Series de Potencias 2015
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Transcript of Series de Potencias 2015
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8/18/2019 Series de Potencias 2015
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y = f (x)
pn(x)
pn(x) = a0 + a1x + a2x2 + ... + anx
n.
n
x
n.
A(0, f (0))
f (x)
(n + 1)
x = 0
y = p0(x)
A a0 A
a0 = f (0) x = 0.
y = a0 + a1x
A
A
a0 = f (0) y a1 = f (0).
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n y = a0 + a1x + a2x2 + ... + anx
n,
A
A
1 2 3 n
y = pn(x) n
x = 0
f (x)
x = 0
f (x)
pn(0) = f (0), p
n(0) = f (0), pn(0) = f
(0) ... pnn(0) = f n(0)
f (0) = a0, f (0)
1! = a1,
f (0)
2! = a2 ...
f n(0)
n! = an
f
n
x0
p(x) = f (x0) + f (x0)(x − x0)
1! +
f (x0)(x − x0)2
2! + ... +
f n(x0)(x − x0)n
n!
f
x0
x0 = 0
p(x) = f (0) + f (0)x
1! +
f (0)x2
2! + ... +
f n(0)xn
n!
f
f (x) = sen(x)
sen(x)
cos(x), −sen(x), −cos(x), .... x = 0
1, 0, −1, 0, .... x x − x3
3!
f (x) = sen(x) x = 0
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f (x) = sen(x)
sen(x) ∼ x
f (x) = log(x + 1)
f (x) = log(x + 1)
x = 0
f (x)
n + 1
I
x0 Rn(x)
f (x) P n(x) Rn(x) = f (x) − P n(x)
f (x) = P n(x) + Rn(x)
f (x) = f (x0) + f (x0)
1! (x − x0) +
f (x0)
2! (x − x0)
2 + ... + f n(x0)
n! (x − x0)
n + Rn.
Rn(x)
Rn(x) = f n+1(ξ )
(n + 1)!(x − x0)
n+1 ξ x0 y x
f (x) = f (x0) + f (x0)
1! (x − x0) +
f (x0)
2! (x − x0)
2 + ... + f n(x0)
n! (x − x0)
n + f n+1(ξ )
(n + 1)!(x − x0)
n+1.
f (x)
x0 = 0
f (x) = f (0) + f (0)
1! x +
f (0)
2! x2 + ... +
f n(0)
n! xn +
f n+1(ξ )
(n + 1)!xn+1.
ξ
x0 x
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Rn(x) I
Rn(x) = f (x) − f (x0) + f (x0)
1! (x − x0) + f (x0)
2! (x − x0)2 + ... + f n(x0)
n! (x − x0)n
∗
x
x0 g(t) I
g(t) = f (x) − f (t) − f (t)
1! (x − t) −
f (t)
2! (x − t)2 − ... −
f n(t)
n! (x − t)n − Rn(x)
(x − t)n+1
(x − x0)n+1
g(x) = 0
g(x0) = [f (x)−f (x0)−f (x0)
1! (x − x0)−
f (x0)
2! (x−x0)
2−...−f n(x0)
n! (x−x0)]−Rn(x)
(x − x0)n+1
(x − x0)n+1 = 0
∗ Rn(x)
x
x0 I g(x) = 0 g(x0) = 0
ξ
x
x0 g(ξ ) = 0
g(t)
g(t) = 0 − f (t) − [−f (t)
1! +
f (t)
1! (x − t) − 2
f (t)
2! (x − t) +
f (x0)
2! (x − x0)
2 − ...
−nf n(t)
n! (x − t)n−1 +
f n+1(t)
n! (x − t)n] + (n + 1)Rn(x)
(x − t)n
(x − x0)n+1
g(t) = −f n+1(t)
n! (x − t)n + (n + 1)Rn(x)
(x − t)n
(x − x0)n+1
g(ξ ) = 0
g(ξ ) = −f n+1(ξ )
n!
+ Rn(x) n + 1
(x − x0)n+1
= 0
Rn(x) = f n+1(ξ )
(n + 1)!(x − x0)
n+1.
f (x) = ex
e
ex
= 1 + x + x2
2! + ...
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P 5(x) = 1 + x + x2
2! +
x3
3! +
x4
4! +
x5
5!
ex
e ∼ 1 + 1 + 1
2 +
1
6 +
1
20+
1
120 + R5(1)
e ∼ 2, 716 + R5(1)
R5(1) = eξ
(5 + 1)!(1)5+1
0 < ξ
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∞n=0
anxn
0
∞n=0
anxn
x
x = 0.
x
(−R, R)
|x| > R.
∞n=0
xn = 1 + x + x2 + ... + xn + ...
(−1, 1)
x ∞n=0
xn = 1
1 − x
(−1, 1) ∞n=0
xn
1
1 − x.
R
∞n=0
anxn
x ∈ (−R, R)
x |x| > R.
(−R, R)
x
x = 0
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∞n=0
anxn
L = ĺımn→∞
an+1an 0 ≤ L ≤ ∞
L = ∞ x = 0
L = 0 x
0 < L < ∞ R = 1
L |x| < R
−R < x < R |x| > R
∞n=0
xn
n! x
x = 0 1
x = 0 L = ĺımn→∞
1
(n + 1)!1
n!
= lı́m
n→∞1
n + 1 = 0
L = 0
R = ∞ x
x = 0
∞n=0
n! xn x = 0 L = ĺımn→∞
(n + 1)!n! = lı́mn→∞(n + 1) = ∞
x = 0
∞n=0
2nxn
n
−
1
2, 1
2
(x − x0)
∞n=0
an(x − x0)n = a0 + a1(x − x0) + a2(x − x0)
2 + ... + an(x − x0)n + ...
x0
x0 = 0 x
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x
x − x0
∞n=0
an(x − x0)n
|x − x0| < R (x0 − R, x0 + R)
|x − x0| > R
∞n=0
(x − 1)n
n
L = ĺımn→∞
1
n + 11
n
= 1
|x − 1| < 1 x ∈ (0, 2) |x − 1| > 1
x = 0, ∞n=0
(−1)n
n
x = 2
∞n=0
(1)n
n
∞
n=0xn
n [0, 2)
(x − x0)
x0.
f (x)
n=0
an(x − x0)n
(x0 − R , x0 + R) R > 0
(x0 − R , x0 + R) f (x)
(x0 − R , x0 + R)
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f (x) =∞n=0
an(x − x0)n
R
f |x − x0| < R
f
f (x) =∞n=1
ann(x − x0)n−1 = a1 + 2a2(x − x0) + ... + ann(x − x0)
n−1 + ... |x − x0| < R
f
ˆ f (x)dx =
∞n=0
´ an(x − x0)
ndx =∞n=0
an(x − x0)
n+1
n + 1 + C
|x − x0| < R
f (x) =∞n=0
xnn!
x
x0 = 0 R = ∞
f (x) = f (x)
x
f (x) = ex
L = 0 x.
x.
f (x) = d
dx
1 + x +
x2
2! +
x3
3! ...
= 0 + 1 +
2x
2! +
3x2
3! + ... = 1 + x +
x2
2! + ... = f (x)
f (x) = f (x)
f (x) = C ex
f (0) = 1
C = 1 f (x) = ex
∞n=0
xn 1
1 − x
|x| < 1 1
(1 − x)2
1
(1 − x)2
f (x) = 1
1 − x
f (x) = 1
(1 − x)
2 = 1 + 2x + 3x2 + 4x3 + ... + nxn−1 + ... , |x| < 1
∞n=0
xn+1
n + 1 = −ln(1−x) −1 < x
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ˆ x0
1
1 − u du = −ln(1 − x)
−ln(1 − x) =
ˆ x0
1
1 − u du =
∞n=0
xn+1
n + 1
ln(x + 1) =∞n=0
(−1)n x
n+1
n + 1 (−1, 1)
f
x0 f
x − x0 x0
ε > 0 / f (x) =∞n=0
an(x − x0)n ∀x ∈ (x0 − ε , x + ε)
P (x) = a0 + a1x + a2x2 + ... + anxn
f (x) =
1
1 − x
x = 0
f (x) =
1
1 − x =
∞
n=0x
n
(3)
f
∞n=0an(x − x0)
n
f
I
f (x) =∞n=0
an(x − x0)n ∀x ∈ I .
f
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f
f (x) =
∞n=0
an(x − x0)n en I = (x0 − R, x0 + R)
an = f n(x0)
n! para n = 0, 1, 2,...
f (x) = a0 + a1(x − x0) + a2(x − x0)2 + ... + an(x − x0)
n + ... ; f (x0) = a0
f (x) = 1· a1 + 2a2(x − x0) + ... + nan(x − x0)n−1 + ... ; f (x0) = 1· a1
f (x) = 1· 2· a2 + ... + n(n − 1)an(x − x0)n−2 + ... ; f (x0) = 1· 2.a2
f (x) = 1· 2· 3· a3 + ... + n(n − 1)(n − 2)an(x − x0)n−3 + ... ; f (x0) = 1· 2· 3· a3
an = f (n)(x0)
n!
f (x) = f (x0) + f (x0)
1! (x − x0) +
f (x0)
2! (x − x0)
2 + ... + f n(x0)
n! (x − x0)
n + ...
f
f
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f
I
x0 ∈ R
f
x0
∞n=0
f (n)(x0)
n! (x − x0)
n = f (x0) + f (x0)(x − x0)
1! +
f (x0)(x − x0)2
2! + ... +
f (n)(x0)(x − x0)n
n! + ...
f
x0 = 0
∞n=0
f (n)(0)n!
xn = f (0) + f
(0)1!
x + f (0)2!
x2 + ... + f (n)(0)
n! xn + ...
Rn(x) = f (x) − P n(x), P n(x)
n ĺımn→∞
Rn(x) =
0 ∀ x ∈ I f (x) P n(x)
Rn(x).
f
(x0 − R, x0 + R
f (x) = f (x0) + f (x0)
1! (x − x0) +
f (x0)
2! (x − x0)
2 + ... + f n(x0)
n! (x − x0)
n + ...
f
lı́mn→∞
Rn(x) = 0
Rn(x) = f n+1(ξ )
(n + 1)!(x − x0)
n+1 ξ (x0 − R, x0 + R
ĺımn→∞
Rn(x) = 0
f (x) = P n(x) + Rn(x) n → ∞,
f (x)
ξ
ξ
f
|Rn(x)|
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f −R < x − x0 < R
f
x = x0
f
x0
f
g
f −R < x − x0 < R
f
x = 0
f (n)(0) = 0 , ∀n
0 + 0
1x +
0
2!x2 +
0
3!x3 + ... = 0
f (x) = 0 ∀x f (x) x = 0
f (x) =
e− 1
x2 x = 0
0 x = 0
R − {0}
x = 0
x = 0
x = 0
Rn(x) 0 n → ∞.
sen(x) = x − x3
3! +
x5
5! + ... + (−1)
n x2n+1
(2n + 1)! + ... =
∞n=0
(−1)n
(2n + 1)!x2n+1
x
cos(x) = 1 − x2
2! +
x4
4! + ... + (−1)n
x2n
(2n)! + ... =
∞n=0
(−1)n
(2n)! x2n x
ex = 1 + x + x2
2! +
x3
3! +
x4
4! + ... +
xn
n! + ... =
∞n=0
xn
n!
x
e =∞n=0
1
n!
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f (x) = 1
1 − x
∞n=0
xn = 1 + x + x2 + ... + xn + ..., ∀x ∈
(−1, 1) .
f (x) = 1(1 − x)2
f (x) = ln(x + 1)
∞n=0
(−1)n xn+1
n+1 , ∀x ∈ (−1, 1)
ln(1 − x) −x x
ln(x + 1)
ex = 1 + x + x2
2! +
x3
3! +
x4
4! + ... +
xn
n! + ... =
∞n=0
xn
n! x
lı́mn→∞
xn
n! = 0
f (x) = sen(x)
sen(x) = x − x3
3! + x5
5! + ... + (−1)n
x2n+1
(2n + 1)! + ... =
∞n=0
(−
1)n
(2n + 1)! x2n+1 x
x
ĺımn→∞
Rn(x) = ĺımn→∞
f n+1(ξ )
(n + 1)!(x − x0)
n+1
| f n+1(x) |=| cos(x) | | f n+1(x) |=| sen(x) | | Rn(x) |≤ | x |n+1
(n + 1)!
ĺımn→∞
xn
n! = 0
ĺımn→∞
Rn(x) = 0
f
,
e−x2
e
x
=
∞n=0
xn
n!
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´ e−x
2
dx =∞n=0
(−1)n x2n+1
(2n+1)n! + C