**Unformatted text preview: **Calculus I
Lecture 8 February 3, 2018 Problem set 8 Section 11.9: 4, 10, 12, 40
Section 11.10: 16, 18, 20, 55, 68, 70, 72. Outline 1 Power series representations (11.9) 2 Taylor series (11.10) Power series representations
1
We have shown that 1−x has a power series expansion in powers of
1
1−x = ∞
X x n, x ∀x ∈ (−1, 1). n=0 From this we can derive other power series expansions by as follows.
Substitute −x 2 for x in the series above, we obtain Substitution: 1
1 + x2 = ∞
X 5 + 3x = 1
2 + 3u This holds for
x2 = 1 ∀x ∈ (−1, 1). n=0 1
To expand 5+3x in powers of
1 (−1)n x 2n
x + 1, 1 2 1 + 3u
2 − 32 u ∈ (−1, 1) Example: 2x+3 in powers of =
i.e. 1
2 let ∞
X
n=0 u = x + 1.
3 (− u)n =
2 x ∈ ( −35 , −31 ). x − 1. Then ∞
X
(−3)n
n=0 2n+1 (x + 1)n . Integration:
get 1
Integrating the power series for 1−x term by term, we − ln(1 − x) = C + ∞
X
xj
j=1 j ∀x ∈ [−1, 1). Let x = 0 we get C = 0.
1
Integrate the power series of 1+x 2 we get
arctan x = ∞
X
(−1)j x 2j+1
j=0 2j +1 ∀x ∈ (−1, 1]. (note the change in the interval of convergence) Dierentiation: Example:
1 (1 − x)2 = ∞
X
j=0 (j + 1)x j ∀x ∈ (−1, 1). Taylor polynomials
Recall that when x is close to a, f (x) ≈ f (a) + f 0 (a)(x − a).
The RHS is called the linearization of f at a. To get better approximation we use higher order polynomials. Tn (x) = n
X
f (k) (a)
k=0 Note that at the point a, f and Tn k! (x − a)k . agree to the
(n) nth order derivatives: Tn (a) = f (a), Tn0 (a) = f 0 (a), . . . , Tn (a) = f (n) (a).
These are the Taylor polynomials of f at a. Taylor remainder
Using integration by parts and induction, we can prove that Z
f (x) − Tn (x) = x (x − t)n
dt.
n! f (n+1) (t) a n = 0: (1) This is none other than the fundamental theorem of calculus Z x f 0 (t)dt. f (x) − f (a) = (2) a n = 1: Integrating by part, using Z x 0 Z f (t)dt =
a u =f0 x udv = uv |xa a = f 0 (a)(x − a) + v = −(x − t) and x Z
−
Z x we have vdu
a f 00 (t)(x − t)dt. a Putting this into formula (2) we get f (x) = f (a) + f 0 (a)(x − a) + Z
a x f 00 (t)(x − t)dt (3) v = − 12 (x − t)2 we have
Z x
Z x
x
00
x
f (t)(x − t)dt =
udv = uv |a −
vdu
a
a
a
Z x
(x − t)2
f 00 (a)
2
(x − a) +
f 000 (t)
dt.
= n = 2:
Z Integrating by part, using u = f 00 2 and 2 a Putting this into formula (3) we get
0 f (x) = f (a) + f (a)(x − a) + f 00 (a)
2 2 Z x (x − a) + f 000 (t) a Exercises:
I Follow this procedure, prove the formula (1) when I Use induction to prove (1) for all n. n = 3. (x − t)2
2 dt Taylor Series
The Taylor series centered at a of a function ∞
X
f (n) (a)
n=0 n! f is (x − a)n . For many, but not all, functions, this series converges to
near a. (To check whether the Taylor series of f f (x) for x converges to f, one use the Taylor remainder formula.)
When the Taylor series does converges to
representation for f in powers of f (x) =
(When a=0 we obtain a x −a ∞
X
f (n) (a)
n=0 f, n! (x − a)n . this is also known as Maclaurin series.) Examples:
x e =
sin x =
(1 + x)k = ∞
X
xk
k=0
∞
X
k=0
∞
X
j=0 k!
(−1)k x 2k+1
(2k + 1)!
k(k − 1) . . . (k − j + 1) j
x
j! ∀x ∈ (−1, 1). (binomial series)
(4)
SOME EXAMPLES OF BINOMIAL
Taylor series can also be used to compute limit. E.g., sin(x 2 ) − x 2
,
x→0 1 − cos(x 3 )
lim 2 ex − 1
.
x→0 ln(1 + x sin x)
lim ...

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