5.2 Sigma Notation and Limits of Finite
Sums
Trig Example
Example
Estimate the area under the curve f (x) = sin x between x = 0
and x = π with n = 3.
Trig Example
Example
Estimate the area under the curve f (x) = sin x between x = 0
and x = π with n = 3.
What do we need to do first?
Trig Example
Example
Estimate the area under the curve f (x) = sin x between x = 0
and x = π with n = 3.
What do we need to do first? find ∆x
Trig Example
Example
Estimate the area under the curve f (x) = sin x between x = 0
and x = π with n = 3.
What do we need to do first? find ∆x
∆x =
π −0
3
=
π
3
Trig Example
Example
Estimate the area under the curve f (x) = sin x between x = 0
and x = π with n = 3.
What do we need to do first? find ∆x
∆x =
π −0
3
=
π
3
2π
Our intervals, therefore, will be [0, π3 ], [ π3 , 2π
3 ], and [ 3 , π ].
Trig Example
LHS:
π
π
2π
sin(0) + sin
+ sin
3
3
3
Trig Example
LHS:
π
π
2π
sin(0) + sin
+ sin
3
3
3
"
√
√ #
√
3
3
π 3
π
0+
+
=
=
3
2
2
3
RHS:
π
2π
π
sin
+ sin
+ sin(π )
3
3
3
Trig Example
LHS:
π
π
2π
sin(0) + sin
+ sin
3
3
3
"
√
√ #
√
3
3
π 3
π
0+
+
=
=
3
2
2
3
RHS:
π
2π
π
sin
+ sin
+ sin(π )
3
3
3
"√
#
√
√
π
π 3
3
3
=
+
+0 =
3
2
2
3
Trig Example
So, since the LHS and RHS sum are the same, we are pretty
confident that this is the exact area.
Trig Example
So, since the LHS and RHS sum are the same, we are pretty
confident that this is the exact area.
But, we would be wrong ... why?
Trig Example
So, since the LHS and RHS sum are the same, we are pretty
confident that this is the exact area.
But, we would be wrong ... why?
This is because of the small n. Even though the estimates are
equal, that is not enough to be guaranteed that we have the
exact value.
Trig Example
So, since the LHS and RHS sum are the same, we are pretty
confident that this is the exact area.
But, we would be wrong ... why?
This is because of the small n. Even though the estimates are
equal, that is not enough to be guaranteed that we have the
exact value.
Z π
0
π
sin(x) dx = −cos(x) = −(−1 − 1) = 2
0
Sigma Notation
We have a better way to express sums then to write everything
out. This is called Sigma notation.
n
∑ ak
k =1
where ak is the general term and k is the index.
Sigma Notation
We have a better way to express sums then to write everything
out. This is called Sigma notation.
n
∑ ak
k =1
where ak is the general term and k is the index.
So, in the last example,
LHS =
Sigma Notation
We have a better way to express sums then to write everything
out. This is called Sigma notation.
n
∑ ak
k =1
where ak is the general term and k is the index.
So, in the last example,
2
LHS =
∑ sin
k =0
kπ
3
Sigma Notation
We have a better way to express sums then to write everything
out. This is called Sigma notation.
n
∑ ak
k =1
where ak is the general term and k is the index.
So, in the last example,
2
LHS =
∑ sin
k =0
RHS =
kπ
3
Sigma Notation
We have a better way to express sums then to write everything
out. This is called Sigma notation.
n
∑ ak
k =1
where ak is the general term and k is the index.
So, in the last example,
2
kπ
LHS = ∑ sin
3
k =0
3
kπ
RHS = ∑ sin
3
k =1
Sigma Notation Examples
Example
List the terms of
4
∑ k2
k =1
Sigma Notation Examples
Example
List the terms of
4
∑ k2
k =1
4
∑ k2 = 1 + 4 + 9 + 16
k =1
Sigma Notation Examples
Example
List the terms of
3
1
∑ k+2
k =1
Sigma Notation Examples
Example
List the terms of
3
1
∑ k+2
k =1
3
1
1
1
1
∑ k+2 = 3 + 4 + 5
k =1
Rules for Summations
1
Sums
∑(an + bn ) = ∑ an + ∑ bn
Rules for Summations
1
2
Sums
Differences
∑(an + bn ) = ∑ an + ∑ bn
∑(an − bn ) = ∑ an − ∑ bn
Rules for Summations
1
2
3
Sums
Differences
∑(an + bn ) = ∑ an + ∑ bn
∑(an − bn ) = ∑ an − ∑ bn
Constant multiples
∑ kan = k ∑ an
for k 6= 0
Rules for Summations
1
2
3
Sums
Differences
∑(an + bn ) = ∑ an + ∑ bn
∑(an − bn ) = ∑ an − ∑ bn
Constant multiples
∑ kan = k ∑ an
for k 6= 0
4
Constant value
n
∑ c = cn
k =1
Important Summations
n
∑k=
k =1
n(n + 1)
2
Important Summations
n
∑k=
k =1
n
∑ k2 =
k =1
n(n + 1)
2
n(n + 1)(2n + 1)
6
Important Summations
n
∑k=
k =1
n
∑ k2 =
k =1
n
n(n + 1)
2
n(n + 1)(2n + 1)
6
n(n + 1)
∑k =
2
k =1
3
2
Using these rules and summations
Example
10
∑
k =1
2k2 − k
Using these rules and summations
Example
10
∑
k =1
10
2k2 − k
= 2 ∑ k2 +
k =1
10
∑k
k =1
Using these rules and summations
Example
10
∑
k =1
10
2k2 − k
= 2 ∑ k2 +
k =1
=2
10(11)(21)
6
10
∑k
k =1
10(11)
−
2
Using these rules and summations
Example
10
∑
k =1
10
2k2 − k
= 2 ∑ k2 +
k =1
=2
10(11)(21)
6
10
∑k
k =1
10(11)
−
2
= 715
Using these rules and summations
Example
20
∑ k2
k =4
Using these rules and summations
Example
20
∑ k2
=
20
k =4
3
k =1
k =1
∑ k2 − ∑ k2
Using these rules and summations
Example
20
∑ k2
=
=
20
k =4
3
k =1
k =1
∑ k2 − ∑ k2
20(21)(41) 3(4)(7)
−
6
6
Using these rules and summations
Example
20
∑ k2
=
=
20
k =4
3
k =1
k =1
∑ k2 − ∑ k2
20(21)(41) 3(4)(7)
−
6
6
= 2870 − 14
= 2856
Using these rules and summations
Example
7
∑
k =3
3k3 + 2k
Using these rules and summations
Example
7
∑
7
∑
k =1
3k3 + 2k −
k =3
2 ∑
k =1
3k3 + 2k
3k3 + 2k
Using these rules and summations
Example
7
∑
7
∑
k =1
7
3k3 + 2k −
7
k =3
2 ∑
k =1
2
3k3 + 2k
3k3 + 2k
2
3 ∑ k3 + 2 ∑ k − 3 ∑ k3 − 2 ∑ k
k =1
k =1
k =1
k =1
Using these rules and summations
Example
7
∑
7
∑
3k3 + 2k −
k =1
7
7
k =3
2 ∑
3k3 + 2k
3k3 + 2k
k =1
2
2
3 ∑ k3 + 2 ∑ k − 3 ∑ k3 − 2 ∑ k
k =1
"
=3
7(8)
2
2 #
+2
7(8)
2
k =1
"
−3
2(3)
2
k =1
2 #
k =1
−2
2(3)
2
Using these rules and summations
Example
7
∑
7
∑
3k3 + 2k −
k =1
7
7
k =3
2 ∑
3k3 + 2k
3k3 + 2k
k =1
2
2
3 ∑ k3 + 2 ∑ k − 3 ∑ k3 − 2 ∑ k
k =1
"
=3
7(8)
2
2 #
+2
7(8)
2
k =1
"
−3
2(3)
2
k =1
2 #
k =1
−2
2(3)
2
= 2375
More Properties
Theorem
lim
n→ ∞
1 n
1=1
n k∑
=1
More Properties
Theorem
lim
1 n
1=1
n k∑
=1
lim
1
n2
n→ ∞
n→ ∞
n
1
∑k= 2
k =1
More Properties
Theorem
lim
1 n
1=1
n k∑
=1
lim
1
n2
n→ ∞
n→ ∞
lim
n→ ∞
1
n3
n
1
∑k= 2
k =1
n
1
∑ k2 = 3
k =1
More Properties
Theorem
lim
1 n
1=1
n k∑
=1
lim
1
n2
n→ ∞
n→ ∞
lim
n→ ∞
lim
n→ ∞
n
1
∑k= 2
k =1
n
1
n3
1
k =1
1
n4
∑ k3 = 4
∑ k2 = 3
n
k =1
1
Riemann Sums
To find the approximate area under a curve f (x), we break the
region into rectangular regions (since we can easily find the
area of a rectangle) and we add their areas together.
x0 x1
x2 x3 x4
x5
Riemann Sums
We partition the interval [a, b] into n rectangles with endpoints
a = x0 , x1 , . . . , xn−1 , xn = b
Riemann Sums
We partition the interval [a, b] into n rectangles with endpoints
a = x0 , x1 , . . . , xn−1 , xn = b
This gives the subintervals
∆x1 = x1 − x0 , ∆x2 = x2 − x1 , . . . , ∆xn = xn − xn−1
Riemann Sums
We partition the interval [a, b] into n rectangles with endpoints
a = x0 , x1 , . . . , xn−1 , xn = b
This gives the subintervals
∆x1 = x1 − x0 , ∆x2 = x2 − x1 , . . . , ∆xn = xn − xn−1
Note: Nothing says we have to partition the interval into
a
subintervals of the same size. When we do, ∆xi = ∆x = b−
n
and we say we have a regular partition.
Riemann Sums
If we did use a consistant ∆x, we would have something like
x0 x1 x2 x3 x4 x5 x6 x7 x8
Riemann Sums
In general, we have
n
RHS = f (x1 )∆x + · · · + f (xn )∆x =
∑ f (xk )∆x
k =1
LHS = f (x0 )∆x + · · · + f (xn−1 )∆x =
n−1
∑ f (xk )∆x
k =0
where each [xk−1 , xk ] is the kth subinterval. We are dividing
a ≤ x ≤ b into n subintervals, each of size n1 .
Riemann Sums
If we have an infinite number of subintervals, then we get the
overestimates and the underestimates to be the same, we get
the over and under estimates to become the same.
n−1
lim
n→ ∞
∑
k =0
f (xk )∆x = lim
n→ ∞
n
Z b
k =1
a
∑ f (xk )∆x =
f (x) dx
Riemann Sums
If we have an infinite number of subintervals, then we get the
overestimates and the underestimates to be the same, we get
the over and under estimates to become the same.
n−1
lim
n→ ∞
∑
k =0
f (xk )∆x = lim
n→ ∞
n
Z b
k =1
a
∑ f (xk )∆x =
f (x) dx
We call a, b the limits of integration and we call f (x) the
integrand.
Formal Definition of the Integral
Let f (x) be defined on [a, b]. The number I is the definite
integral of f over [a, b] and I is the limit of the Riemann sum
provided:
Given e > 0, there exists δ > 0 such that for all partitions
p = {x0 , x1 , . . . , xn } of [a, b] with ||ρ|| < δ and any choice
of ck in {xk−1 , xk }, we have
n
∑ f (xk )∆xk − I < e
k =1
Formal Definition of the Integral
Let f (x) be defined on [a, b]. The number I is the definite
integral of f over [a, b] and I is the limit of the Riemann sum
provided:
Given e > 0, there exists δ > 0 such that for all partitions
p = {x0 , x1 , . . . , xn } of [a, b] with ||ρ|| < δ and any choice
of ck in {xk−1 , xk }, we have
n
∑ f (xk )∆xk − I < e
k =1
||ρ|| is the norm of the partition, defined as the largest of the
subinterval lengths.
Riemann Sums
That is,
n
∑ f (xk )∆xk
lim
||ρ||→∞ k=1
is the same as
Z b
a
f (x) dx
Sums and Integrals
Example
For f (x) = cos(x2 ) on [π, 3π ], the area bound by the curve and
the x-axis could be expressed as
n
lim
n→ ∞
or
∑ cos(xk )2 ∆xk
k =1
Z 3π
π
cos(x2 ) dx
Using n Subdivisions to Find a Sum
Example
Find the RHS for f (x) = x2 − 2 by dividing [0, 1] into n equal
subintervals. Then, take the limit as n → ∞ to calculate the area
between the x-axis and the curve.
Using n Subdivisions to Find a Sum
Example
Find the RHS for f (x) = x2 − 2 by dividing [0, 1] into n equal
subintervals. Then, take the limit as n → ∞ to calculate the area
between the x-axis and the curve.
Since we want n equal subintervals, ∆x = n1 .
Using n Subdivisions to Find a Sum
Example
Find the RHS for f (x) = x2 − 2 by dividing [0, 1] into n equal
subintervals. Then, take the limit as n → ∞ to calculate the area
between the x-axis and the curve.
Since we want n equal subintervals, ∆x = n1 .
1
Our partition would be [0, n1 ], [ n1 , n2 ], . . . , [ n−
n , 1]
Using n Subdivisions to Find a Sum
Then we get the following:
n 1
1 1
2 1
RHS = f
+f
+···+f
n n
n n
n n
Using n Subdivisions to Find a Sum
Then we get the following:
n 1
1 1
2 1
RHS = f
+f
+···+f
n n
n n
n n
n
k 1
= ∑f
n n
k −1
Using n Subdivisions to Find a Sum
Then we get the following:
n 1
1 1
2 1
RHS = f
+f
+···+f
n n
n n
n n
n
k 1
= ∑f
n n
k −1
!
n
k 2
1
=∑
−2
n
n
k −1
Using n Subdivisions to Find a Sum
Then we get the following:
n 1
1 1
2 1
RHS = f
+f
+···+f
n n
n n
n n
n
k 1
= ∑f
n n
k −1
!
n
k 2
1
=∑
−2
n
n
k −1
=
1
n3
n
1
n
∑ k2 − n ∑ 2
k =1
k =1
Using n Subdivisions to Find a Sum
Then we get the following:
n 1
1 1
2 1
RHS = f
+f
+···+f
n n
n n
n n
n
k 1
= ∑f
n n
k −1
!
n
k 2
1
=∑
−2
n
n
k −1
=
1
n3
n
1
n
∑ k2 − n ∑ 2
k =1
k =1
n(n + 1)(2n + 1)
= lim
−2
n→ ∞
6n3
Using n Subdivisions to Find a Sum
Then we get the following:
n 1
1 1
2 1
RHS = f
+f
+···+f
n n
n n
n n
n
k 1
= ∑f
n n
k −1
!
n
k 2
1
=∑
−2
n
n
k −1
=
1
n3
n
1
n
∑ k2 − n ∑ 2
k =1
k =1
n(n + 1)(2n + 1)
= lim
−2
n→ ∞
6n3
2n3 + 3n2 + n
= lim
−2
n→ ∞
6n3
Using n Subdivisions to Find a Sum
When we total this, we get RHS =
1
3
− 2 = − 53 .
Using n Subdivisions to Find a Sum
When we total this, we get RHS =
1
3
− 2 = − 53 .
Notice that the estimate of the sum is negative. What does this
mean?
Using n Subdivisions to Find a Sum
When we total this, we get RHS =
1
3
− 2 = − 53 .
Notice that the estimate of the sum is negative. What does this
mean?
It means that the region bound by the curve and the x-axis is
below the x-axis. Also note that we were asked to find the RHS
and not to find the area of the region ...
Example
Write the Riemann sum for f (x) = 2x over the interval from
[1, 4] with n equal subdivisions.
Example
Write the Riemann sum for f (x) = 2x over the interval from
[1, 4] with n equal subdivisions.
n
∑f
k =1
3k
1+
n
3
n
Example
Write the Riemann sum for f (x) = 2x over the interval from
[1, 4] with n equal subdivisions.
n
3k 3
∑ f 1+ n n
k =1
n
3k 3
= ∑ 2 1+
n n
k =1
Using n Subdivisions to Find a Sum
If we wanted to continue so that we could solve, we would
have
3 n
6k
lim ∑ 2 +
n→ ∞ n
n
k =1
Using n Subdivisions to Find a Sum
If we wanted to continue so that we could solve, we would
have
3 n
6k
lim ∑ 2 +
n→ ∞ n
n
k =1
"
#
3 n
3 n 6k
= lim
2+ ∑
n→ ∞
n k∑
n k =1 n
=1
Using n Subdivisions to Find a Sum
If we wanted to continue so that we could solve, we would
have
3 n
6k
lim ∑ 2 +
n→ ∞ n
n
k =1
"
#
3 n
3 n 6k
= lim
2+ ∑
n→ ∞
n k∑
n k =1 n
=1
= 3(2) + lim
n→ ∞
18 n
k
n2 k∑
=1
Using n Subdivisions to Find a Sum
If we wanted to continue so that we could solve, we would
have
3 n
6k
lim ∑ 2 +
n→ ∞ n
n
k =1
"
#
3 n
3 n 6k
= lim
2+ ∑
n→ ∞
n k∑
n k =1 n
=1
18 n
∑k
n→∞ n2
k =1
18 n(n + 1)
= 3(2) + lim 2
n→ ∞ n
2
= 3(2) + lim
Using n Subdivisions to Find a Sum
If we wanted to continue so that we could solve, we would
have
3 n
6k
lim ∑ 2 +
n→ ∞ n
n
k =1
"
#
3 n
3 n 6k
= lim
2+ ∑
n→ ∞
n k∑
n k =1 n
=1
18 n
∑k
n→∞ n2
k =1
18 n(n + 1)
= 3(2) + lim 2
n→ ∞ n
2
1
= 6 + 18
= 15
2
= 3(2) + lim
Solving with Geometry
Example
Z 1 p
−1
1 − x2 dx
Solving with Geometry
Example
Z 1 p
−1
1 − x2 dx
We can solve this, even though we don’t know what the
antiderivative is, by using geometry. What does this integrand
represent?
Solving with Geometry
Example
Z 1 p
−1
1 − x2 dx
We can solve this, even though we don’t know what the
antiderivative is, by using geometry. What does this integrand
represent?
Solving with Geometry
We need to note that this is the top half of the unit circle.
Solving with Geometry
We need to note that this is the top half of the unit circle.
What is the area of a circle?
Solving with Geometry
We need to note that this is the top half of the unit circle.
What is the area of a circle? A(r) = πr2
Solving with Geometry
We need to note that this is the top half of the unit circle.
What is the area of a circle? A(r) = πr2
The circle has a radius of 1 and we only need half the circle,
so ...
Solving with Geometry
We need to note that this is the top half of the unit circle.
What is the area of a circle? A(r) = πr2
The circle has a radius of 1 and we only need half the circle,
so ...
Z 1 p
−1
1 − x2 dx =
π
2