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In Section Sequences, we showed how the formula for compound interest is a geometric sequence. In retirement planning, it is seldom
the case that an investor deposits a set amount of money into an account and waits for it to grow. Usually, additional payments
of principal are made at regular intervals and the value of the investment grows accordingly. This kind of investment is called
an annuity and will be discussed in later in this section once we have developed more mathematical machinery that enables
us to add sequences.
In the previous section, we introduced sequences. Each of the numbers in the sequence is called a ‘term’ which implies these
numbers are meant to be added. To that end, we introduce the following notation which is used to describe the sum of (some
of the) terms of a sequence.
Summation Notation: Given a sequence \(\left \{ a_{n} \right \}_{n=k}^{\infty }\) and numbers \(m\) and \(p\) satisfying \(k \leq m \leq p\), the summation from \(m\) to \(p\) of the sequence \(\left \{a_{n}\right \}\) is
written
The variable \(n\) is called the index of summation. The number \(m\) is called the lower limit of summation while the number \(p\) is
called the upper limit of summation.
In English, Definition sigmanotation is simply defining a short-hand notation for adding up the terms of the sequence \(\left \{ a_{n} \right \}_{n=k}^{\infty }\) from \(a_{m}\) through \(a_{p}\). The
symbol \(\Sigma \) is the capital Greek letter sigma and is shorthand for ‘sum’. The lower and upper limits of the summation tells us
which term to start with and which term to end with, respectively. For example, using the sequence \(a_{n} = 2n-1\) for \(n \geq 1\), we can write \(a_{3} +a_{4} + a_{5} + a_{6}\)
as
The index variable is considered a ‘dummy variable’ in the sense that it may be changed to any letter without affecting the
value of the summation. For instance,
One place you may encounter summation notation is in mathematical definitions. For example, summation notation allows us
to define polynomials as functions of the form
for real numbers \(a_{k}\), \(k = 0, 1, \ldots n\). The reader is invited to compare this with what is given in Definition polynomialfunction. Summation notation is particularly
useful when talking about matrix operations. For example, we can write the product of the \(i\)th row \(R_{i}\) of a matrix \(A = [a_{ij}]_{m \times n}\) and the \(j^{\text {th}}\) column \(C_{j}\)
of a matrix \(B = [b_{ij}]_{n \times r}\) as
\[ Ri \cdot Cj = \displaystyle {\sum _{k=1}^{n} a_{ik}b_{kj}} \]
Again, the reader is encouraged to write out the sum and compare it to Definition rowcolumnproduct. Our next example gives us practice with
this new notation.
Proceeding as in (a), we replace every occurrence of \(n\) with the values \(0\) through \(4\). We recall the factorials, \(n!\)
as defined in number Example seqex1, number factorialintroex and get:
The key to writing these sums with summation notation is to find the pattern of the terms. To that end, we make good
use of the techniques presented in Section Sequences.
The terms of the sum \(1\), \(3\), \(5\), etc., form an arithmetic sequence with first term \(a = 1\) and common difference \(d = 2\).
Using Equation arithgeoformula, we get \(a_{n} = 1 + (n-1)2 = 2n-1\), \(n \geq 1\).
At this stage, we have the formula for the terms, namely \(2n-1\), and the lower limit of the summation, \(n=1\). To
finish the problem, we need to determine the upper limit of the summation. In other words, we need to
determine which value of \(n\) produces the term \(117\). Setting \(a_{n} = 117\), we get \(2n-1=117\) or \(n = 59\). Our final answer is
The numerators, \(1\), \(-1\), etc. can be described by the geometric sequence (This is indeed a geometric
sequence with first term \(a = 1\) and common ratio \(r=-1\).)\(C_{n} = (-1)^{n-1}\) for \(n \geq 1\), while the denominators are given by the arithmetic
sequence (It is an arithmetic sequence with first term \(a=1\) and common difference \(d=1\).)\(D_{n} = n\) for \(n \geq 1\). Hence, we get
the formula \(a_{n} = \frac {(-1)^{n-1}}{n}\) for our terms, and we find the lower and upper limits of summation to be \(n=1\) and \(n = 117\), respectively.
Thus
Thanks to Example seqex2, we know that one formula for the \(n^{\text {th}}\) term is \(a_{n} = \frac {9}{10^{n}}\) for \(n \geq 1\). This gives us a formula for the
summation as well as a lower limit of summation.
To determine the upper limit of summation, we note that to produce the \(n-1\) zeros to the right of the decimal
point before the \(9\), we need a denominator of \(10^{n}\). Hence, \(n\) is the upper limit of summation.
Since \(n\) is used in the limits of the summation, we need to choose a different letter for the index of
summation. (To see why, try writing the summation using ‘\(n\)’ as the index.) We choose \(k\) and get
The following theorem presents some general properties of summation notation.
Properties of Summation Notation: Suppose \(\left \{a_{n}\right \}\) and \(\left \{b_{n}\right \}\) are sequences so that the following sums are defined.
Distributive Property:\(\displaystyle {\sum _{n=m}^{p} c \, a_{n} = c \sum _{n=m}^{p} a_{n}}\), for any real number \(c\).
Additive Index Property:\(\displaystyle {\sum _{n=m}^{j} a_{n} + \sum _{n=j+1}^{p} a_{n} = \sum _{n=m}^{p} a_{n}}\), for any natural number \(m \leq j < j+1 \leq p\).
Re-indexing:\(\displaystyle {\sum _{n=m}^{p} a_{n} = \sum _{n=m+r}^{p+r} a_{n-r}}\), for any integer \(r\).
There is much to be learned by thinking about why the properties hold, so we leave the proof of these properties to the reader. (To get started, remember the mantra “When in doubt, write it out!”)
There are at least two ways to approach this problem. By definition, \(\displaystyle { \sum _{n=1}^{21} a_{n} = a_{1} + a_{2} + \ldots + a_{21}}\). That is, we add up the first \(21\) terms of the
sequence \(a_{n}\). Similarly, \(\displaystyle { \sum _{n=1}^{20} a_{n} = a_{1} + a_{2} + \ldots + a_{20}}\) means we add up the first \(20\) terms of the sequence. Hence, \(a_{21} = \displaystyle { \sum _{n=1}^{21} a_{n} - \sum _{n=1}^{20} a_{n} = 7-(-3) = 10.}\)
Alternatively, we can use the Additive Index Property:
which gives \(a_{21} = \displaystyle { \sum _{n=1}^{21} a_{n} - \sum _{n=1}^{20} a_{n} = 7-(-3) = 10}\) as well.
To re-index \(\displaystyle { \sum _{n=2}^{437} n (n-1) x^{n-2}}\) so \(n\) starts at \(0\), we follow the formula in Theorem summationpropsex with \(r=-2\):
We leave it to the reader to check by writing
out the first few, and last few, terms.
Alternatively, to better see why the re-indexing works in this way, we can introduce a new counter, \(k\). We want
this new counter to start at \(k =0\) whereas the current counter starts at \(n=2\), so we want \(k = n-2\). When \(n=2\), \(k=0\), as required, and
when \(n=437\), \(k = 435\).
Moreover, \(n = k+2\), so substituting this into the sum, we get
which is the same sum we had before, just with a different dummy variable. □
We now turn our attention to the sums involving arithmetic and geometric sequences. Given an arithmetic sequence \(a_{k} = a + (k-1) d\) for \(k \geq 1\), we
let \(S\) denote the sum of the first \(n\) terms. To derive a formula for \(S\), we write it out in two different ways
\[ \begin{array}{ccccccccccc}
S & = & a & + & (a + d) & + & \ldots & + & (a + (n-2)d) & + & (a + (n-1)d) \\
S & = & (a + (n-1)d) & + & (a + (n-2)d) & + & \ldots & + & (a + d) & + & a \\
\end{array}\]
If we add these two
equations and combine the terms which are aligned vertically, we get
The right hand side of this equation contains \(n\) terms, all of which are equal to \((2a + (n-1)d)\) so we get \(2S = n(2a + (n-1)d)\). Dividing both sides of this equation
by \(2\), we obtain the formula
\[S = \frac {n}{2} (2a + (n-1)d)\]
If we rewrite the quantity \(2a + (n-1)d\) as \(a + (a + (n-1)d) = a_{1} + a_{n}\), we get the formula
\[ S = n \left (\frac {a_{1} + a_{n}}{2}\right )\]
A helpful way to remember this last formula is to recognize that we have expressed the sum as the product of the number of
terms \(n\) and the average of the first and \(n^{\text {th}}\) terms.
To derive the formula for the geometric sum, we start with a geometric sequence \(a_{k} = ar^{k-1}\), \(k \geq 1\), and let \(S\) once again denote the sum of the
first \(n\) terms. Comparing \(S\) and \(rS\), we get
\[ \begin{array}{ccccccccccccccc}
S & = & a & + & ar & + & ar^2 & + & \ldots & + & ar^{n-2} & + & ar^{n-1} & & \\
r S & = & & & ar & + & ar^2 & + & \ldots & + & ar^{n-2} & + & ar^{n-1} & + & ar^{n} \\
\end{array}\]
Subtracting the second equation from the first forces all of the terms except \(a\) and \(ar^{n}\) to cancel out and we get \(S - rS = a - ar^{n}\). Factoring, we get \(S(1-r) = a \left (1-r^{n}\right )\).
Assuming \(r \neq 1\), we can divide both sides by the quantity \((1-r)\) to obtain
\[S = a \left ( \frac {1-r^n}{1-r}\right )\]
If we distribute \(a\) through the numerator, we get \(a - ar^{n} = a_{1} - a_{n+1}\) which yields the formula
\[S = \frac {a_{1}-a_{n+1}}{1-r}\]
In the case when \(r=1\), we get the formula
\[ S = \underbrace {a + a + \ldots +a }_{\text {$n$ times}} = n \, a\]
Our results are summarized below. (Alternatively, we can use Exercise geoseriespreview in Section Polydivision.)
Sums of Arithmetic and Geometric Sequences:
The sum \(S\) of the first \(n\) terms of an arithmetic sequence \(a_{k}= a + (k-1)d\) for \(k \geq 1\) is
\(S = \displaystyle {\sum _{k=1}^{n} a_{k} = \sum _{k=1}^{n} a =n a}\), if \(r =1\).
While we have made an honest effort to derive the formulas in Equation arithgeosum, formal proofs require the machinery in Section
Induction.
\(~\)
Find the sum: \(1 + 3 + 5 + \ldots + 117\)
Find a formula for the sum \(\displaystyle { \sum _{k=1}^{n} k }\).
The classic wheat and chessboard problem asks the following question. Given a chessboard with its squares
numbered \(1\) to \(64\), suppose on the first square was placed one grain of wheat, the second square, two grains, the third
square, four grains, and so on, each square receiving twice the number of grains as its predecessor. How many total
grains of wheat would end up on the chessboard?
Solution.
Recognizing the terms of \(1 + 3 + 5 + \ldots + 117\) as \(1\), \(3\), \(5\), and so on, we see we have an arithmetic sequence with \(a=1\) and \(d=2\). Using
Equation arithgeoformula, we get a formula for the terms \(a_{n} = 1 + 2(n-1) = 2n-1\) for \(n \geq 1\). In order to use the formula in Equation arithgeosum, we need to
determine the number of terms being added, \(n\). Setting \(2n-1 = 117\), we find \(n = 59\). Feeding in all of our data into Equation arithgeosum,
we get \(1 + 3 + 5 + \ldots + 117 =59 \left (\frac {1+117}{2}\right ) = 3481\).
Applying the adage ‘when in doubt, write it out,’ we have \(\displaystyle { \sum _{k=1}^{n} k = 1 + 2 + 3 + 4 + \ldots + n}\). We see the terms here form an arithmetic
sequence with \(a = d = 1\). Moreover, we are adding exactly \(n\) terms, so Equation arithgeosum gives \(\displaystyle { \sum _{k=1}^{n} k = 1 + 2 + 3 + 4 + \ldots + n = \frac {n(n+1)}{2}}\).
As a side note, the special case: \(1 + 2 + 3 + \ldots + 100\) was allegedly given to Carl Friedrich Gauss while he was in elementary
school. Instead of computing the sum in a brute force method, he arrived at the answer by grouping \(1+99 = 100\), \(2+98 = 100\),
etc. so that he had \(50\) groups of \(100\) with \(50\) left over for a total of \(5050\). This is the exact same methodology we used
to prove the sum of the arithmetic sequence formula in Equation arithgeosum.
Since we are doubling the number of grains of wheat as we move from one square to the next, a geometric sequence
with \(r=2\) describes the number of grains on each individual square.
Since we start with one grain on the first square, the number of grains on the \(k\)th square is \(a_{k} = (1) (2)^{k-1} = 2^{k-1}\) for \(k \geq 1\).
Adding up the number of grains on each square gives:
in accordance with Equation arithgeosum. (The weight of these grains would total approximately \(2.6 \times 10^{15}\) pounds which is approximately \(15\)
times the entire biomass of the planet.) □
An important application of the geometric sum formula is the investment plan called an annuity. Annuities differ from the kind
of investments we studied in Section ExpLogApplications in that payments are deposited into the account on an on-going basis, and this
complicates the mathematics a little. (The reader may wish to re-read the discussion on compound interest in Section ExpLogApplications
before proceeding.)
Suppose you have an account with annual interest rate \(r\) which is compounded \(n\) times per year. We let \(i = \frac {r}{n}\) denote the interest rate
per period. Suppose we wish to make ongoing deposits of \(P\) dollars at the end of each compounding period. Let \(A_{k}\) denote the
amount in the account after \(k\) compounding periods.
Then \(A_{1} = P\), because we have made our first deposit at the end of the first compounding period and no interest
has been earned. During the second compounding period, we earn interest on \(A_{1}\) so that our initial investment
has grown to \(A_{1}(1+i) = P(1+i)\) in accordance with Equation simpleinterest. Adding our second payment at the end of the second period, we
get
\[A_{2} = A_{1}(1+i) + P = P(1+i) + P = P(1+i)\left (1 + \frac {1}{1+i}\right )\]
The reason for factoring out the \(P(1+i)\) will become apparent in short order. During the third compounding period, we earn
interest on \(A_{2}\) which then grows to \(A_{2}(1+i)\). We add our third payment at the end of the third compounding period to
obtain
The sum in the parentheses above is the sum of the first \(k\) terms of a geometric sequence with \(a = 1\) and \(r = \frac {1}{1+i}\). Using Equation arithgeosum, we
get
If we let \(t\) be the number of years this investment strategy is followed, then \(k = nt\), and we get the formula for the future value of an
ordinary annuity.
Future Value of an Ordinary Annuity: Suppose an annuity offers an annual interest rate \(r\) compounded \(n\) times per year. Let \(i = \frac {r}{n}\)
be the interest rate per compounding period. If a deposit \(P\) is made at the end of each compounding period, the amount \(A\) in the
account after \(t\) years is given by
\[A = \frac {P\left ((1+i)^{nt} - 1\right )}{i}\]
The reader is encouraged to substitute \(i = \frac {r}{n}\) into Equation fvannuity and simplify. Some familiar equations arise which are cause for pause
and meditation. One last note: if the deposit \(P\) is made a the beginning of the compounding period instead of at the end, the
annuity is called an annuity-due. We leave the derivation of the formula for the future value of an annuity-due as an exercise
for the reader.
An ordinary annuity offers a \(6 \%\) annual interest rate, compounded monthly.
If monthly payments of \(\$50\) are made, find the value of the annuity in \(30\) years.
How many years will it take for the annuity to grow to \(\$100,\! 000\)?
Solution.
We have \(r = 0.06\) and \(n = 12\) so that \(i = \frac {r}{n} = \frac {0.06}{12} = 0.005\). With \(P=50\) and \(t=30\),
To find how long it will take for the annuity to grow to \(\$100,\!000\), we set \(A = 100000\) and solve for \(t\). We isolate the exponential and
take natural logs of both sides of the equation.
This means that it takes just over \(40\) years for the investment to grow to \(\$100,\!000\). Comparing this with our answer to
part 1, we see that in just \(10\) additional years, the value of the annuity nearly doubles. This is a lesson worth
remembering. □
1 Geometric Series
As defined in Section ??, sequences are an infinite list of numbers. So far in this section, we have concerned
ourselves with adding only finitely many terms. In Calculus, infinite sums, called series are studied at great
length. While we do not have the mathematical machinery to embark upon an exhaustive study here, we can
nevertheless focus our attention on what is arguably one of the most prevalent and useful types of series, geometric
series.
As a motivating example, consider the number \(0.\overline {9}\). We can write this number as
It stands to reason that we should define \(0.\overline {9} = \ds {\lim _{n \rightarrow \infty }}\) \(\left (1 - \frac {1}{10^{n}}\right )\). Passing to a continuous variable along with our knowledge of exponential functions
gives \(\ds {\lim _{n \rightarrow \infty }}\) \(\left (1 - \frac {1}{10^{n}} \right ) =\) \(\ds {\lim _{x \rightarrow \infty }}\) \(\left (1 - \frac {1}{10^{x}} \right ) = 1 - 0 = 1\).
We have just argued that \(0.\overline {9} = 1\), which may shock some readers. (To make this more palatable, it is usually accepted that \(0.\overline {3} = \frac {1}{3}\) so
that \(0.\overline {9} = 3\left (0.\overline {3}\right ) = 3\left (\frac {1}{3} \right ) = 1\).)
Note that in this manner, any non-terminating decimal can be thought of as an infinite sum whose denominators are the
powers of \(10\), so the phenomenon of adding up infinitely many terms and arriving at a finite number is not as foreign of a
concept as it may appear. We have the following theorem.
Geometric Series: Given the sequence \(a_{k} = ar^{k-1}\) for \(k \geq 1\), where \(|r| < 1\),
If \(|r| \geq 1\), the sum \(a + ar + ar^2 + \ldots \) does not exist.
The justification of the result in Theorem 2 comes from taking the formula in Equation 0.1 for the sum of the first \(n\) terms of a
geometric sequence and taking the limit as \(n \rightarrow \infty \).
Assuming \(|r|<1\) means \(-1 < r < 1\), so per Theorem ??, \(\ds {\lim _{n \rightarrow \infty }}\) \(r^{n} = 0\). Using this fact along with the Limit Properties listed in Theorem
??
:
The interested reader is invited to
research this sum as it relates to Zeno’s Dichotomy Paradox.
To use Theorem 2 as it applies to the repeating decimal \(4.2\overline {17}\), we first need to rewrite this decimal in terms of a
geometric series. Expanding \(4.2\overline {17} = 4.2 + 0.017 + 0.00017 + 0.0000017 + \ldots ,\) we see the series \(0.017 + 0.00017 + 0.0000017 + \ldots \) is geometric with \(a = 0.017\) and \(r = 0.01\). Hence, we can apply Theorem 2 to
that part of the decimal to get:
We note that another popular method for converting repeating decimals to fractions goes something like this: let \(x =4.2\overline {17}\). Then, \(100x = 421.7\overline {17}\).
Hence, \(99x = 100x - x = 421.7\overline {17} - 4.2\overline {17} = 417.5\). Hence, \(x = \frac {417.5}{99} = \frac {835}{198}\). While this procedure results in the same (correct!) answer, the manipulations involved (such as the
multiplication and subtraction) are actually using some of the properties listed in Theorem 1 extended to infinite sums via
Theorem ??.
2 Area
One of the (two) major geometric problems studied in Calculus is finding the area under a curve (The other is the
concept of tangent lines which we touch on (awesome pun!) in Section ??.) (more specifically, the area between the graph of
a function and the \(x\)-axis.) (The area can actually represent a wide variety of things such as displacement,
probability, or, as odd as it sounds, volume.) In this section, we explore how summation notation is used to
help better formulate this problem, and, as with our study of Geometric Series, sneak a peak into Calculus
itself.
Suppose we wish to determine the area between the graph of a continuous function \(y = f(x)\) over the interval \([a,b]\) and the \(x\)-axis as shown
below.
Figure 1: Area under the graph of \(y=f(x)\).
Since we don’t know any area formulas for arbitrary regions, we stick to what we know - rectangles. To keep things simple, we
divide \([a,b]\) into \(n\) equal pieces (subintervals), and use the right-endpoints of each piece to determine the height of the rectangles.
(In Calculus, you’ll also use left endpoints and midpoints ….) We let \(x_{k}\) represent the right endpoint of the \(k\)th subinterval, so
the height of the \(k\)th rectangle is \(f(x_{k})\).
The width of the \(k\)th rectangle is the length of the \(k\)th subinterval. Since the interval itself is \(b-a\) units long and we are dividing the
interval into \(n\)equal pieces, each piece is \(\frac {b-a}{n}\) units long. For brevity, we’ll call this length ‘\(\Delta x\).’ Below on the right is a depiction of \(RS_{\, 7}\), a
‘right endpoint sum’ using \(7\) (equally spaced) subintervals. (On intervals over which the function is increasing, we find the
area of rectangles overestimates the area we want; on intervals over which the function is decreasing, we find the area of the
rectangles underestimates the area we want.)
Figure 2: Visualizing \(RS_7\), a ‘right endpoint sum.’
The idea here is to approximate the area of the shaded region by the sum of the areas of the rectangles. In symbols:
\[ \text {Area} \approx f(x_{1}) \Delta x + f(x_{2}) \Delta x + f(x_{3}) \Delta x + \ldots + f(x_{7}) \Delta x = \sum _{k = 1}^{7} f(x_{k}) \Delta x \]
Our ultimate goal is to find a formula for the area approximation as described above as a function of the number of rectangles \(n\)
and look to see what happens as \(n \rightarrow \infty \).
We first note that the right endpoints \(x_{k}\), are terms in an arithmetic sequence: the first right endpoint, \(x_{1}\) is \(\Delta x\) to the
right of \(a = x_{0}\), so \(x_{1} = x_{0} + \Delta x\); the second right endpoint, \(x_{2}\) is \(\Delta x\) units to the right of \(x_{1}\), so \(x_{2} = x_{1} + \Delta x\); the third right endpoint \(x_{3} = x_{2} + \Delta x\) and so on. In
general, \(x_{k} = x_{k-1} + \Delta x\), proving the \(x_{k}\) are terms of an arithmetic sequence with common difference \(d = \Delta x\). It follows that \(x_{k}\), the \(k\)th right
endpoint is \(k \Delta x\) units to the right of \(x_{0} = a\), so that \(x_{k} = a + k \Delta x\). We summarize the notation and formulas for right endpoint sums
below.
Summary of Formulas for Right Endpoint Sums, \(RS_{n}\)
Number of rectangles: \(n\)
Width of each rectangle: \(\Delta x = \frac {b-a}{n}\)
Right endpoint: \(x_{k} = a + k \Delta x\)
Height of \(k\)th rectangle: \(f(x_{k})\)
\(\text {Area} \approx RS_{n} = \text {the sum of the area of the rectangles} = \displaystyle {\sum _{k = 1}^{n} f(x_{k}) \Delta x_{k}}\)
Below we summarize some common summation formulas we’ll need when actually computing these sums. Formal proofs of
these require the machinery of Section ?? and are found there.
Summation Formulas
\(\displaystyle {\sum _{k=1}^{n} c = c n}\)
\(\displaystyle {\sum _{k=1}^{n} k = \frac {n(n+1)}{2}}\)
Consider \(f(x) = 4x-x^2\) over the interval \([0,4]\).
Graph \(f\) over this interval and shade the area between the graph of \(f\) and the \(x\)-axis.
Compute \(RS_{n}\) for \(n=4\) and \(n=8\). Interpret your results graphically.
Find a formula for \(RS_{n}\) in terms of \(n\) and determine \(\ds {\lim _{n \rightarrow \infty } RS_{n}}\).
Solution.
The graph of \(f(x) = 4x-x^2\) is a parabola with intercepts \((0,0)\) and \((4,0)\) with a vertex at \((2,4)\). Using the GeoGebra interactive below, we
can input \(f(x) = 4x-x^2\) and set the bounds of the interval \([a,b] = [0,4]\). GeoGebra numerically (Curious as to how does GeoGebra
does this? Research Quadrature Rules. ) determines the area to be (what appears to be) \(10.\overline {6}\) square units.
To find \(RS_{4}\), we begin by chopping up the interval \([0,4]\) into \(4\) equal pieces so each subinterval has length \(\Delta x = \frac {4}{4} = 1\) unit. Our right
endpoints are: \(x_{1} = 1\), \(x_{2} = 2\), \(x_{3} = 3\), and \(x_{4} = 4\). We find \(f(1) = 3\), \(f(2) = 4\), \(f(3) = 3\), and \(f(4) = 0\). Hence,
\[ RS_{4} = \sum _{k=1}^{4} f(x_{k}) \Delta x = f(x_{1}) \Delta x + f(x_{2}) \Delta x + f(x_{3}) \Delta x + f(x_{4}) \Delta x = (3)(1) + (4)(1) + 3(1) + (0)(1) = 10.\]
The GeoGebra interactive below shows us what this sum represents geometrically. After inputting our function \(f(x) = 4x- x^2\) and the
bounds \([a,b]=[0,4]\) we can adjust a slider to select the number of rectangles, \(n\), used to calculate the right endpoint sum, \(RS_{n}\). A
different slider allows us to select each rectangle, in sequence, and computes the area of the selected rectangle. The
value of \(RS{n}\) is the sum of the areas of these rectangles.
To find \(RS_{8}\), we divide the interval \([0,4]\) into \(8\) equal pieces, so each has length \(\Delta x = \frac {4}{8} = 0.5\) units. This produces the right endpoints: \(x_{1} = 0.5\), \(x_{2} = 1\), \(x_{3} = 1.5\), \(x_{4} = 2\), \(x_{5} = 2.5\), \(x_{6} = 3\), \(x_{7} = 3.5\), \(x_{8} = 4\).
In addition to the function values we used to compute \(RS_{4}\), we need \(f(0.5) = 1.75\), \(f(1.5) = 3.75\), \(f(2.5) = 3.75\), and \(f(3.5) = 1.75\). Hence,
\[ \begin{array}{rcl}
RS_{8} & = & \displaystyle {\sum _{k=1}^{8} f(x_{k}) \Delta x} \\ & = & f(x_{1}) \Delta x + f(x_{2}) \Delta x + f(x_{3}) \Delta x + f(x_{4}) \Delta x \\ & & + f(x_{5}) \Delta x + f(x_{6}) \Delta x + f(x_{7}) \Delta x + f(x_{8}) \Delta x \\ & = & (1.75)(0.5) + (3)(0.5) + (3.75)(0.5) + (4)(0.5) \\ & & + (3.75)(0.5) + (3)(0.5) + (1.75)(0.5) + (0)(0.5) \\ & = & 10.5 \\ \end{array} \]
Hence, the area under the graph \(f\) is approximately \(10.5\) square units as approximated by the sum of the rectangles above
on the right. (Again, since \(f(x_{8}) = f(4) = 0\), the eighth ‘rectangle’ has \(0\) height.) Once again, the GeoGebra interactive above is very
useful in visulizing this sum by adjusting the slider to \(n=8\).
To find a formula for \(RS_{n}\), we imagine dividing the interval \([0,4]\) into \(n\) equal pieces each of length \(\Delta x = \frac {4}{n}\). We have \(n\) right endpoints, \(x_{1}\), \(x_{2}\), …\(x_{n}\)
where \(x_{k} = 0 + k \Delta x = \frac {4k}{n}\). Since \(f(x) = 4x-x^2\),
Note we can partially check our answer at thus point by substituting \(n=4\) and \(n=8\) to our formula to \(RS_{n}\) to see if we recover our
answers from above. We get \(RS_{4} = \frac {32(4)^2 - 32}{3(4^2)}= \frac {480}{48} = 10\) and \(RS_{8} = \frac {32(8)^2 - 32}{3(8)^2} = \frac {2016}{192} = 10.5\), as required.
To find \(\ds {\lim _{n \rightarrow \infty } RS_{n}=\lim _{n \rightarrow \infty }}\) \(\frac {32n^2-32}{3n^2}\), we compare the leading term of the numerator and denominator. As \(n \rightarrow \infty \), \(\frac {32n^2-32}{3n^2} \approx \frac {32n^2}{3n^2} = \frac {32}{3}\). Hence, \(\ds {\lim _{n \rightarrow \infty } RS_{n}}=\) \(\frac {32}{3}\). Hence, as we use more and
more rectangles, (even though they become skinnier and skinner and hence, individually have smaller and
smaller areas …) the sum total of the area of those rectangles approaches \(\frac {32}{3} = 10.\overline {6}\) square units, which matches with what
GeoGebra had predicted in part 1. In Calculus, we more or less define the area under \(f\) to be \(\frac {32}{3}\) square
units. □
The reader is encouraged to revisit the GeoGebra interactives showcased in Example 6 and input different functions, different
bounds, and different numbers of rectangles for \(RS_{n}\). For example, what happens when you have a function which dips below the
\(x\)-axis? What does \(RS_{n}\) and \(\ds {\lim _{n \rightarrow \infty } RS_{n}}\) calculate then?
It is worth noting that, as with other examples in the text, Example 6 is more or less lifted straight out of a Calculus lecture.
That being said, the vast majority of the mechanics here involve precalculus notions. (The only Calculus bit is the limit
concept which I guess is precalculus now …) In general, the machinations in Calculus amount to applying the limit concept to
the mechanics of precalculus.
PreCalculus![[Picture]](Summation.online-3323eace4d315adbc5bdc7eb1cccb5d5.svg)
![[Picture]](Summation.online-704d27b8527387149df2e90789824f61.svg)