Download MAC 2312 RIEMANN SUMS 1.) By taking the limit of right Riemann

MAC 2312
RIEMANN SUMS
1.) By taking the limit of right Riemann sums with n equal subintervals, calculate the Riemann integral of f (x) = sin x over [0, π2 ].
(i) Begin by writing down the right Riemann sum Rn for f (x) = sin x
on [0, π2 ] with n equal subintervals. Explain why the following answer
is correct:
µ ¶
n
³π´ X
jπ
Rn =
·
sin
.
2n
2n
j=1
In the above formula, we used j as our summation index because we
are going to use i for something else.
µ ¶
n
X
jπ
The hardest part of the problem is to simplify the sum
sin
.
2n
j=1
You will do this by using De Moivre’s formula for complex numbers, a
special summation formula not covered in class, and some trigonometry.
√
(ii) De Moivre’s formula: Let θ be any real number, let i = −1
denote the imaginary unit, and let k be any integer (k = 0, ±1, ±2, . . . ).
Then the following formula holds:
(cos θ + i sin θ)k = cos(kθ) + i sin(kθ).
(1)
You will learn how to fully prove this formula when you study mathematical induction. For now, first observe that the formula holds for
k = 1. Additionally, using trigonometry (namely, double angle formulas) and multiplication of complex numbers, show that the formula (1)
holds for k = 2.
(iii) Summing a finite number of terms of a geometric sequence: Let q be a complex number such that q 6= 1 and let n be
a positive integer. Then the following formula holds:
(2)
n
X
µ
j
2
3
n
q = q + q + q + ··· + q = q ·
j=1
qn − 1
q−1
¶
.
Explain why the formula (2) holds for n = 1, n = 2 and n = 3. The
general case can be proven by using mathematical induction, as you
will see in more advanced classes. There are other ways of proving this
this formula; we will explain one proof when we get to Chapter 9.
1
2
(iv) Instead of trying to do
n
X
µ
sin
j=1
jπ
2n
¶
directly, you will do this sum
as a part of the following sum:
µ ¶
µ ¶¶
n µ
X
jπ
jπ
(3)
cos
+ i sin
,
2n
2n
j=1
√
where i = −1 is the imaginary unit.
The sum (3) adds n complex numbers of the form cos(θj ) + i sin(θj ),
jπ
where θj = 2n
. The value of the sum (3) is a complex number, which
we will call z. Explain why
µ ¶
n
X
jπ
(4)
sin
= Im z,
2n
j=1
where Im z denotes the imaginary part of the complex number z (for
instance, if z = 3 + i · 5, then Im z = 5).
(v) We will first find a simple formula (with no Σ symbols) for the
complex number
µ ¶
µ ¶¶
n µ
X
jπ
jπ
z=
cos
+ i sin
,
2n
2n
j=1
and then see what Im z looks like.
First, use a fact mentioned earlier in this handout to explain why
n µ
X
µ
cos
j=1
jπ
2n
¶
µ
+ i sin
jπ
2n
¶¶
=
n ³
X
j=1
cos
³π´
³ π ´´j
+ i sin
.
2n
2n
Next, labeling
³π´
³π´
q = cos
+ i sin
2n
2n
and using our special sum from part (iii) together with De Moivre’s
formula, carefully show that
n
X
−(cos α + sin α) + i(cos α − sin α)
qj =
,
(cos α − 1) + i sin α
j=1
where α =
π
2n
.
(vi) Now, using what you know about complex numbers, find the imaginary part of
−(cos α + sin α) + i(cos α − sin α)
.
z=
(cos α − 1) + i sin α
3
If you did everything correctly, you should get (you may want to do
some trig to get the formula that looks like what you see below):
Im z =
where α =
π
.
2n
1 − cos α + sin α
,
2 − 2 cos α
Now, by doing some trig, show that
1 − cos α + sin α
1
sin α
1
1
¡ ¢.
(5)
= +
= +
2 − 2 cos α
2 2 − 2 cos α
2 2 tan α2
π
Next, using formulas (4), (5) and plugging in α = 2n
, conclude that
µ ¶
n
X
jπ
1
1
¡ π ¢.
(6)
sin
= +
2n
2 2 tan 4n
j=1
(vii) By formula (6) and by part (i) you will get
µ ¶
n
π
³π´ X
jπ
π
4n
¡ π ¢.
Rn =
·
sin
=
+
2n
2n
4n
tan
4n
j=1
Calculate lim Rn . Explain your work carefully. To deal with the
n→∞
second term in Rn , you may want to use Example 9 on page 86 of your
textbook.
R π/2
(viii) Based on your calculations, what is 0 sin x dx?
2.) Using the same method as in 1.), calculate the Riemann integral
of f (x) = cos x over [0, π2 ].