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f
f(x) = a0 + a1x+ a2x2 + · · ·+ anxn,
ai ai fn
f ain
-
aai
ai,j
ff
fx a n
a0, a1, a2, . . . , an n
g tb n = 3 b0, b1, b2, b3 = 2, 0, 4,−1
g
g(t) = 2 + 0t+ 4t2 + (−1)t3.
g g(t) = 2+4t2− t3g t = 2
2 t
g(2) = 2 + 4(22)− 23 = 10.
-
g(t) = (t− 1)(t+ 6)2
t1, t2, t3, . . . , tn
f(x) = x+ 1g(t) = 1 + t
t f(t)(t, f(t))
f(x) = x5 − x− 1
3 6
-
(x, y) = xy
(x) = 1− x(x, y) = 1− (1− x)(1− y)
-
f(x) = 0
g(x) = 12
h(x) = 1 + x+ x2 + x3
i(x) = x1/2
j(x) =1
2+ x2 − 2x4 + 8x8
k(x) = 4.5 +1
x− 5
x2
l(x) = π − 1ex5 + eπ3x10
m(x) = x+ x2 − xπ + xe
n! = 1 ·2 · · · · ·n 0! = 1
f(x) = 0f
n = 0 a0 = 0f n = 1 a0 = 0 a1 = 0 n = 2
f(x) = 0
anf(x) = 0
f(x) = 0−1
-
f : A → B
f : A → BA B f
f : A→ B
-
R
f : R → R
ZN
∈q ∈ N qq q
p, p+ 2 11 13 23
-
n ≥ 0 n + 1(x1, y1), (x2, y2), . . . , (xn+1, yn+1) R2 x1 < x2 < · · · < xn+1
p(x) n p(xi) = yi i
R2Z3 N10
nn + 1
n = 0 n+ 1 = 1(7, 4)
a0+a1x+a2x2+ · · ·+adxdd n d = 0d = 0 f f(x) = a0
f(7) = 4 f(x) = 4
(7, 4)
n = 1 n + 1 = 2(2, 3), (7, 4) f f(2) = 3f(7) = 4
f(x) = a0 + a1x.
f(2) = 3, f(7) = 4
a0 + a1 · 2 = 3a0 + a1 · 7 = 4
f(x) (a, b) f(a) = bf
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a0 a0 = 3− 2a1(3 − 2a1) + a1 · 7 = 4 a1 = 1/5
a0 = 3− 2/5
f(x) =
(3− 2
5
)+
1
5x =
13
5+
1
5x.
x1 < x2 < · · · < xn+1 x1 < x2x1 = x2
(2, 3), (2, 5)
a0 + a1x
f(2) = 3 f(2) = 5
a0 + a1 · 2 = 3a0 + a1 · 2 = 5
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n = 2
(x1, y1) n = 0 x1y1
f(x1) = y1 ff y1
f(x) = y1
(x1, y1), (x2, y2) xx1, x2, . . .
f(x) = y1x− x2x1 − x2
+ y2x− x1x2 − x1
f x1 x1 − x1 =0y1
x1−x2x1−x2 = y1 · 1 x1 y1
x1 = x20/0f(x2)
y2 f(x1) = y1 f(x2) = y2f
f
f(x) =y1
x1 − x2(x− x2) +
y2x2 − x1
(x− x1),
f(x) =x1y2 − x2y1x1 − x2
+
(y1 − y2x1 − x2
)x
xf x
f(x1) = y1(x1, y1), (x2, y2), (x3, y3)
x1
x2
-
f(x) = y1(x−x2)(x−x3)
(x1−x2)(x1−x3) + y2(x−x1)(x−x3)
(x2−x1)(x2−x3) + y3(x−x1)(x−x2)
(x3−x1)(x3−x2)
f x1y1
x2, x3f 2
n (x1, y1), . . . , (xn, yn)i yi
x−xj j i(xi − xj) j
f(x) =n∑
i=1
yi ·
⎛
⎝∏
j ̸=i
x− xjxi − xj
⎞
⎠
∑,∏
∑ni=1( )
∑ni=0
n+ 1
∑
+ ∑ ∏
-
∏j ̸=i i
jj
ji i
j xx 1 n j
j ̸= i jj = i
∏j ̸=i
j i
f(x) =n∑
i=1
(i)
⎛
⎝∏
j ̸=i(i, j)
⎞
⎠
∑ ∏
-
(x1, y1), . . . , (xn+1, yn+1) n + 1 xif(x)
f(x) =n+1∑
i=1
yi∏
j ̸=i
x− xjxi − xj
f(x) nn i xi ii yi
!
f : R→ R z f(z) = 0
R nn
-
f g n(x1, y1), . . . , (xn+1, yn+1)
f gf, g (f −g)(x)(f − g)(x) = f(x) − g(x) f − g
f ai g bi f − gci = ai − bi f g ci ai −bi
f, g f − gf − g n
x7
x5 (f − g)(xi) = 0 i xxi
(x1, y1), . . . , (xn+1, yn+1) f g if(xi) = g(xi) = yi
d f − g d ≤ nf−g n
n+1 xi f − g f − gf g
n ≥ 0 n + 1(x1, y1), (x2, y2), . . . , (xn+1, yn+1) R2 x1 < x2 < · · · < xn+1
p(x) n p(xi) = yi i
(x1, y1), . . . , (xn+1, yn+1) xif(x)
f(x) =n+1∑
i=1
yi
⎛
⎝∏
j ̸=i
x− xjxi − xj
⎞
⎠
f(x) ≤ nn i xi ii yi
g(x)f = g f−g
n n+ 1 xif − g f = g
-
nn+ 1
-
x
(x − xj)/(xi − xj)a0 = −xj/(xi − xj) a1 = 1/(xi − xj)
-
nk ≤ n
-
s f(x) f(0) =s d f(x) d f
a0, . . . , ad f a0 = sad ̸= 0 n f(x)
f(1), f(2), . . . , f(n) i (i, f(i))k
k − 1 g(x)g(x) f(x)
g(0) = f(0) dk
d = k − 1 kg(x) g(x) = f(x)
n = 5k = 3 f(x)
d = k − 1 = 2 109 f
f(x) = 109 + · x+ · x2
(1, f(1)), (2, f(2)), (3, f(3)), (4, f(4)), (5, f(5))
f(x) = 109− 55x+ 271x2,
f(0) = 109
(1, 325), (2, 1083), (3, 2383), (4, 4225), (5, 6609).
f(x)
-
f(x) f(0) kf(0)
f f(0) k
f d d
d dx x = 0 y (x, y)
y
df(0) y
y f(0)s =
f(x) f(0)s f(1), . . . , f(10) y
10f(1), f(2), . . . , f(10)y = f(0) 10
109
f(x) = 109− 55x+ 271x2
(2, 1083), (5, 6609)533 f(2) = 1083 f(5) =6609 f(0) = 533
-
(2, 1083), (5, 6609)
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f 2 g 1f · g 3
f n g mf · g n+m
f g−1
a, bn ϕ(n) n > 1
n nn > 1
an 1
f g f(x) =g(x)h(x) h f g h
g hf
f, gf g
a, naϕ(n) n
x
√2 π e
φ = 1+√5
2
√2 +√3
π
-
e π + e πe
f(x) = a0 + a1x+ · · ·+ anxn n nr1, . . . , rn
n∑
i=1
ri = −an−1an
n∏
i=1
ri = (−1)na0an
.
r f(x) f(x) = (x−r)g(x)g(x)
x y
ff
w(x) =20∏
i=1
(x− i)
-
x19 w(x) −210 2−230.5
p
M > 1M = m1 ·m2 · · ·mk mi > 1
i, j mi mj r1, . . . , rk0 ≤ ri < mi ri mi
x 0 ≤ x < M x = ri mi i
kp(x) = (x − a1)(x − a2) · · · (x − ak)
ai
2
p(x) = (x− a1)(x− a2) · · · (x− am)(x2 + bm+1x+ am+1) · · · (x2 + bkx+ ak),
-
f(x) g(x) h(x) j(x) l(x) i√x = x1/2 k(x)
m(x)π, e π e
p, p + 2
M pq p q − p ≤M
M 26
100 M 70
M246
6
f
S
d d(x, y)x, y S
-
S S
dd(x, y) =
d(y, x)
f
fd
S f d
f (S, d) d
-
n p np p
a = n p an/p n < p n p = n
≡
a ≡ n p.
pp
n p k (n · k) ≡ 1 p
kf(x) x p
0 p
f(x) x pf
xf(0)
(d+ 1)f(0) 0 p
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