James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Problems Plus
Before you look at the solution of the
example, cover it up and first try to
solve the problem yourself.
sx 1 2d
Example Find the sum of the series ò
n
n−0 sn 1 3d! .
SolutION The problem-solving principle that is relevant here is recognizing something
familiar. Does the given series look anything like a series that we already know?
Well, it does have some ingredients in common with the Maclaurin series for the exponential
function:
e x − ò
n−0
x n
n! − 1 1 x 1 x 2
2! 1 x 3
3! 1 ∙ ∙ ∙
We can make this series look more like our given series by replacing x by x 1 2:
sx 1 2d
e x12 − ò
n
n−0 n!
− 1 1 sx 1 2d 1
sx 1 2d2
2!
1
sx 1 2d3
3!
1 ∙ ∙ ∙
But here the exponent in the numerator matches the number in the denominator
whose factorial is taken. To make that happen in the given series, let’s multiply and
divide by sx 1 2d 3 :
sx 1 2d
ò
n
n−0 sn 1 3d! − 1 sx 1 2d
ò
n13
sx 1 2d 3 n−0 sn 1 3d!
sx 1 2d3
− sx 1 2d
23F 1
3!
sx 1 2d4
4!
1 ∙ ∙ ∙G
We see that the series between brackets is just the series for e x12 with the first three
terms missing. So
sx 1 2d
ò
n
n−0 sn 1 3d! − sx 1 2d23Fe sx 1 x12 2d2
2 1 2 sx 1 2d 2 G n
2!
Problems
1. If f sxd − sinsx 3 d, find f s15d s0d.
2. A function f is defined by
8
P¢
4
P£
2
P
1
A 1 P¡
Where is f continuous?
3. (a) Show that tan 1 2 x − cot 1 2 x 2 2 cot x.
(b) Find the sum of the series
f sxd − lim
n l `
x 2n 2 1
x 2n 1 1
ò
n−1
1
2 n tan x 2 n
P∞
FIGURE for problem 4
4. Let hP nj be a sequence of points determined as in the figure. Thus| AP1 | − 1,
| PnPn11 | − 2n21 , and angle AP nP n11 is a right angle. Find lim n l ` /P n AP n11.
787
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