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Jack Koki
Mrs. Tallman
AP Calculus
13 April 2015
Sequences, Series, and Approximations of Transcendental Functions
In order to earn credit for an entry-level college calculus course, students must
demonstrate a conceptual understanding of infinite power series as a method of approximating
the values of various transcendental functions; written in this expanded form, various tests can
also be applied to determine whether a particular infinite series converges or diverges, and error
can also be calculated in various ways. When the time comes for me to take the AP Calculus BC
Exam this May, I plan on demonstrating my personal understanding of infinite power series, the
convergence tests,and error to exempt myself from the entry-level calculus classes offered at
Wayne State University. Therefore, to demonstrate my understanding and mastery of this
content, I will thoroughly compare and contrast sequences and series, convergence and
divergence, Taylor and Maclaurin series, and the three methods of computing error when a
transcendental function is approximated using a power series.
A sequence is formally defined as an infinite ordered set of numbers. Although a
sequence can be any ordered set of numbers, there are two cases in particular that are
mathematically significant. An arithmetic sequence exists if all the terms have some common
difference (𝑑) that is added to each consecutive term.
2, 5, 8, 11, 14, 17, 20, 23, 26, 29, … 𝑎𝑛 = 𝑎𝑛−1 + 3
In the sequence above, each term is found by adding a common difference of 3 to the previous
term; therefore, the series is considered arithmetic, where 𝑑 = 3. Additionally, if a sequence has
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some common ratio (𝑟) that exists between all of its terms, the sequence is said to be geometric.
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, … 𝑔𝑛 = 𝑔𝑛−1 ∗ 2
In this sequence, each term is found by multiplying the previous term by a common ratio of 2;
therefore, the sequence is considered geometric, where 𝑟 = 2.
With this in mind, a series is the sum of a certain number of terms in a particular
sequence. A series is said to converge if and only if its sequence of partial sums converge; in
other words, a series converges if the𝑛𝑡ℎ term of its sequence of partial sums approaches a finite
limit 𝐿 as the number of terms (𝑛) approaches infinity. Furthermore, a geometric series
converges if the absolute value of its common ratio is less than 1. Consider the following
geometric sequence below:
1 1 1 1
1, , , , , …
2 4 8 16
To find the corresponding sequence of partial sums, find the sum of each of these terms.
1 3 7 15
1, 1 , 1 , 1 , 1 , …
2 4 8 16
1
Note that the common ratio of this geometric series is 𝑟 = 2. As more terms are added to this
sequence of partial sums, it becomes evident that the sequence converges to 2 as 𝑛 approaches
infinity. Therefore, this particular geometric series converges.
Alternatively, a series is said to diverge if the 𝑛𝑡ℎ term of its sequence of partial sums
approaches an infinite limit 𝐿 as 𝑛 approaches infinity. Consider the sequence of prime numbers
below:
2, 3, 5, 7, 11, 13, 17, 19, 23, …
To find the corresponding sequence of partial sums, find the sum of each of these terms.
2, 5, 10, 17, 28, 41, 58, 77, 100, …
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As more terms are added to this sequence of partial sums, it becomes evident that the terms
become larger and larger as 𝑛 approaches infinity. Therefore, since the limit of this sequence of
partial sums is infinity as 𝑛 approaches infinity, the series of primes diverges.
Be wary of convergence and divergence; it is possible for the terms of a sequence to
converge when the same series diverges. For instance, consider the harmonic sequence below:
1 1 1 1 1 1 1
1, , , , , , , , …
2 3 4 5 6 7 8
The individual terms of this harmonic sequence seem to converge to zero as more and more
terms are listed. While this may be true, do not equate the convergence of the sequence’s terms
to the convergence of the same series.
1, 1.5, 1.83, 2.08, 2.28, 2.45, 2.59, 2.71, …
As more terms are added to this sequence of partial sums, its terms become larger and larger as 𝑛
approaches infinity. Just like the series of primes, since the limit of this sequence of partial sums
is infinity as 𝑛 approaches infinity, the harmonic series diverges.
There are a handful of transcendental functions that can be modeled by various infinite
power series; that is, each of these series continue on forever, and each term contains a power of
the variable 𝑥. These estimators can be derived using either a Taylor or Maclaurin series. Given a
function with differentiable derivatives, a Taylor series expansion of a transcendental function
about 𝑥 = 𝑎 can be derived by evaluating the function at 𝑎, finding the 𝑛𝑡ℎ derivative of the
function at 𝑎 over the 𝑛𝑡ℎ factorial times the quantity of 𝑥 − 𝑎 to the 𝑛𝑡ℎ power, and summing
up these quantities.
𝑓(𝑥) = 𝑓(𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎) +
𝑓′′(𝑎)
𝑓′′′(𝑎)
𝑓 (𝑛) (𝑎)
(𝑥 − 𝑎)𝑛 + ⋯
(𝑥 − 𝑎)2 +
(𝑥 − 𝑎)3 + ⋯
2!
3!
𝑛!
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Consider the transcendental function 𝑓(𝑥) = 𝑒 𝑥 . Using the methods described above, a Taylor
series for this function can be derived to expand the function about 𝑥 = 1.
Since 𝑒 𝑥 is its own derivative, this makes expanding the function about 𝑥 = 1 a quick and easy
task.
𝑒 𝑥 = 𝑒 + 𝑒(𝑥 − 1) +
𝑒
𝑒
𝑒
𝑒
(𝑥 − 1)2 + (𝑥 − 1)3 + (𝑥 − 1)4 + (𝑥 − 1)5 + ⋯
2!
3!
4!
5!
Similarly, a Maclaurin series can be derived by the same process; the only difference
between the two expansions is that 𝑎, the horizontal translator of x, is equal to 0. Therefore, the
general form of a Maclaurin series is as follows:
𝑓(𝑥) = 𝑓(0) + 𝑓 ′ (0)(𝑥) +
𝑓′′(0)
𝑓′′′(0)
𝑓 (𝑛) (0)
(𝑥)𝑛 + ⋯
(𝑥)2 +
(𝑥)3 + ⋯
2!
3!
𝑛!
Using the same transcendental function 𝑓(𝑥) = 𝑒 𝑥 , a Maclaurin series for this function can be
derived to expand the function about 𝑥 = 0.
𝑒 𝑥 = 𝑒 0 + 𝑒 0 (𝑥) +
𝑒 𝑥 = 1 + (𝑥) +
𝑒0
𝑒0
𝑒0
𝑒0
(𝑥)2 + (𝑥)3 + (𝑥)4 + (𝑥)5 + ⋯
2!
3!
4!
5!
1
1
1
1
(𝑥)2 + (𝑥)3 + (𝑥)4 + (𝑥)5 + ⋯
2!
3!
4!
5!
Not only do they model various transcendental functions,these infinite power series can
also be used to approximate their values.To demonstrate the accuracy of the previously derived
Taylor series, let’s compare the actual value of 𝑒 1.3 to the value given by the power series. Using
a calculator,𝑒 1.3 is equal to 3.6692966676. Now, plug in 1.3 for 𝑥 in the Taylor series:
𝑒 1.3 = 𝑒 + 𝑒(1.3 − 1) +
𝑒
𝑒
𝑒
𝑒
(1.3 − 1)2 + (1.3 − 1)3 + (1.3 − 1)4 + (1.3 − 1)5 + ⋯
2!
3!
4!
5!
𝑒 1.3 ≈ 𝟑. 𝟔𝟔𝟗𝟐𝟗𝟑𝟕𝟗𝟐𝟖
|3.6692966676 − 3.6692937928| = 𝟎. 𝟎𝟎𝟎𝟎𝟎𝟑
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Using the 6th partial sum, the Taylor series gives a value nearly equal to the true value of 𝑒 1.3 ; in
fact, the approximation is accurate to five decimal places. Let’s also demonstrate the accuracy of
the previously derived Maclaurin series for 𝑒 𝑥 by comparing the actual value of 𝑒 0.3 to the value
given by the power series. Using a calculator,𝑒 0.3 is equal to 1.3498588076. Now, plug in 0.3 for
𝑥 in the Maclaurin series:
𝑒 0.3 = 1 + (0.3) +
1
1
1
1
(0.3)2 + (0.3)3 + (0.3)4 + (0.3)5 + ⋯
2!
3!
4!
5!
𝑒 0.3 ≈ 𝟏. 𝟑𝟒𝟗𝟖𝟓𝟕𝟕𝟓𝟎𝟎
|1.3498588076 − 1.3498577500| = 𝟎. 𝟎𝟎𝟎𝟎𝟎𝟏
Using the 6th partial sum, the Maclaurin series gives a value nearly equal to the true value of 𝑒 0.3 ;
this approximation is also accurate to five decimal places.
The Taylor and Maclaurin series give considerably accurate approximations for
transcendental functions; as higher partial sums are found, the estimates become even more
accurate. Nevertheless, always remember that these power series are tools for estimation; no
matter how high a partial sum you use to approximate the value of a transcendental function,
there will always be slight error from the true value. One way of computing this error is by
manually calculating the absolute value of the difference between the predicted and true values
(like in the previous two examples). A second way of doing so only applies when you are dealing
with an alternating series. If this is the case, the error (𝑅𝑛 ) is no greater than the absolute value of
the first term of the tail, or the terms of a series remaining after finding a particular partial sum.
Consider the 6th partial sum estimation for cos(0.3) using the Maclaurin series for 𝑓(𝑥) =
cos(𝑥):
cos(𝑥) = 1 −
1 2 1 4 1 6 1 8
1 10
𝑥 + 𝑥 − 𝑥 + 𝑥 −
𝑥 +⋯
2!
4!
6!
8!
10!
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cos(0.3) = 1 −
1
1
1
1
1
(0.3)2 + (0.3)4 − (0.3)6 + (0.3)8 −
(0.3)10 + ⋯
2!
4!
6!
8!
10!
1
Using the 6th partial sum, the first term of the power series’ tailis 12! (0.3)12. Therefore, using the
previously stated information, the error is less than or equal to the absolute value of this term
evaluated at𝑥 = 0.3.
𝑅𝑛 ≤ |
1
(0.3)12 |
12!
𝑅𝑛 ≤ 1.11 ∗ 10−15
In other words, this approximation for cos(0.3) is accurate to at least14 decimal places.
A third method of computing error involves calculating Lagrange Error. This method is
especially useful when either a Taylor of Maclaurin series is involved. Recall that the general
term of a Taylor series expansion is given by the following formula:
𝑓 (𝑛) (𝑎)
(𝑥 − 𝑎)𝑛
𝑡𝑛 =
𝑛!
If the first term of the tail is notated 𝑡𝑛+1, the Mean Value Theorem states that for any value of 𝑥
in the interval of convergence, there is a value 𝑥 = 𝑐 between 𝑎 and 𝑥 such that the following
condition is true:
𝑓 (𝑛+1) (𝑐)
(𝑥 − 𝑎)𝑛+1
𝑅𝑛 (𝑥) =
(𝑛 + 1)!
Simply put, the Lagrange Error Bound says that the error of a particular approximation is equal
to the (𝑛 + 1)𝑡ℎ derivative evaluated at 𝑐. Since finding this derivative can be tough to do, this
can be done by finding an upper bound for the (𝑛 + 1)𝑡ℎ derivative and using that as the value
for 𝑓 (𝑛+1) (𝑐) in the equation above. By using this maximum value (𝑀) of |𝑓 (𝑛+1) (𝑐)|, the
Lagrange Error is no greater than the previous formula using 𝑀as the value for the (𝑛 + 1)𝑡ℎ
derivative.
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𝑅𝑛 (𝑥) ≤
𝑀
|𝑥 − 𝑎|𝑛+1
(𝑛 + 1)!
Let’s apply the Lagrange Error formula to the 6th partial sum estimation for 𝑒 0.3 using the
Maclaurin series for 𝑓(𝑥) = e𝑥 :
𝑒 0.3 = 1 + (0.3) +
1
1
1
1
(0.3)2 + (0.3)3 + (0.3)4 + (0.3)5 + ⋯
2!
3!
4!
5!
Using this example, the approximation ends when 𝑛 = 5 in the sigma notation, so Lagrange says
the error can be computed using the following formula with substitutions:
30.3
𝑅𝑛 (𝑥) ≤
|0.3|6
(6)!
𝑅𝑛 (𝑥) ≤ 0.00001
Because𝑒 𝑥 is its own derivative, 3 was used as the maximum value (𝑀) in the formula since 3 is
close to the value of 𝑒 ≈ 2.72. The Lagrange Error Bound, therefore, says that this
approximation for 𝑒 0.3 is accurate to at least 4 decimal places.
With all these concepts relating to power series and approximations fully explained, let’s
apply this newfound knowledge to solve some calculus problems. Consider the function 𝑓
defined by the following power series for all real numbers 𝑥 for which the series converges:
∞
2
𝑛
𝑓(𝑥) = 1 + (𝑥 + 1) + (𝑥 + 1) + ⋯ + (𝑥 + 1) + ⋯ = ∑(𝑥 + 1)𝑛
𝑛=0
To find the interval of convergence, simply recognize that this is a geometric power series with a
common ratio 𝑟 = 𝑥 + 1. In order for a geometric series to converge, the absolute value of its
common ratio must be less than 1.
|𝑥 + 1| < 1
−1 < 𝑥 + 1 < 1
−2 < 𝑥 < 0
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After conducting minor algebraic cleanup, the interval of convergence is found to be (-2, 0).
Now that the interval of convergence for the series is known, the sum of this geometric
power series within this interval can be found by dividing the first term of the series by the
quantity of one less than the common ratio 𝑟.
𝑆∞ =
𝑆∞ =
𝑔1
1−𝑟
1
1 − (𝑥 + 1)
𝑆∞ = −
1
𝑥
Therefore, as long as 𝑥 remains within the interval of convergence, the sum of this infinite
geometric power series can be found by taking the negative reciprocal of the value of 𝑥.
𝑥
1
Now, consider the function defined by ∫−1 𝑓(𝑡) 𝑑𝑡. To evaluate 𝑔(− 2), the question is
1
really asking what the area under the curve is from −1 to − 2. Since both of these values fall
1
within the previously calculated interval of convergence, the value of 𝑔(− 2) exists and can be
calculated by substituting the appropriate variables into the integral expression and evaluating its
antiderivative at the two limits of integration.
1
−
2
1
1
𝑔 (− ) = ∫ − = − ln|𝑥|
2
𝑥
−1
𝑥=−
1
2
𝑥 = −1
1
1
𝑔 (− ) = − ln |− | + ln|−1| = ln 2
2
2
Therefore, using the Fundamental Theorem of Calculus, the area under the curve between these
two values of 𝑥 that fall within the interval of convergence is equal to 𝑙𝑛 2 ≈ 0.6931 square
units.
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Consider another scenario where ℎ is the function defined by ℎ(𝑥) = 𝑓(𝑥 2 − 1). To find
the first three nonzero terms and the general term of the Taylor series for ℎ about 𝑥 = 0, simply
substitute 𝑥 2 − 1 in for all the 𝑥 variables in the function 𝑓.
∞
2
2
4
ℎ(𝑥) = 𝑓(𝑥 − 1) = 1 + 𝑥 + 𝑥 + ⋯ + 𝑥
2𝑛
+ ⋯ = ∑ 𝑥 2𝑛
𝑛=0
1
1
After deriving the new power series, the value of ℎ(2) can be evaluated since 2 falls within the
interval of convergence previously calculated.
1
𝑔1
ℎ( ) =
2
1−𝑟
1
1
1
ℎ( ) =
=
2
1 − (𝑥 2 ) 1 − (1)2
2
1
1
1 𝟒
ℎ( ) =
= 3 =
1
2
𝟑
1−
4
4
1
4
After performing some rudimentary algebra, the value of ℎ(2) is found to be 3.
Convergence and divergence can be determined for any infinite series using a multitude
of tests. For example, the Direct Comparison test can be used to determine whether the following
series converges or diverges as the number of terms approach infinity:
∞
3
𝑛 ⁄2 + 1
∑ 2
5𝑛 + 7
𝑛=0
Since the translators are trivial when determining convergence and divergence, this series can be
directly compared to the following series:
∞
3
𝑛 ⁄2
1
∑ 2 = 0.5
𝑛
5𝑛
𝑛=0
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By the p-Series test, since 𝑝 = 0.5, this series is known to diverge. In order to determine if the
original series diverges as well, its tail must be bound below the terms of the series known to
diverge by the p-Series test. See Figure 1 below for a visual of this concept.
1
Figure 1. Direct Comparison to 5𝑛0.5
Figure 1 shows both series graphed on the same set of axes. As you can see, the tail of
𝑓1(𝑥), the series we are interested in, seems to be bound below that of 𝑓2(𝑥), the series known
to diverge by the p-Series test, while viewing the graphs from a standard window. However, after
zooming in around the 106th term, it is clear that the tail of series we are interested in lies above
3
𝑛 ⁄2 +1
the series known to diverge. Therefore, the infinite series ∑∞
𝑛=0 5𝑛2 +7 diverges by Direct
Comparison.
Next, consider the following infinite series:
∞
∑(−1)𝑛
𝑛=2
1
ln(𝑛)
Since the series alternates, the Alternating Series test can be used to determine if the series
converges. In addition to alternating, the series must also meet the criterion that its limit is 0 as
the number of terms approach infinity.
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1
1
1
=
= =𝟎
𝑛→∞ ln(𝑛)
ln(∞) ∞
lim
Since this is true, the last condition is that the(𝑛 + 1)𝑡ℎ term must be greater than 0 and less than
or equal to the 𝑛𝑡ℎ term.
0 < 𝑎𝑛+1 ≤ 𝑎𝑛
0<
1
1
≤
ln(𝑛 + 1) ln(𝑛)
Since the (𝑛 + 1)𝑡ℎ term will always have a larger value in the denominator than the 𝑛𝑡ℎ term,
the reciprocal of that value will always make the (𝑛 + 1)𝑡ℎ term smaller than the 𝑛𝑡ℎ term. Since
1
𝑛
it meets all three of the conditions, the series∑∞
𝑛=2(−1) ln(𝑛) converges by the Alternating
Series test.
Consider another alternating infinite series:
∞
4
∑(−1)𝑛 ( )𝑛
3
𝑛=0
Rather than implementing the Alternating Series test again to test for convergence, the nth-Term
test can be applied to this series to show that the series diverges. In order to do so, simply show
that its limit is not equal to 0 as the number of terms approach infinity.
4
4
lim ( )𝑛 = ( )∞ = ∞ ≠ 𝟎
𝑛→∞ 3
3
Therefore, since its limit approaches infinity as the number of terms approach infinity, the series
4
𝑛
𝑛
∑∞
𝑛=2(−1) (3) diverges by the nth-Term test.
One unique quality about the Ratio test is that it allows you to find the interval of
convergence for an infinite series; like a geometric series, the absolute value of its “ratio” as the
number of terms approach infinity must be less than one. After the interval of convergence is
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acquired, the endpoints must also be tested for convergence; this can be done using a suitable test
after substituting the values of the endpoints of the interval for 𝑥. Consider the following power
series:
∞
∑
𝑛=0
(2𝑥)𝑛
𝑛+1
Using the Ratio test, the interval of convergence can be found by finding the limit as the number
of terms approach infinity of the ratio of the absolute value of the (𝑛 + 1)𝑡ℎ term to the 𝑛𝑡ℎ term.
lim |
𝑛→∞
𝑎𝑛+1
|
𝑎𝑛
(2𝑥)𝑛+1
lim |
𝑛→∞
𝑛+2
(2𝑥)𝑛
|
𝑛+1
lim |
𝑛→∞
(2𝑥)𝑛+1 𝑛 + 1
∗
|
(2𝑥)𝑛 𝑛 + 2
lim |2𝑥 ∗ 1| = |𝟐𝒙|
𝑛→∞
|2𝑥| < 1
−1 < 2𝑥 < 1
−
𝟏
𝟏
<𝑥<
𝟐
𝟐
After applying l’Hospital’s rule, finding the limit, and doing some algebraic manipulation, the
1
1
2
2
interval of convergence for this series is found to be − < 𝑥 < . As previously mentioned, the
1
endpoints must also be tested for convergence. When 𝑥 = − 2, the series becomes the following:
∞
∑
𝑛=0
1 𝑛
(2 ∗ − 2)
𝑛+1
∞
(−1)𝑛
=∑
𝑛+1
𝑛=0
Since the series alternates, the Alternating Series test can be used to determine if the series
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converges. In addition to alternating, the series must also meet the criterion that its limit is 0 as
the number of terms approach infinity.
1
1
1
=
= =𝟎
𝑛→∞ n + 1
∞+1 ∞
lim
Since this is true, the last condition is that the (𝑛 + 1)𝑡ℎ term must be greater than 0 and less
than or equal to the 𝑛𝑡ℎ term.
0 < 𝑎𝑛+1 ≤ 𝑎𝑛
0<
1
1
≤
n+2 n+1
Since the (𝑛 + 1)𝑡ℎ term will always have a larger value in the denominator than the 𝑛𝑡ℎ term,
the reciprocal of that value will always make the (𝑛 + 1)𝑡ℎ term smaller than the 𝑛𝑡ℎ term. Since
it meets all three of the conditions, the series ∑∞
𝑛=2
(−1)𝑛
𝑛+1
converges by the Alternating Series test;
1
hence, − 2 is inclusive in the interval of convergence.
1
When 𝑥 = 2, the series becomes the following:
∞
1 𝑛
(2 ∗ 2)
∞
∞
𝑛=0
𝑛=0
(1)𝑛
1
∑
=∑
=∑
𝑛+1
𝑛+1
𝑛+1
𝑛=0
The Limit Comparison test can be used to determine whether the series converges or diverges as
the number of terms approach infinity. Since the translators are trivial when determining
convergence and divergence, this series can be directly compared to the harmonic series:
∞
∑
𝑛=0
1
𝑛
By the p-Series test, since 𝑝 = 1, the harmonic series is known to diverge. In order to determine
if the original series diverges as well, the limit as the number of terms approach infinity of the
ratio of the previous series to this harmonic series must be equal to a positive real number.
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𝑎𝑛
𝑛→∞ 𝑏𝑛
lim
1
lim 𝑛+1
𝑛→∞ 1
𝑛
𝑛
𝑛→∞ 𝑛 + 1
lim
1
=𝟏
𝑛→∞ 1
lim
After applying l’Hospital’s rule and finding the limit, the test finds the ratio of the translated
harmonic series the limit as 𝑛 approaches infinity of the ratio of the translated harmonic series to
1
the parent harmonic series to be 1, a positive real number. Therefore, the infinite series ∑∞
𝑛=0 𝑛+1
1
diverges by the Limit Comparison test; hence, 2 is not inclusive in the interval of convergence.
So, after doing all the calculus, the interval of convergence for ∑∞
𝑛=0
(2𝑥)𝑛
𝑛+1
1
1
is − 2 ≤ 𝑥 < 2.
Recall that always power series can be used as tools for approximating various
transcendental functions; no matter how high a partial sum you use to estimate its value, there
will always be slight error from the true value. Let’s demonstrate this concept by using the 5th
partial sum of the power series for 𝑒 𝑥 to approximate 𝑒 0.1 whose true value is equal to
1.1051709181.
𝑒 0.1 = 1 + (0.1) +
1
1
1
(0.1)2 + (0.1)3 + (0.1)4 + ⋯
2!
3!
4!
𝑒 0.3 ≈ 𝟏. 𝟏𝟎𝟓𝟏𝟕𝟎𝟖𝟑𝟑𝟑
|1.1051709181 − 1.1051708333| = 𝟖. 𝟒𝟕𝟒 ∗ 𝟏𝟎−𝟖
Using the 5th partial sum, the Maclaurin series gives a value nearly equal to the true value of 𝑒 0.3 ;
this approximation is accurate to seven decimal places.
Koki 15
Let’s also apply the Lagrange Error formula to the 5th partial sum estimation for 𝑒 0.1
using the Maclaurin series for 𝑓(𝑥) = e𝑥 :
𝑅𝑛 (𝑥) ≤
𝑀
|𝑥 − 𝑎|𝑛+1
(𝑛 + 1)!
Using this example, the approximation ends when 𝑛 = 4 in the sigma notation, so Lagrange says
the error can be computed using the following formula with substitutions:
30.1
𝑅𝑛 (𝑥) ≤
|0.1|5
(5)!
𝑅𝑛 (𝑥) ≤ 0.000000093
Because 𝑒 𝑥 is its own derivative, 3 was used as the maximum value (𝑀) in the formula since 3 is
close to the value of 𝑒 ≈ 2.72. The Lagrange Error Bound, therefore, says that this
approximation for 𝑒 0.1 is accurate at least 7 decimal places.
In the time that exists between now and May, we will learn a few more topics that will
likely appear on my AP Calculus Exam, especially physics-related subjects. Nevertheless, my
knowledge regarding sequences and series, convergence and divergence, Taylor and Maclaurin
series, and the three methods of computing the error of a power series approximation will surely
pay off if questions of this nature do in fact make an appearance on the BC Calculus Exam this
spring. After completing this final essay and learning a myriad of calculus concepts throughout
the year, my chances of bypassing the entry-level calculus courses offered at Wayne State
University next fall are in my favor, thanks to the rigors of the AP Calculus curriculum that Mr.
Acre and Mrs. Tallman provide to their intellectual students.