294 CHAPTER 5 Series and Differential Equations

Example 6. (A handy trick)In the above series we may substitute
−x^2 forxand get the Maclaurin series fore−x
2
:

e−x

2

∼ 1 −x^2 +

x^4

2!

−

x^6

3!

+···=

∑∞

n=0

(−1)n

x^2 n

n!

.

Note how easy this is compared to having to calculate d
n
dxn(e

−x^2 ) (where

the chain and product rules very much complicate things!).

Example 7. (A handy trick)Let’s find the Maclaurin series expan-

sion of the functionf(x) =

sinx

x

. In this case, we certainly wouldn’t

want to compute successive derivatives of the quotient

sinx

x

. However,

remembering the Maclaurin series expansion of sinxand then dividing
byx will accomplish the same thing much more easily; the resulting
series is

sinx

x

∼ 1 −

x^2

3!

+

x^4

5!

−···=

∑∞

n=0

(−1)n

x^2 n

(2n+ 1)!

.

Example 8. The Taylor series expansion of cosx about x =

π

2

.

We have, where f(x) = cosx, that f

(π

2

)

= cos

(π

2

)

= 0, f′

(π

2

)

−sin

(π

2

)

= − 1 , f′′

(π

2

)

= −cos

(π

2

)

= 0, f′′′

(π

2

)

= sin

(π

2

)
= 1, after
which point the cycle repeats. Therefore, the Taylor series is

cosx ∼ −

(

x−π 2

)

+

(

x−π 2

) 3

3!

−

(

x−π 2

) 5

5!

+···=

∑∞

n=1

(−1)n

(

x−π 2

) 2 n− 1

(2n−1)!

.

Example 9. (A handy trick)We know that

1 +x+x^2 +x^3 +···=

∑

n=0

xn=

1

1 −x

valid for|x|< 1.

we can get further valid sums by differentiating the above:

1 + 2x+ 3x^2 + 4x^3 +··· =

∑∞

n=0

(n+ 1)xn

=

d

dx

(

1

1 −x

)

=

1

(1−x)^2

, valid for|x|< 1.