◦ f ) = (f ◦ g).
f : Z → Z g : Z → Z f (x) = x2 + 1 g(x) = 2x − 3. (g ◦ f )(4),
f (4) = (4)2 + 1 = 17
g(f (4)) = 2 ∗ 17 − 3 = 31
(g ◦ f ) = 2(x2 + 1) − 3 = 2x2 + 2 − 3 = 2x2 − 1.
(f ◦ g) = (2x − 3)2 + 1 = 4x2 − 12x + 10
f : A → B g : B → C h : C → D
h ◦ (g ◦ f ) = (h ◦ g) ◦ f
A = {1, 2, 3, 4, 5} B = {6, 7, 8, 9} C = {10, 11, 12, 13, 14}. f : A → B
g : B → C
f : {(1, 6), (2, 6), (3, 9), (4, 7), (5, 7)}
g : {(6, 10), (7, 11), (8, 12), (9, 13)}
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f g
a
f(a) g(f(a))
g o f
(g ◦ f ) = {(1, 10), (2, 10), (3, 13), (4, 11), (5, 11)}
(1, 6)
f
(a, b)
(b, c)
(a, (, c))
b = 6
f
(1, 6)
(2, 6)
g
(1, (, 10))
(2, (, 10))
A
A
idA A
idA = {(a, a)∀a ∈ A}
A
B
f : A → B
f ◦ idA = idB ◦ f = f
A B f : A → B
f ◦ f −1 = idB
f −1 ◦ f = idA
A =
{1, 2, 3
}
B =
{2, 4, 6
}
f =
{(1, 2), (2, 4), (3, 6)
}
f = {(x, y) : x ∈ A, y ∈ B, y = 2x}
idA = {(1, 1), (2, 2), (3, 3)}
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idB = {(2, 2), (4, 4), (6, 6)}
f −1 = {(2, 1), (4, 2), (6, 3)} = {(y, x) : y ∈ B, x ∈ A, x = y2}.
f −1 ◦ f = {(1, 1), (2, 2), (3, 3)} = idA
f ◦ f −1 = {(2, 2), (4, 4), (6, 6)} = idB
f ◦ idA = {(1, 2), (2, 4), (3, 6)}
idB ◦ f = {(1, 2), (2, 4), (3, 6)}.
idA = {(x, x) : x ∈ A, x}
idB = {(y, y) : y ∈ B, y}
f ◦ idA = f (x) = 2xidB ◦ f = y ◦ (2x) = 2x
f −1 ◦ f = 12
(2x) = idA
f ◦ f −1 = 2( y2
) = idB.
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RN
RN RN
(2, 1) R2 (1, 2, 3) R3
RN
RN
RN = {u = (u1, u2, u3,.....,uN ) : ui ∈ R, i = 1,...,N }
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RN
V
• u + v = u + v, ∀u,v ∈ V
• (u + v) + w = u + (v + w), ∀u,v,w ∈ V
• ∃0 ∈ V : 0 + u = u, ∀u ∈ V
• ∀u ∈ V, ∃v,u + v = 0
• ∃1 ∈ V : 1u = u, ∀u ∈ V
• (α + β )u = αu + β u, ∀α, β ∈ R, ∀u ∈ V
• α(β u) = (αβ )u∀α, β ∈ R, ∀u ∈ V • α(u + v) =αu + αv, ∀α ∈ R, ∀u,v ∈ V
R2
RN
V V × V → R u,v → u · v
• u · u > 0, ∀u ∈ V,u = 0
• u · v = v · u, ∀u, v ∈
• (αu) · v = α(u · v), ∀α ∈ R, ∀u,v ∈ V
• (u + v)
·w = u
·w + v
·w,
∀u,v,w
∈ V
RN
u · v = u1v1 + ... + unvn
uAv
A
N × N
(u1, u2)
5 −1−1 5
v1
v2
= 5v1u1 − u1v2 − u2v1 + 5u2v2
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u · u = 5u21 − 2u1u2 + 5u22
u = (0, 0) A
V
V → R u → u
• u + v ≤ u + v , ∀u,v ∈ V
• αu = α u ∀u ∈ V, ∀α ∈ R
• u > 0, ∀u ∈ V,x = 0
R2
u = u21 + u22
uP =
N i=1
uP i
1P
, p ∈ N∗,ppar
2 p = 2
p → ∞
max
u∞ = max1≤i≤N ⌋ui⌊
u v
u v
d(u,v) = u− v
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RN
V ∈ RN
u2 =√ u · u
|u · v| ≤ u2 v2
V ⊆ RN u u
u =N i=1
αivi
αi ∈ R vi ∈ V u
u v ∈ V ⊆ RN u
v
u
v
u
v, u v
V ⊆ RN S m V S
S
α1u1 + .... + αmum = 0
α1 = α2 = ... = αm = 0
V ⊆ RN S = {u1,u2, ...,un} V
V
S
S
V
S
V v =
n
i=1 αiui
v ∈V
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B V
R2
B = {(1, 0), (0, 1)} (1, 0) = α(0, 1)
α ∈ R R2 B
• (2, 4) = 2(1, 0) + 4(0, 1)
• (33.5, −100) = 33.5(1, 0) − 100(0, 1).
B ⊆ R2 R2 B
R2
V ⊆ RN B V B V
• B
• B
V
•
•
B
•
R2
B1 = {(1, 0), (0, 1)}B2 =
{(3, 1), (
−2, 1)
}
(1, 0) = α(0, 1), ∀α ∈ R
(3, 1) = α(−2, 1), ∀α ∈ R R2
B1 B2 (7, 9)
(7, 9) = α1(1, 0) + α2(0, 1)
(7, 9) = β 1(3, 1) + β 2(−2, 1)
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RN
7 = α1 9 = α2
7 = 3β 1 − 2β 29 = β 1 + β 2
β 1 = 5, β 2 = 4
RN
RN
RN
r u
Br(u) = {v ∈ RN : d(u,v) < r} RN
r
u
Br(u) = {v ∈ RN : d(u,v) ≤ r} RN
r
u
Br(u) = {v ∈ RN : d(u,v) = r}
u
v
G ⊆ RN RN u ∈ G ǫ > 0 Bǫ(u) ⊆ G
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RN
ǫ
A =
{x : x
∈ (0, 1)
}
ǫ
x
Bǫ(x) = (x − ǫ, x + ǫ) ǫ min{x
2, 1− x
2}
Bǫ(x → 0) = (x − x2 , x + 1 − x2 ) → (0, 1) ∈ A I = {x : x ∈ [0, 1]}. ǫ
x
Bǫ(x) = (x − ǫ, x + ǫ) x = 0
ǫ
(−ǫ, ǫ)
I
RN
RN
∅
∁(V )
∁(V ) = RN \V = {u ∈ RN : u = F }
∁(I ) = (−∞, 0) ∪ (1, ∞)
I = [0, 1]
∁(A) = (−∞, 0] ∪[1, ∞)
A = (0, 1) ∁(∅) = RN ∁(RN ) = ∅.
F ⊆ RN RN
A 0] [1
A
A
I
I
I
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RN
∅ RN
E = {x : x ∈(0, 1]}
1
E
E
∁(E ) = (−∞, 0] ∪ (1, ∞)
0
E
E
u ∈ RN
A ⊆ RN
:
• u u
• u A u A
• u A u A
• u A u ∁(A)
I
•
• I
• I
F
• F
u ∈ RN S ⊆ RN Bǫ(u)
S u
•
S ⊆ R u = supS /∈ S u
S
ǫ > 0
v ∈ S
v ∈ (u− ǫ,u + ǫ)
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RN
• A (0, 1) [0, 1]
A
• N = {0, 1, 2, 3, 4, 5, ....}
0.00001 N
•
• Z N
• (2, 3) ∪{4} [2, 3]
{4}
4
F ⊆ RN RN
RN
S ⊆ R u = supS /∈ S u
S
inf S
• Z
• A = (0, 1) ∪ {2, 3, 4, 5, .....}
[0, 1]
• A = {1, 2, 3}
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RN
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RN
R2 R3
P = {1 + x2, 3x + 4x3, 6 − 3x + 2x2 + x3}.
P
a + bx+ cx2 + dx3
P
P
B = {1, x , x2, x3}
1 + x2 = 1(1, 0, 0, 0) + 0(0, x, 0, 0) + 1(0, 0, x2, 0) + 0(0, 0, 0, x3)
4
d2u
dx2 − k2u = 0, k > 0
x ∈ [0, 1].
u(x)
∀x ∈ [0, 1]
u(x)
V
[0, 1]
B = {ekx, e−kx}
V
u(x) = α1ekx
+ α2e−kx
RN
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u p =
|u| p 1
p
u p ≥ 1
um,p =
Ω
|α|≤m
|Dαu| p
1
p
u1,2 Ω ∈ [0, 1] × [0, 1]
um,p = 1
0
10
|u|2 +
du
dx
2
+
du
dy
2
dxdy
12
m = 0
R2 L p p 1 2 ∞
1
p = ∞
− ddx
a
du(x)
dx
= f (x)
a f (x) u(x)
T (u) = f
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T : U → V
T = − ddx
a d
dx
u, U, f V
T
U
V
u ∈ U v ∈ V
• T (αu) = αT (u) ∀u ∈ U, ∀α ∈ R
• T (u1 + u2) = T (u1) + T (u2), ∀u1,u2 ∈ U
T 1 T 2 T 1T 2
u v
V
ℵ(T ) = {u : u ∈ U, T (u) = 0}
T : U → V
T (u) = u1 + u2
ℵ(T ) = {u ∈ U : u1 + u2 = 0}
(−u2, u2)
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v = (−1, 1) T
T : U
→ V
ℵ(T ) = {0}
T : U
→ V
T (u) = 7u
ℵ(T ) = {u ∈ U : u = 0}
ℜ(T ) = I m (T ) ⊆ V T : U → V
U e V ⊆ RN N V
T
U
T
N
T : U → V
T (u) = u1 + u2 rank
1
1
a
U
R2
U
V
T : U → V
{φ1,φ2, ....,φn} U {ϕ1,ϕ2, .....,ϕm} V
U
V
u =n
i=1 αiφiv =
m j=1
β jϕ j
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u v T (u) = v
T (u) =n
i=1
αiT (φi) =m
j=1
β jϕ j,
T (φi) ∈ V ,
T (φi) =m
j=1
t jiϕ j
m j=1
β jϕ j −n
i=1
αi
m j=1
t jiϕ j
= 0
m j=1
β j − ni=1
t jiαiϕ j = 0
β = Tα
T m × n t ji
•
T
•
T
T
U
V
φ = {1, x , x2, x3}ϕ =
{1, x
}
U u =
4i=1 αiφi = α1 + α2x + α3x
2 + α4x3
n = 4)
V
v =2
j=1 β jϕ j = β 1 + β 2x m = 2
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U V
D = d2
dx2
D : U → V
D
4i=1
αiφi
=
2 j=1
β jϕ j
D
i
4
i=1αiD(φi) =
2
j=1β jϕ j
D(φi) ∈ V,
D(φi) =2
j=1
d jiϕ j
V
D(φi)
i = 1..4 {1, x , x2, x3}
D(φ1) = D(1) = 0
D(φ2) = D(x) = 0
D(φ3) = D(x2) = 2
D(φ4) = D(x3) = 6x
V
D(φ1) = d11 ∗ 1 + d21 ∗ x = 0 ∗ 1 + 0 ∗ xD(φ2) = d12 ∗ 1 + d22 ∗ x = 0 ∗ 1 + 0 ∗ xD(φ3) = d13 ∗ 1 + d23 ∗ x = 2 ∗ 1 + 0 ∗ xD(φ4) = d14 ∗ 1 + d24 ∗ x = 0 ∗ 1 + 6 ∗ x
4i=1
αiD(φi) =2
j=1
β jϕ j
α1 (0 ∗ 1 + 0 ∗ x) + α2 (0 ∗ 1 + 0 ∗ x) + α3 (2 ∗ 1 + 0 ∗ x) + α4 (0 ∗ 1 + 6 ∗ x) = β 1 ∗ 1 + β 2 ∗ x
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V
ϕ1 → α1 ∗ 0 + α2 ∗ 0 + α32 + α40 = β 1ϕ2 → α1 ∗ 0 + α2 ∗ 0 + α30 + α46 = β 2
0 0 2 0
0 0 0 6
α1
α2
α3
α4
=
β 1
β 2
.
D =
0 0 2 0
0 0 0 6
U V
p(x) = 7 + 2x + 15x2 − 0.5x3
D : U → V,
0 0 2 0
0 0 0 6
7
2
15
−0.5
=
30
−3
v = 30 − 3x
u = (x1, x2, x3) R3
3
v = (x
′
1, x
′
2, x
′
3)
R : R3 → R3
R(x1, x2, x3) = (x1cos(θ) − x2sin(θ), x1sin(θ) + x2cos(θ), x3).
R3 {(1, 0, 0), (0, 1, 0), (0, 0, 1)}
R(1, 0, 0) = (cos(θ),sin(θ), 0)R(0, 1, 0) = (−sin(θ),cos(θ), 0)R(0, 0, 1) = (0, 0, 1)
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R =
cos(θ) sin(θ) 0−sin(θ) cos(θ) 0
0 0 1
.
T =
1 0 0
0 1 0
0 0 0
T : R3 → R3. kernel
p(x) = a0 + a1x + a2x2 + a3x
3
4
U = {1, x , x2, x3}.
P
p = [1, x , x2, x3]
a0
a1
a2
a3
= Ua
a
U
p ∈ P p(x) x
p(s) = C(s)p
C(s) s P
p
p(s) = C(s)Ua = [1, s , s2, s3]
a0
a1
a2
a3
.
p(s)
p
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4 s p(s)
[1, si, s2i , s
3i ]
a0
a1
a2
a3
= p (si) , i = 1..4
E(S )a = v
S = {s1, s2, s3, s4} v = { p (s1) , p (s2) , p (s3) , p (s4)} E
a = E−1(S )v
E−1
T : u∆ → uΦ, u∆,uΦ ∈V
u∆ =
N i=1 αiδi ∆ uΦ =
N j=1 β jφ j Φ
T
N i=1
αiδi
=
N j=1
β jφ j
N i=1
αiT (δi) =N
j=1
β jφ j
T (δi) Φ
N i=1
αi
N j=1
r jiφ j =N
j=1
β jφ j
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r11 ... r1N ...
. . . ...
rN 1 ... rNN
T
α1...
αN
=
β 1...
β N
R
u∆ u∆ ∈ V V ⊆ R2 R ∆
Φ
P : u∆ → uΦ
u∆ = α1δ1 + α2δ2
uΦ = β 1φ1 + β 2φ2
u∆ = uΦ
Φ V ∆
δ1 = r11φ1 + r12φ2
δ2 = r21φ1 + r22φ2
u∆ = α1δ1 + α2δ2 = α1 (r11φ1 + r12φ2) + α2 (r21φ1 + r22φ2)
β 1φ1 + β 2φ2 = α1 (r11φ1 + r12φ2) + α2 (r21φ1 + r22φ2)
φ
φ1 → β 1 = r11α1 + r21α2φ2 → β 2 = r12α1 + r22α2
r11 r21r12 r22
α1α2
= β 1β 2
N = 2
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∆ = {(1, 0), (0, 1)} Φ = {(3, 1), (−2, 1)}
(1, 0) = r11(3, 1) + r12(−2, 1)(0, 1) = r21(3, 1) + r22(−2, 1)
1 = 3r11 − 2r120 = 1r11 + 1r12
0 = 3r21 − 2r22
1 = 1r21 + 1r22
r11 = 0.2 r12 = −0.2 r21 = 0.4 r22 = 0.6
R =
0.2 0.4
−0.2 0.6
0.2 0.4−0.2 0.6 α1α2 = β 1β 2 .
(7, 9)
∆
(5, 4)
Φ
0.2 0.4
−0.2 0.6
7
9
=
5
4
7(1, 0) + 9(0, 1) = 5(3, 1) + 4(−2, 1).
u∆ u∆ ∈ V V ⊆ R2
P
Φ
∆
P : uΦ → u∆
u∆ = α1δ1 + α2δ2
uΦ = β 1φ1 + β 2φ2
u∆ = uΦ
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∆ V Φ
φ1 = p11δ1 + p12δ2
φ2 = p21δ1 + p22δ2
uΦ = β 1φ1 + β 2φ2 = β 1 ( p11δ1 + p12δ2) + β 2 ( p21δ1 + p22δ2)
α1δ1 + α2δ2 = β 1 ( p11δ1 + p12δ2) + β 2 ( p21δ1 + p22δ2)
δ
δ1 →α1 = p11β 1 + p21β 2δ2 → α2 = p12β 1 + p22β 2
p11 p21
p12 p22
β 1
β 2
=
α1
α2
.
β
β = P−1α,
R = P−1 ∆
p11 p21
p12 p22
=
φ11 φ12
φ21 φ22
.
∆ = {(1, 0), (0, 1)}
Φ = {(3, 1), (−2, 1)}
(3, 1) = p11(1, 0) + p12(0, 1)
(−2, 1) = p21(1, 0) + p22(0, 1)
3 = p11
1 = p12
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−2 = p211 = p22
p11 = 3 p12 = 1 p21 = −2 p22 = 1
P =
3 −21 1
,
3 −21 1
5
4
=
7
9
.
R2 P R
x
φi
x =n
i=1
αiφi
β j δ j
x′
=n
j=1β jδ j .