Corollary to Taylor’s theorem
We will only focus on Taylor series centered at x = 0:
P
f (n) (0) n
Def: T0 (f (x)) = ∞
n=0
n! x
(n)
P
Taylor’s theorem: f (x) = kn=0 f n!(0) x n + Errk with
Errk =
f (k+1) (c) k+1
(k+1)! x
for some c between 0 and x
P
f (n) (0) n
Corollary (to Taylor’s theorem): f (x) = ∞
n=0
n! x if
limk→∞ ErrP
k = 0 for all c between 0 and x.
1 n
< x < ∞ because
Eg: e x = ∞
n=0 n! x for −∞
ec
limk→∞ Errk = limk→∞ (k+1)!
x k+1 = 0
Non-Eg for the Corollory: See example 10.8.3.
(limk→∞ Errk 6= 0)
Question What does a series converge to? (Memorize series
from
Link)
P∞ here:
1
Eg Pn=0 n! computes what?
∞ 2n
Eg
n! =?
Pn=0
∞ (−1)n+1
Eg
=?
n=1
n
P∞
(−1)n
Eg
n=0 2n+1 =?
Useful facts from 10.7
Ans: e 1 , e 2 , ln 2, arctan 1 = π4
Compose: T0 (f (p(x))) = Tp(0) (f (u))|u=p(x) for −R < x < R
P
P∞ 1 2n
2
1 n
Eg: T0 (e x ) = T0 (e u )|u=x 2 = ∞
for
n=0 n! u |u=x 2 =
n=0 n! x
P∞ 1
2
1
−R < x < R. By the corollary, e = n=0 n! .
df
d
Diff: T0 ( dx
) = dx
T0 (f (x)) for −R <x < R P
P∞
1
1
n
d
n−1
T0 ( 1−x
) = T0 (1−x)
Eg:
so ∞
is dx
2
n=0 2n−1
n=0 nx
1
computes (1−x)
2 |x= 1 = 4
´
´ 2
Integ: fdx = T0 (f (x))dx for −R < x < R
´1 2
´1
´ 1 P x 2n
P∞
2
Eg: 0 e x dx = 0 T0 (e x )dx = 0
n=0
n! dx =
P∞
P∞
P∞
12n+1
02n+1
1
n=0 n!(2n+1) −
n=0 n!(2n+1) =
n=0 n!(2n+1)
x 2n+1 1
n!(2n+1) ]0
=
Less useful
Multiplication/Division T0 (f (x)g (x)) = T0 (f (x))T0 (g (x)) for
−R < x < R
Eg T0 (tan x) can be obtained by long division of
T0 (sin x) by T0 (cos x).