Apostila Fundamentos Matemática

8/17/2019 Apostila Fundamentos Matemática

1/266


2/266


3/266

RN

RN


4/266


5/266

p


6/266

QR


7/266

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∗}


8/266

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]


9/266

•

•

•

•

•

•

•

A

B

A

B

✞✝ ☎ ✆ x y x + y A x y B x+y

A ⇒ B

A B A ⇒ B

A

B


10/266

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


11/266

A B

A

B

A B

B

∧

A B A e B

A ⇒ not(B)

não A

¬

x

A

x

B

B


12/266

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,


13/266

⇐ 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.

∼


14/266


15/266

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


16/266

⊆

∀

∀x ∈ A,

x

•

•

•

•

x

x

x

∃

∃x ∈ A, x

•

• N x

2

¬(∃x ∈ Z, x

∀x ∈ Z, ¬(x


17/266

¬(∀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}


18/266

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}


23/266

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


24/266

a ∈ R a

|a| = +a se a > 0−a se a


25/266

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),...}.


26/266

✬

✫

✩

✪

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


27/266

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


28/266

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


29/266

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|


30/266

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)}


31/266

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)}


32/266

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.


33/266

RN

RN RN

(2, 1) R2 (1, 2, 3) R3

RN

RN

RN = {u = (u1, u2, u3,.....,uN ) : ui ∈ R, i = 1,...,N }


34/266

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


35/266

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


36/266

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


37/266

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)


38/266

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


39/266

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


40/266

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)


41/266

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}


42/266

RN


43/266

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


44/266

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


45/266

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)


46/266

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


47/266

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


48/266

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


49/266

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)


50/266

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


51/266

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


52/266

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


53/266

∆ = {(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


54/266

∆ 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


55/266

−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 .

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