SECTION 9.6: Taylor and Maclaurin Series

SECTION 9.6: Taylor and Maclaurin Series

Exercise 1

We make use of the Maclaurin series of e^x:

e^x=

∞

X

n=0

1

n!xⁿ = 1 + x +1 2x²+ 1

3!x³+ · · · . The series above holds for all real number x.

Substituting 3x for x, and then multiply by e, we get:

e^3x+1= e + 3ex +3²e

2 x²+3³e

3! x³+ · · · =

∞

X

n=0

3ⁿe n! xⁿ.

The resulting Maclaurin series is valid for any real x since the original Maclaurin series holds for any real x.

Exercise 3

Solution 1:

Using the relations between trigonometric functions and the exponential function, we obtain:

sin x − π

4

= 1

2i(e^i(x−π4)− e^{−i(x−π4)})

= 1 2i

√ 2 2 −

√ 2 2 i

! e^ix− 1

2i √

2 2 +

√ 2 2 i

! e^−ix

= −

√2 4 −

√2 4 i

! e^ix−

√2 4 −

√2 4 i

! e^−ix

=

∞

X

n=0

−

√2 4 −

√2 4 i

!iⁿ n!xⁿ−

∞

X

n=0

√2 4 −

√2 4 i

!(−i)ⁿ n! xⁿ

=

∞

X

k=0

−√ 2 2

i^2k (2k)!x^2k+

∞

X

k=0

−√ 2i 2

i^2k+1 (2k + 1)!x^2k+1

=

√2 2

∞

X

k=0



−(−1)^k

(2k)!x^2k+ (−1)^k (2k + 1)!x^2k+1



(The fifth equality holds since i^k= (−i)^k when k is even; and i^k = −(−i)^k when k is odd).

The Maclaurin series holds for all real value x since only exponential func- tions are involved, and they are valid for all real value.

Solution 2:

Observe that sin(x − ^π₄) =

√ 2

2 (sin x − cos x), we immediately get:

sin x − π

4



=

√2 2

∞

X

k=0



−(−1)^k

(2k)!x^2k+ (−1)^k (2k + 1)!x^2k+1

 .

Exercise 5

Substitute ^x₃ for x in the Maclaurin series of sin x, then multiply by x², we obtain:

x²sinx

3 = x² x 3 − 1

3!

x³ 3³ + 1

5!

x⁵ 3⁵ − · · ·



=

∞

X

n=0

(−1)ⁿ

3²ⁿ⁺¹(2n + 1)!x²ⁿ⁺³. The resulting Maclaurin series is valid for every real x since the Maclaurin series of the sin x holds for every real x.

Exercise 8

Substitute 5x² for x in the Maclaurin series of tan⁻¹x, we obtain:

tan⁻¹(5x²) = 5x²−(5x²)³

3 +(5x²)⁵

5 − · · · =

∞

X

n=0

(−1)ⁿ5²ⁿ⁺¹ 2n + 1 x⁴ⁿ⁺². As we know the Maclaurin series of tan⁻¹x converges for all −1 ≤ x ≤ 1, the Maclaurin series of tan⁻¹(5x²) converges if and only if −1 ≤ 5x² ≤ 1, equivalently, if 0 ≤ x²≤ ¹₅ ⇐⇒ −

√5 5 ≤ x ≤

√5 5 .

Exercise 12

Substitute 2x² for x in the Maclaurin series of e^x, we have:

e^2x2= 1 + 2x²+(2x²)²

2! +(2x²)³

3! + · · · =

∞

X

n=0

2ⁿ n!x²ⁿ Therefore,

e^2x2− 1 x² = 1

x²

∞

X

n=1

2ⁿ n!x²ⁿ=

∞

X

n=1

2ⁿ n!x²ⁿ⁻².

Note that the function f (x) = ^e2x2_x2⁻¹ is not defined at x = 0, but it has a limit at x = 0 (this can be confirmed by examining the constant term of the Maclaurin series of f (x) or by using the l’Hˆopital’s rule):

x→0lim

e^2x2− 1 x² = lim

u→0

e^2u− 1 u = lim

u→0

2e^2u 1 = 2.

If we define f (0) = 2, the Maclaurin series converges for any real value x, as the Maclaurin series of e^x does.

Exercise 17

Solution 1:

Compute the nth derivative of f (x), we have f (π) = −1 and:

f⁰(x) = − sin x, f⁰(π) = 0 f⁰⁰(x) = − cos x, f⁰⁰(π) = 1 f⁽³⁾(x) = sin x, f⁽³⁾(π) = 0 f⁽⁴⁾(x) = cos x, f⁽⁴⁾(π) = −1 f⁽⁵⁾(x) = − sin x, f⁽⁵⁾(π) = 0

... ...

Thus, the Taylor series for f (x) = cos x in powers of x − π is

cos x = −1 + 1

2!(x − π)²− 1

4!(x − π)⁴+ · · · =

∞

X

n=0

(−1)ⁿ⁺¹

(2n)! (x − π)²ⁿ. The series converges for all real value x by the ratio test:

n→∞lim      

(−1)ⁿ⁺²

(2(n+1))!x²⁽ⁿ⁺¹⁾

(−1)ⁿ⁺¹ (2n)! x²ⁿ

= lim

n→∞

(2n)!

(2n + 2)!|x|²= lim

n→∞

|x|²

(2n + 1)(2n + 2) = 0.

Solution 2:

Let u = x − π. The by the Maclaurin series of cos u we have:

cos x = − cos u = −1 +u² 2! −u⁴

4! + · · · =

∞

X

n=0

(−1)ⁿ⁺¹u²ⁿ (2n)!

=

∞

X

n=0

(−1)ⁿ⁺¹

(2n)! (x − π)²ⁿ

Exercise 19

Let t = ^x−2₄ . Then we have:

ln(2 + x) = ln(4 + (x − 2)) = ln 4 + ln



1 + x − 2 4



= ln 4 + ln(1 + t).

So we can use the Maclaurin series of ln(1 + t) to compute the required Taylor series as follows:

ln(2 + x) = ln 4 + ln(1 + t)

= 2 ln 2 + t −t² 2 +t³

3 − · · ·

= 2 ln 2 +

∞

X

n=1

(−1)ⁿ⁻¹ n tⁿ

= 2 ln 2 +

∞

X

n=1

(−1)ⁿ⁻¹

n4ⁿ (x − 2)ⁿ

Since the Maclaurin series of ln(1 + t) converges when −1 ≤ t ≤ 1, and x = 4t + 2, the resulting Taylor series of ln(2 + x) converges when −2 ≤ x ≤ 6.

Exercise 22

We use a trigonometric identity to express cos²x in terms of cos 2x, and then use addition formula for cosine to compute the Taylor series of cos 2x at

π

4 = 2 ×^π₈.

cos²x = 1 + cos 2x 2

= 1 2+1

2cos 2x −π

4 +π 4



= 1

2+ cos 2x −π

4

 cosπ

4 − sin 2x −π

4

 sinπ

4

= 1 2+

√2 2

 1 − 2²

2!

 x − π

8

2

+2⁴ 4!

 x −π

8

4

− · · ·



−

√2 2

 2

x − π 8

−2³ 3!

 x − π

8

3

+2⁵ 5!

 x − π

8

5

− · · ·



= 1 +√ 2 2 +

√ 2 2

∞

X

n=1

(−1)ⁿ

 2²ⁿ⁻¹ (2n − 1)!

 x −π

8

²ⁿ⁻¹ + 2²ⁿ

(2n)!

 x −π

8

²ⁿ .

The resulting Taylor series is valid for every real x since the Maclaurin series of the sine and cosine function both hold for every real x.

Exercise 24

We have

x

1 + x = 1 − 1

1 + x = 1 − 1

2 + (x − 1) = 1 −1 2

1 1 + ^x−1₂

= 1 − 1 2

"

1 − x − 1

2 + x − 1 2

²

− · · ·

#

= 1 2 +

∞

X

n=1



−1 2

n+1

(x − 1)ⁿ.

From the resulting Taylor series, we immediately see that the Taylor series converges when |x − 1| < 2 ⇐⇒ −1 < x < 3.

Exercise 28

We first need to compute the first three nonzero terms (excluding the con- stant term) in the Maclaurin series for sec x, then we can compute the first three nonzero terms in the Maclaurin series for sec x tan x, since (sec x)⁰= sec x tan x.

sec x = (cos x)⁻¹= 1

1 − ¹₂x²+₂₄¹x⁴−₇₂₀¹ x⁶+ · · ·

= 1 + 1 2x²− 1

24x⁴+ 1

720x⁶− · · ·



+ 1 2x²− 1

24x⁴+ 1

720x⁶− · · ·

2

+ 1 2x²− 1

24x⁴+ · · ·

3

+ · · ·

= 1 +1 2x²+



−1 24+1

4

 x⁴+

 1

720 − 2 2 × 24+1

8



x⁶+ · · ·

= 1 +1 2x²+ 5

24x⁴+ 61

720x⁶+ · · · . Hence,

sec x tan x = (sec x)⁰ = x +5

6x³+ 61

120x⁵+ · · · .

Exercise 29

Using the Maclaurin series of tan⁻¹u and e^x, we have:

tan⁻¹(e^x− 1) = tan⁻¹

 x + 1

2x²+1

6x³+ · · ·



=

 x + 1

2x²+1

6x³+ · · ·



−1 3

 x +1

2x²+1

6x³+ · · ·

3

+ · · ·

= x + 1

2x²+ 1 6 −1

3× 1



x³+ · · · = x +1 2x²−1

6x³+ · · · .