Thomas Calculus 13th [Solutions]

Section 10.1 Sequences 711
133. a2k
L given an 0 there corresponds an N 1 such that 2 k N1 a2k
L . Similarly,
2 1 .
a2k
1 L k N2 a2k
1 L Let 1 2
is even or odd, and hence a L.
n
N max{ N , N }. Then n N an
L whether n
134. Assume a n 0. This implies that given an 0 there corresponds an N such that n N a n 0
an an an 0 a n 0. On the other hand, assume a n 0. This implies that given
an 0 there corresponds an N such that for n N,
a 0 a a a 0
a n
0.
n n n n
135. (a)
2
2 2
2
2
a
xn xn a
xn
n
n
x n
2xn 2xn 2xn
2
2
x a x a
f ( x) x a f ( x) 2x xn 1 xn xn
1
(b) x1 2, x2 1.75, x3 1.732142857, x4 1.73205081, x 5 1.732050808; we are finding the positive
2
2
number where x 3 0; that is, where x 3, x 0, or where x 3.
136. x1 1, x2 1 cos(1) 1.540302306, x3 1.540302306 cos(1 cos(1)) 1.570791601,
x 4 1.570791601 cos(1.570791601) 1.570796327 to 9 decimal places. After a few steps, the arc
2
x n 1 and line segment cos x n 1 are nearly the same as the quarter circle.
137-148. Example CAS Commands:
Mathematica: (sequence functions may vary):
Clear[a, n]
_ a[n ]:
n
1/ n
first25 Table[N[a[n]],{n, 1, 25}]
Limit[a[n], n 8]
The last command (Limit) will not always work in Mathematica. You could also explore the limit by enlarging
your table to more than the first 25 values.
If you know the limit (1 in the above example), to determine how far to go to have all further terms within 0.01
of the limit, do the following.
Clear[minN, lim]
lim 1
Do[{diff Abs[a[n] lim], If[diff .01, {minN n, Abort[]}]}, {n, 2, 1000}]
minN
For sequences that are given recursively, the following code is suggested. The portion of the command
a[n _]: a[n] stores the elements of the sequence and helps to streamline computation.
Clear[a, n]
a[1] 1;
a[n _]: a[n] a[n 1] (1/5)
n 1
first25 Table[N[a[n]],{n, 1, 25}]
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2014 Pearson Education, Inc.