مواضيع المحاضرة: TAYLOR AND MACLAURIN SERIES
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Lecture no .2-8 by Hussein J. Ab dulHussein

Advanced Calculus II Al Muthanna University, College of Science

1

TAYLOR AND MACLAURIN SERIES

In the last two sections we used the fact that within its interval of convergence, the function
f (x)= ∑ ak (x− x0)k

k=0

Is differentiable and integrable In this section we look more closely at the coefficients a nd show that
they can be represented in terms of derivatives of the function f .We begin with the case and assume
that so that theorem on power series differentiation applies We have
f (x)= ∑ ak xk

k=0
= a0+ a1x+ a2x2+ anxn+ ⋯ . (1)
And clearly,
f(0)= a0+ 0+ 0+ ⋯ + 0+ ⋯ = a0 (2)
If we differentiate (1), we obtain
f′(x)= ∑ ak xk−1

k−1
= a1+ 2a2x+ a3x2+ ⋯ + na nxn−2+ ⋯ (3)
And f′(0)= a1 (4)
Continuing to differentiate, w e obtain
f′′(x)= ∑ k(k− 1) akxk−2

k=1

= 2a2+ 3∙2a3x+ 4∙3a4x2+ ⋯ + n(n− 1)na nxn−2+ ⋯

And f′′(0)= 2a 2 Or
a2= f′′(0)
2 = f′′(0)
2! (5)
Similarly, f′′′(x)= ∑ k(k− 1)(k− 2) ak xk−3 ∞k=3
= 3∙2a3x+ 4∙3a4x+ 5∙4∙3a5x2+ ⋯ + n(n− 1)(n− 2)anxn−3+ ⋯
And f′′′(0)= 3∙2a3 , Or

Lecture no .2-8 by Hussein J. Ab dulHussein

Advanced Calculus II Al Muthanna University, College of Science

2

a3= f′′′(0)

3∙2 = f′′′(0)
3! (6)
It is not difficult to see that this pattern continues and that for every positive integer n
an= f(n)(0)
n! (7)
For n = 0 we use the convention 0!= 1 and f(0)(x)= f(x) Then formula (7) holds for
every n , and we have the following:
If f (x)= ∑ ak xk ∞k=0 then f(x)= ∑ f(k)(0)
k!
∞k=0 xk
= f(0)+ f′(0)x+ f′′(0)x2
2!+ ⋯ f(n)(0)xn
n!+ ⋯ (8)
for every x in the interval of convergence
In the general case, if
f (x)= ∑ ak( x− x0)k

k=0

=a0+ a1(x− x0)+ a2(x− x0)2+ ⋯ + an(x− x0)n+ ⋯ . (9)

Then f(x0)= a0 .
And differentiating as before, we find that
an= f(n)(x0)
n! (10 )
Thus we have the following If

f (x)= ∑ ak( x− x0)k


k=0

Then f (x)= ∑ f(k) (x0)

k! ( x− x0)k ∞k=0
= f(x0)+ f′(x0)(x− x0)+ f′′(x0)( x− x0)2
2! + ⋯ + f(n)(x0) ( x− x0)n
n! + ⋯ (11)

Lecture no .2-8 by Hussein J. Ab dulHussein

Advanced Calculus II Al Muthanna University, College of Science

3

for every x in the in terval of convergen ce

Definition 1 TAYLOR AND MACLAURIN SERIES The series in (11) is called the Taylor
series of the function f at x0 The special case x0= 0 in (8) is called a Maclaurin series ‡ We
see that the first n terms of the Taylor series of a function a re simply the Taylor polynomial
described in Section 13 .1.
WARNING: We have shown here that if f (x)= ∑ ak( x− x0)k ∞k=0 . then f is infinitely
differentiable (i ∙e∙.f has derivatives of all orders) and that the series for f is the Taylor
series or Maclaurin series if x0= 0) of f What we have not shown is that if f is infinit ely
differentiable at x0. then f has Taylor series expansion at x0 In general , this last
statement is false, as we will see in Example 3 on page 749
EXAMPLE 1 Find the Maclaurin series for ex
Solution If f (x)= ex.then f (0)= f′(0)= ⋯ = f (k)(0)= 1.and ́
ex= ∑ xk
k!

k=0
= 1+ x+ x2
2!+ x3
3!+ ⋯ + xn
n!+ ⋯ (12 )

What this example shows is that if has a Macla urin series expansion, then the series must be the
series (12) It does not show that actually does have such a series expansion To prove that the
series in (12) is really equal to we differentiate, as in Example 14.9.5, and use the fact that the
only conti nuous function that satisfies
f′(x)= f (x). f (0)= 1
Is the function ex
EXAMPLE 2 Assuming that the function can be written as a Maclaurin series, find that series
Solution If f (x)= cos x . then f (0)= 1. f′(0)= 0.f′′(0)= −1 f′′′(0)= 0.
f(4)(0)= 1.and so on . so that if cos x= ∑ akxk ∞k=0 .
Then cos x= f(0)+ f′(0)+ f′′(0)x2
2! + f′′′(0)x3
3! + f(4)(0)x4
4! + ⋯ .
Or
cos x= 1− x2
2!+ x4
4!− x6
6!+ ⋯ = ∑ (−1)kx2k
(2k )!

k=0
(13 )

Lecture no .2-8 by Hussein J. Ab dulHussein

Advanced Calculus II Al Muthanna University, College of Science

4

NOTE Again, this does not prove that the equality in (13) is correct It only shows that if has a
McLaurin expansion, then the expansion must be given by (13) We w ill show that has a
McLaurin series in Example 5
EXAMPLE 3 Let
f (x)= {e−1/x2. if x≠ 0
0. if x= 0
Find a Maclaurin expansion for f if one exists
Solution. First, we note that since limx→∞e−1/x2= 0.fis continuous Now recall from Example
12.3.5, that limx→∞xae−bx = 0.if b> 0 Let y= 1
x2 Y Then as x→ 0.y→ ∞ Also 1
xn =
(1
x2)n/2 so that limx→0 (e−1x2
xn) = limy→∞ yn/2e−y= 0 by Example 12.3.5.
Now for as so that is continuous at 0 Similarly, which also approaches 0 as by the limit result
ab ove In fact, every derivative of f is continuous and for every n , Thus f is infinitely
differentiable, and if it had a Maclaurin series that represented the function, then we would have
f (x)= f(0)+ f′(0)+ f′′(0)x2
2!+ ⋯
But f(0)+ f′(0)+ f′′(0)= ⋯ = 0 . so that the Maclaurin series would be the zero series But
since f is obviously not the zero function, we can only conclude that there is no Maclaurin series
that represents f at any point other than 0
Example 3 illustrates that infinite differentiability is not sufficient to guarantee that a given
function can be represented by its Taylor series Something more is needed
Definition 2 ANALYTIC FUNCTION We say that a function f is analy tic at x0 if f can be
represented by a Taylor series in some neighborhood of x0
We see that the function and are analytic at 0, while the function
f (x)= {e−1/x2. if x≠ 0
0. if x= 0
Is not, A condition that guarantees analytic of an infinitely differentiable function is given below
Theorem 1. Suppose that the function f has continuous derivatives of all orders in a neighborhood
N(x0) of t he number x0.Then f is analytic at x0 if and only if

Lecture no .2-8 by Hussein J. Ab dulHussein

Advanced Calculus II Al Muthanna University, College of Science

5

limn→∞Rn(x)= limn→∞

f(n+1)(Cn)
(n+ 1)! (x− x0)n+1= 0 (14 )
For every x in N (x0) where Cn is between x0 and x
REMARK The expression between the equal sings in (14) is simply the remainder term given
by Taylor 's theo rem
Proof :The hypotheses of Taylor 's theorem apply, so that can write, for any n,
f(x)= Pn(x)+ f(x)= Rn(x). (15 )
Where Pn(x) is the nth degree Taylor Polynomial for f To show that f is analytic , we
must show that
limn→∞Pn(x)= f (x) (16 )
For every x in N(x0) But if x is in N(x0) we obtain , from (14)and (15) ,
limn→∞Pn(x)= limn→∞[f (x)− Rn(x)]= f (x)− limn→∞Rn(x)= f (x)− 0= f (x)
Conversely , if is analytic, then f (x)= limn→∞Pn(x)so Rn(x)→ 0 as n → ∞
EXAMPLE 4 If f (x)= ex then f(n)(x)= ex.and
0< Cn <| x |
limn→∞|f(n+1)(Cn)
(n+ 1)! (x− x0)n+1| = limn→∞
eCn|x|(n+1)
(n+ 1)! ≤ e|x |limn→∞
|x|(n+1)
(n+ 1)!→ 0 .
Since| x|(n+1)
(n+1)! is the (n+ 2) and term in the conve rging power series ∑| x|k ∞k=0 /k! and the terms in
a converging power series → 0 , Since this result is true for any x∈ ℝ we may take N =
(−∞ .∞ ) to conclude that the series (12) is valid for every real number x
EXAMPLE 5 Let Sin ce all derivatives of cos x are equal to ± sin x or ± cos x . we see
that| f(n+1)(Cn)|≤ 1 Then for x0= 0.|Rn(x)|≤| x|(n+1)/(n+ 1)! which → 0 as n → ∞ .so that
the series(13) is also valid for every real number x
EXAMPLE 6 it is evident that for the function in Example 3, Rn(x)↛ 0 if x ≠ 0 This follows
from the fact that Rn(x)= f (x)− Pn(x)= e−1x2− 0= e−1x2≠ 0 if x≠ 0
EXAMPLE 7 Find the Taylor expansion for f (x)= In x at x= 1

Lecture no .2-8 by Hussein J. Ab dulHussein

Advanced Calculus II Al Muthanna University, College of Science

6

Solution Since f′(x)= 1

x.f′′(x)= − 1
x2.f′′(x)= 2
x3. f(4)(x)= − 6
x4 .⋯ .
f(n)(x)= (−1)n+1(n− 1)!
xn . we find that f(1)= 0.f′(1)= 1.f′′(1)= −1.f′′′(1)= 2 .
f(4)(1)= −6.⋯ . f(n)(1)= (−1)n+1(n+ 1)!Then wherever valid .
In x= ∑ f(k)(1)(k− 1)k
k!

k=0
(17 )
we showed that
In (1+ x)= ∑ (−1)k+1xk
k + Rn+1(x).
n+1
k=1
(18 )
Where Rn+1(x)→ 0 as n→ ∞ whenever −1< x≤ 1 From (18) we see that
In u= In [ 1+ (u− 1)]∑ (−1)k+1(u− 1)k
k + Rn+1(u− 1).
n+1
k=1
(19 )
Where Rn+1(u− 1)→ 0 as n → ∞ whenever −1< x≤ u− 1≤ 1.or 0 < u≤ 2 But this
implies that the series (17) converges to In (x) for 0 < x≤ 2 When x= 2 . we obtain from (17),
In 2= 1− 1
2+ 1
3− 1
4+ 1
5− ⋯ = ∑ (−1)k+1
k

k=1
. (20 )
EXAMPLE 8 Find a Taylor series for f (x)= sin x at x= π
3
Solution Here f (π
3)= √3
2.f′(π
3)= 1
2.f′′(π
3)= − √3
2.f′′′ (π
3)= − 1
2.
and so on, so that

sin x= √3

2 + 1
2(x− π
3)− √3
2
[x− (π
3)]
2
2! − 1
2
[x− (π
3)]
3
3! + √3
2
[x− (π
3)]
4
4! + ⋯
The proof that this series is valid for every real number x is similar to the proof in Example 5 and
is therefore omitted

Lecture no .2-8 by Hussein J. Ab dulHussein

Advanced Calculus II Al Muthanna University, College of Science

7

We provide he re a list of useful Maclaurin series:

ex= ∑ xk
k!

k=0
= 1+ x+ x2
2!+ x3
3!+ ⋯ (21 )
cos x= ∑ (−1)kx2k
(2k )!

k=0
= 1− x2
2!+ x4
4!− x6
6!+ ⋯ (22 )
sin x= ∑ (−1)kx2k+1
(2k + 1)! = x− x3
3!+ x5
5!− x7
7!

k=0
+ ⋯ (23 )
cosh x= ∑ x2k
(2k )!

k=0
= 1+ x2
2!+ x4
4!− x6
6!+ ⋯ (24 )
sinh x= ∑ x2k+1
(2k + 1)!= x− x3
3!+ x5
5!− x7
7!

k=0
+ ⋯ (25 )
BINOMIAL SERIES
We close this section by deriving another series that is quite useful. Let f (x)= (1+ x)r. where r
is a real number not equal to an inte ger. We have
f′(x)= (1+ x)r−1.
f′′(x)= r(r− 1)(1+ x)r−2.
f′′′(x)= r(r− 1)(r− 2)(1+ x)r−3.

f(n)(x)= r(r− 1)(r− 2)⋯ (r− n+ 1)(1+ x)r−n
Not that since r is not an integer , r− n is never equal to 0 , all derivatives exist and are
nonze ro as long as x≠ −1 Then
f(0)= 1 .
f′(0)= r .

Lecture no .2-8 by Hussein J. Ab dulHussein

Advanced Calculus II Al Muthanna University, College of Science

8

f′′(0)= r(r− 1).


f(n)(0)= r(r− 1)⋯ (r− n+ 1) .
And we can write
(1+ x)r= 1+ rx + r(r− 1)
2! x2+ r(r− 1)(r− 2)
3! x3+ ⋯
+ r(r− 1)⋯ (r− n+ 1)
n! xn+ ⋯ (26 )
= 1+ ∑ r(r− 1)⋯ (r− k+ 1)
k! xk

k=0

The series (26) is called the binomial series

PROBLMS
1∙ Find the Maclaurin series sin x for all real x
2. Find the Taylor series for ex at 1
3. Find the Maclaurin series for e−x
4. Find the Taylor series f or cos x at π
4
5.Find the Taylor series for sinh x at In 2
6. Find the Maclaurin series for eax .a is real
7. Find the Maclaurin series for x ex
8. Find the Maclaurin series for x2 e−x2
9. Find the Maclaurin series for (sin x)/x
10 . Find the Taylor series for ex at x= −1
11 . Find the Maclaurin series for sin 2 x[Hint ∶sin 2 x= (1− cos 2x )/2]
12. Find the Taylor series of √x at x= 4. What is its radius of convergence?
13. Find the Maclaurin series of sin −1xWhat is its radius of convergence?
Hint :Expand 1
√1−x2 and in

Lecture no .2-8 by Hussein J. Ab dulHussein

Advanced Calculus II Al Muthanna University, College of Science

9

14. Use the Maclaurin series for sin x to obtain the Maclaurin series for sin x2

15 . Find the Maclaurin series for cos x2
16 . Use Equation (26) to find a power series representation for √1+ x 4
17 . Use the result of problem 26 to find a power series representation for √1+ x3 4


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