Commonly Used Taylor Series
series 1 1−x
when is valid/true
=
1 + x + x2 + x3 + x4 + . . .
=
note this is the geometric series. just think of x as r
∞
xn
x ∈ ( −1, 1)
n=0
e
x
2
=
1 + x + ∞
=
x
=
1 −
=
4
x
+
2!
(−1)n
=
x −
=
x3
3!
x
+
(−1)
(n−1)
n=1
=
x −
=
x2
2
+
∞
(−1)
(n−1)
n=1
x
=
x −
x3
3
+
∞
=
n=1
1 1 1 2! + 3! + 4! + . . . ∞
+ ...
4!
e(17x) =
4!
6
(17x) n=0 n!
∞
n
=
17
n
n
x n!
n=0
−
x
x
+
6!
8!
(−1)(n−1)
x5
− ...
5!
x ∈
−
x7
7!
x9
+
x2n−1 or = (2n − 1)! x3
3
−
xn or = n x5
5
−
x4
4
+
∞
(−1)n
n=0
x5
5
(−1)n+1
n=1
x7
7
x2n−1 or = 2n − 1
+
x9
9
xn n
− ...
∞
(−1)n
n=0
x2n+1 (2n + 1)!
− ...
∞
x 2n+1 2n + 1
1
R
note y = sin x is an odd function (i.e., sin(− sin(−x) = − sin(x)) and the taylor seris of y = sin x has only odd powers.
− ...
9!
R
note y = cos x is an even function (i.e., cos(− cos(−x) = + cos(x)) and the taylor seris of y = cos x has only even powers.
8
x2n (2n)!
∞
tan
+
3!
x ∈
n=0
−1
x
n!
∞
ln(1 + x)
x
n
2
sin x
+
2!
so: e = 1 + 1 +
4
x n=0
cos x
3
x ∈
R
quest question ion:: is y = ln(1 + x) even, odd, or neither? x ∈ ( −1, 1]
quest question ion:: is y = arctan( arctan(x) even, odd, or neither? x ∈ [ −1, 1]
Math 142
Taylor/Maclaurin Polynomials and Series
Prof. Girardi
Fix an interval I in the real line (e.g., I might be ( −17, 19)) and let x0 be a point in I , i.e., x0 ∈ I .
Next consider a function, whose domain is I , f : I → R (n)
and whose derivatives f
: I → R exist on the interval I for n = 1, 2, 3, . . . , N .
Definition 1. The N th -order Taylor polynomial for y = f (x) at x0 is:
pN (x) = f (x0 ) + f (x0 )(x − x0 ) +
f (x0 )
2!
f (N ) (x0 ) (x − x0 ) + · · · + (x − x0 )N , N ! 2
(open form)
which can also be written as (recall that 0! = 1) f (0) (x0 )
f (1) (x0 )
f (2) (x0 )
f (N ) (x0 ) (x−x0 ) +· · ·+ (x−x0 )N N ! 2
← a finite sum, i.e. the sum stops . + (x−x0 )+ 0! 1! 2! Formula (open form) is in open form. It can also be written in closed form, by using sigma notation, as
pN (x) =
N
pN (x) =
f (
n)
(x0 ) (x − x0 )n . n!
n=0
(closed form)
So y = p N (x) is a polynomial of degree at most N and it has the form N
pN (x) =
cn (x − x0 )n
where the constants
n=0
cn =
f (n) (x0 ) n!
are specially chosen so that derivatives match up at x0 , i.e. the constants cn ’s are chosen so that: pN (x0 ) = f (x0 ) (1)
pN (x0 ) = f (1) (x0 ) (2)
pN (x0 ) = f (2) (x0 )
.. . (N )
pN (x0 ) = f (N ) (x0 ) .
The constant c n is the n th Taylor coefficient of y = f (x) about x 0 . The N th -order Maclaurin polynomial for y = f (x) is just the N th -order Taylor polynomial for y = f (x) at x0 = 0 and so it is N
pN (x) =
f (
n)
n=0
Definition 2.
1
(0) n x . n!
The Taylor series for y = f (x) at x0 is the power series:
P ∞ (x) = f (x0 ) + f (x0 )(x − x0 ) +
f (x0 )
2!
(x − x0 )2 + · · · +
f (n) (x0 ) (x − x0 )n + . . . n!
(open form)
which can also be written as f (0) (x0 ) f (1) (x0 ) f (2) (x0 ) f (n) (x0 ) P ∞ (x) = + (x−x0 )+ (x−x0 )2 +· · ·+ (x−x0 )n +. . . ← the sum keeps on going and going . n! 0! 1! 2! The Taylor series can also be written in closed form, by using sigma notation, as ∞
P ∞ (x) =
f (
n)
n=0
(x0 ) (x − x0 )n . n!
(closed form)
The Maclaurin series for y = f (x) is just the Taylor series for y = f (x) at x0 = 0. 1Here we are assuming that the derivatives y = f (n) (x) exist for each x in the interval I and for each n ∈ N ≡ {1, 2, 3, 4, 5, . . . } .
2
Big Questions 3. For what values of x does the power (a.k.a. Taylor) series ∞
P ∞ (x) =
f (
n)
n=0
(x0 ) (x − x0 )n n!
(1)
converge (usually the Root or Ratio test helps us out with this question). If the power/Taylor series in formula (1) does indeed converge at a point x, does the series converge to what we would want it to converge to, i.e., does f (x) = P ∞ (x) ?
(2)
Question (2) is going to take some thought. Definition 4. The N th -order Remainder term for y = f (x) at x 0 is: def
RN (x) = f (x) − P N (x)
where y = P N (x) is the N th -order Taylor polynomial for y = f (x) at x0 . So f (x) = P N (x) + RN (x)
(3)
that is f (x) ≈ P N (x)
within an error of RN (x) .
We often think of all this as: N
f (x) ≈
f (
n)
n=0
(x0 ) (x − x0 )n n!
← a finite sum, the sum stops at N .
We would LIKE TO HAVE THAT ∞
??
f (x) =
f (
n)
n=0
(x0 ) (x − x0 )n n!
← the sum keeps on going and going .
In other notation:
??
f (x) ≈ P N (x)
f (x) = P ∞ (x) and the question is where y = P ∞(x) is the Taylor series of y = f (x) at x0 . ?? Well, let’s think about what needs to be for f (x) = P ∞(x), i.e., for f to equal to its Taylor series.
Notice 5. Taking the lim N →∞ of both sides in equation (3), we see that ∞
f (x) =
f ( n=0
n)
(x0 ) (x − x0 )n n!
← the sum keeps on going and going .
if and only if lim RN (x) = 0 .
N →∞
Recall 6. limN →∞ RN (x) = 0 if and only if limN →∞ |RN (x)| = 0 . So 7. If lim |RN (x)| = 0
(4)
N →∞
then
∞
f (x) =
f ( n=0
n)
(x0 ) (x − x0 )n . n!
So we basically want to show that (4) holds true.
How to do this? Well, this is where Mr. Taylor comes to the rescue!
2
2According to Mr. Taylor, his Remainder Theorem (see next page) was motivated by coffeehouse conversations about works of Newton on planetary motion and works of Halley (of Halley’s comet ) on roots of polynomials.
3
Taylor’s Remainder Theorem 3
Version 1: for a fixed point x ∈ I and a fixed N ∈ N. There exists c between x and x0 so that RN (x)
def
=
f (N +1) (c) (x − x0 )(N +1) . (N + 1)!
theorem
f (x) − P N (x)
=
(5)
So either x ≤ c ≤ x 0 or x0 ≤ c ≤ x . So we do not know exactly what c is but atleast we know that c is between x and x0 and so c ∈ I . Remark: This is a Big Theorem by Taylor. See the book for the proof. The proof uses the Mean Value Theorem. Note that formula (5) implies that N +1)
|RN (x)|
( f
(c) |x − x0 |(N +1) . (N + 1)!
=
(6)
Version 2: for the whole interval I and a fixed N ∈ N.3 Assume we can find M so that
the maximum of f (
N +1)
i.e.,
(x)
max f (
N +1)
c∈I
Then
|RN (x)| ≤
on the interval I ≤ M ,
(c)
≤ M .
M +1 |x − x0 |N (N + 1)!
for each x ∈ I . Remark: This follows from formula (6). Version 3: for the whole interval I and all N ∈ N. 4 Now assume that we can find a sequence { M N }∞ so that N =1
max f (N +1) (c) ≤ M N c∈I
for each N ∈ N and also so that lim
N →∞
M N +1 = 0 |x − x0 |N (N + 1)!
for each x ∈ I . Then, by formula (7) and the Squeeze Theorem, lim |RN (x)| = 0
N →∞
for each x ∈ I . Thus, by So 7, ∞
f (x) =
f ( n=0
n)
(x0 ) (x − x0 )n n!
for each x ∈ I .
3Here we assume that the ( N + 1)-derivative of y = f (x), i.e. y = f (N +1) (x), exists for each x ∈ I . 4Here we assume that y = f (N ) (x), exists for each x ∈ I and each N ∈ N .
4
(7)
