TAYLOR AND MACLAURIN SERIES
345
39.40
Estimate
dx to within two-decimal-place accuracy.
Hence,
For
This is an alternating series. Therefore, we seek n for which IIn 2" <0.005, 200sn 2",
n s 4. Hence, we may use
39.41
Approximate
dx to within two-decimal-place accuracy.
Hence,
and
Since this is an alternating series, we must find n such that
Hence, we use
39.42
Find the Maclaurin series for
From
we obtain
and so
39.43
If
From the Maclaurin expansion found in Problem 39.42,
(because 36 is divisible
by 3); hence, /
<36) (0) = 36!.
39.44
Prove that e is irrational.
Hence,
By the alternating series theorem,
for k 2 2. So
Therefore,
But
that
is an integer.
If e were rational, then k could be chosen large enough so
would be an integer.
Hence, for k large enough and suitably chosen,
would be an integer strictly between 0 and |, which is impossible.
39.45
Find the Taylor series for cos x about -nil.
Since sin x =
we have
39.46
Find the Maclaurin series for In (2 + jc).
ln(2 + x) = ln[2(l + jt/2)] = In2 + ln(l + .x/2). But ln(l + *) =
Hence, In
Thus, In (2 + x) = In 2 +
find/(36)(0).
345
39.40
Estimate
dx to within two-decimal-place accuracy.
Hence,
For
This is an alternating series. Therefore, we seek n for which IIn 2" <0.005, 200sn 2",
n s 4. Hence, we may use
39.41
Approximate
dx to within two-decimal-place accuracy.
Hence,
and
Since this is an alternating series, we must find n such that
Hence, we use
39.42
Find the Maclaurin series for
From
we obtain
and so
39.43
If
From the Maclaurin expansion found in Problem 39.42,
(because 36 is divisible
by 3); hence, /
<36) (0) = 36!.
39.44
Prove that e is irrational.
Hence,
By the alternating series theorem,
for k 2 2. So
Therefore,
But
that
is an integer.
If e were rational, then k could be chosen large enough so
would be an integer.
Hence, for k large enough and suitably chosen,
would be an integer strictly between 0 and |, which is impossible.
39.45
Find the Taylor series for cos x about -nil.
Since sin x =
we have
39.46
Find the Maclaurin series for In (2 + jc).
ln(2 + x) = ln[2(l + jt/2)] = In2 + ln(l + .x/2). But ln(l + *) =
Hence, In
Thus, In (2 + x) = In 2 +
find/(36)(0).
