Calculus Of One Real Variable – By Pheng Kim Ving
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1. Approximations Of Sums Of Series |
In previous sections of this chapter we were concerned
mainly about demonstrating that a series converges or diverges. Other
than some special series like the geometric and telescoping series, we've not
calculated the sum of a series after we've shown
that it converges. For most series, so far we've not developed techniques to
calculate their sums. So at this stage we have to
be happy to simply settle with approximations of these sums. In this section
we're going to calculate approximations of sums of
some series that converge. We'll approximate sums of some convergent series by
their partial sums or by values that involve
their partial sums. We're also going to compute approximations of partial sums
of the harmonic series, which diverges.
Error And Accuracy
Of Approximations
When we try to approximate or estimate the sum s of a convergent series by a value A, it's natural that we want
to know how
accurate the approximate or estimated value A
is, that is, how close the approximate value is to the actual value. The error
of
the approximation of course is the difference s – A between the actual value s
and the approximate value A.
The accuracy is
measured by a bound on the size or magnitude (absolute value) |s
– A| of the error. As a
matter of fact the smaller the size of
the error, the closer A
is to s, and so the more accurate and thus
the better the approximation. Without the knowledge of this
accuracy, the approximation wouldn't be of much use.
If we try to approximate a sum s
by a value A with, for
example, |error| < 0.001 (or |error| < 1/1,000), we have to find A
such that |s – A| < 0.001. This of
course means that A
must be in the interval (s – 0.001, s + 0.001) (|s – A| < 0.001 iff
– 0.001 < s – A < 0.001 iff – 0.001 < A – s
< 0.001 iff s – 0.001 < A < s
+ 0.001).
Number Of
Decimal Digits To Use
Consider this example:
Q = 2.34567...,
A = 2.345a4a5...,
where there's no carrying of 1 from the subtraction 6 – a4 or 6 – (a4 + 1) to add to
the digit 5 in A.
If we use 0 or 1 or 2
decimal digits for A,
(A = 2 or A = 2.3 or A = 2.34) then |E| > 0.001 (|E|
= 0.34567.... or |E| = 0.04567... or |E| =
0.00567...). If we use 3 or more decimal digits (A = 2.345 or A
= 2.345a4 etc), then |E|
< 0.001 (|E| = 0.000d4d5...).
Now consider this example:
Q = 2.34567...,
A = 2.344a4a5...,
where there's a carrying of 1 from the subtraction 6 – a4 or 6 – (a4 + 1) to add to
the 2nd digit 4 in A.
If we use 0 or 1 or 2 or
3 decimal digits for A,
then |E| > 0.001. If
we use 4 or more decimal digits, then |E|
< 0.001.
Because situations as shown in these examples are possible,
we should use at least 4 significant decimal digits for A. Similarly
if Q < A, where |E| = A
– Q and we
consider the subtraction A
– Q.
In general, if |error| < 1/10m,
then the approximate value is accurate to at least m
– 1 decimal places and we should use at
least m + 1 significant decimal digits for
the approximate value.
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2. For Series Whose Convergence Is Determined By The RaT |
Geometric Bound
Here we'll approximate the sums of series whose convergence
is determined by the RaT (Ratio Test). We'll approximate sums
by partial sums.
Example 2.1
It can be shown that:
Notes
Solution
a. Let an =
1/n!. Then:
EOS
Remarks 2.1
d. The convergence of the series is determined by the
RaT. The sum s of the series is approximated by a
partial sum sn. The
bound on the error size is a
geometric series. In general, for series whose convergence is determined by the
RaT, it's
appropriate to approximate its sum
by its partial sum and to try to bound the error size of approximation by a
geometric
series. Such a bound is called a geometric
bound.
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3. For Series Whose Behavior Is Determined By The IT |
Here we'll approximate the sums and partial sums of positive
series whose behavior (convergence or divergence) is determined
by the IT (Integral Test) or the p-series
property.
Interval
Containing Partial Sum Or Sum
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Fig. 3.1 Sum of areas of lower rectangles is less than or equal |
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Fig. 3.2 Area under f over [1, n] is less than or equal to sum of |
(Sum is between integral and integral + first term; see Fig. 3.2.)
Interval
Containing Partial Sum Or Sum
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if the integral and thus the series converge. |
Remark 3.1
Relation [3.3] applies to both convergent and divergent
series, while relation [3.4] of course applies only to convergent series.
Relation [3.3] is useful when we desire to find an interval that has length
equal to the first term of the series and that contains
the partial sum of a divergent series. Relation [3.4] is useful when we desire
to find an interval that has length equal to the
first term of the series and that contains the sum of a convergent series.
Example 3.1
Solution
EOS
Example 3.2
Solution
EOS
Integral Bound
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Fig. 3.3 |
Approximation By sn With Integral Bound On Error
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Remark 3.2
Since the error-size bound in this approximation is an integral, it's called an integral bound.
Example 3.3
Solution
So n must be at least 1,001.
EOS
Term Bound
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Fig. 3.4 |
Approximation
With Term Bound On Error
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Remarks 3.3
a. Approximation [3.6] approximates the sum by a partial
sum plus another quantity, not by a partial sum alone. In this case
that other quantity is an integral,
b. The error-size bound in this approximation is a
term of the series, and thus is called a term bound.
Example 3.4
Note
We'll employ the estimation:
Solution
EOS
Best Approximation With Integral Bound
From approximation [3.5] we have, for example:
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Fig. 3.5 |
Thus:
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Fig. 3.6 Best approximation of a quantity |
Best Approximation With Integral Bound On Error Size
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The approximation of s by sn, a is called the best approximation with integral bound. |
Remarks 3.4
a. Like approximation [3.6], approximation [3.8]
approximates the sum by a partial sum plus another quantity, not by a partial
sum alone. In this case that other
quantity is the average of 2 integrals.
b. The error bound in approximation [3.8] is an expression in terms of the integrals, and thus is called an integral bound.
c. The reason why it's called the “best
approximation” will be seen below under the heading “Comparison Of The
Approximations”.
Example 3.5
Consider the best approximation of the series:
a. Using sn,
determine the smallest interval that you can be sure contains its sum.
b. How large must n be so that
the best approximation of the series has |error| < 0.001?
c. Find this best approximation with |error| < 0.001.
Solution
EOS
Comparison Of
The Approximations
1. In Example 3.3, if s is to be
approximated by sn
with |error| < 0.001, then n must be at least 1,001.
1. In Example 3.4, if s is to be
approximated by sn, i
with |error| < 0.001, then n must be at least 32.
3. In Example 3.5, if s is to be
approximated by sn, a
with |error| < 0.001, then n must be at least 22.
It's easy to check that if we increase n
in each approximation, the error size gets smaller. Clearly the approximation
by sn, a
is
the best. The approximation by sn, i
comes as the relatively close second. The approximation by sn
is the distant third.
Example 3.6
Solution
EOS
Relative Speed Of Convergence
Remark 3.5
Evidently the approximation of s
by sn, i
with the term bound on the error size or by sn
or sn, a
with the integral bound on the error
size is appropriate for series whose convergence is determined by the integral
test or the p-series property.
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4. For Alternating Series Whose Convergence Is Determined By The
AST |
Here we'll approximate the sum of a convergent alternating
series by its partial sums, provided the convergence of the series is
determined by the AST (alternating-series test).
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Fig. 4.1 Partial Sums Of |
Approximation
Of Convergent Alternating Series
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Approximating its sum by the sum of its first n terms generates an error size that is at most the magnitude of its next term. |
Example 4.1
Solution
To get the accuracy of 2 decimal places we need the error size to be less than
0.005. Let s and sn
be the sum and nth partial
sum of the series respectively. We have:
So at least 200 terms are needed.
EOS
This example shows that the alternating harmonic series converges very slowly.
Example 4.2
a. Show that it converges.
b. Estimate its sum to within 0.001.
Solution
a. Clearly the series is alternating, the sequence of the absolute
values of its terms is decreasing, and the sequence of its
terms approaches 0 as n approaches infinity. So, by the AST, it
converges.
b. We need the error size to be less than 0.001. We have:
EOS
Problems & Solutions |
1. Consider the series:
a. Show that it indeed converges.
b. Approximate its sum s by a partial sum sn so that the error size |s – sn| = s – sn is less than 0.0001.
Solution
a. Let an = 2n/(2n)!. Then:
So,
by the RaT, the given series converges.
Note
Solution
Solution
So n must be at least 11.
4. Consider the series:
a. Show that it converges.
b. Estimate its sum with error within 0.01, using a term bound.
Solution
5. Consider the best approximation of the series:
a. Prove that this series converges.
b. Using its nth partial
sum sn, determine the
smallest interval that you can be sure contains its sum.
c. How large must n be so that
the best approximation of the series has |error| < 0.001?
d. Find this best approximation with |error| < 0.001.
Solution
6. Consider the series:
a. Show that it converges.
b. How many terms of its terms are needed to ensure that its partial sum sn approximates its sum s with |error| < 0.001?
c. Calculate that approximate sum.
Solution
a. The series is alternating. The sequence of the absolute values of its terms is decreasing. The sequences of its terms
approaches 0. So, by the AST, it converges.
b. Its (n + 1)th term is (–1)(n+1)+1(n + 1)/2n+1 = (–1)n+2(n + 1)/2n+1. We have:
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