MATH 472 002/992
Final Exam
December 16, 2020
Read the following instructions carefully:
•
There are
seven problems
and you have
120 minutes
.
•
Each submission must be a
single pdf
file including your solutions in the order they were
assigned.
•
Do not use red ink or color
in your submission.
•
Write your name at the top right of each page.
•
Each problem should start on a fresh page in your submission pdf.
•
You may use
one doublesided sheet
(
8
.
5
×
11
inches) of notes in your
own handwriting
.
No other notes or books are allowed.
•
You may use a calculator, but all intermediate steps in your calculations must be compre
hensible from your notes. Simplify your results as far as possible.
•
All results have to be
justified
by appropriate
intermediate steps
and/or
arguments
.
You can, however, use standard formulas and results known from the lecture without deriva
tion (unless explicitly stated otherwise).
•
You will have
15 minutes to upload your submission to Canvas after the end of
the exam
.
Good luck!
Sheet 1 of 15
Problem 1
(12 Points)
Name:
Assume that you are to use Chebyshev interpolation to find a degree 2 interpolating polynomial
Q
2
(
x
)
that approximates the function
f
(
x
) =
x
1
/
3
(=
3
√
x
)
on the interval
[3
,
4]
.
(a)
(5 Points) Write down the
(
x, y
)
points that will serve as interpolation nodes for
Q
2
.
(b)
(4 Points) Find a worstcase estimate for the error

f
(
x
)

Q
2
(
x
)

that is valid for all
x
in
the interval
[3
,
4]
.
(c)
(3 Points) How many digits after the decimal point will be correct when
Q
2
is used to
approximate
f
?
Solution:
(a)
For a
2

degree interpolating polynomial, we need
n
= 3
nodes.
(1P)
The Chebyshev inter
polation nodes on
[

1
,
1]
are given by
¯
x
1
:= cos
(
π
6
)
=
√
3
2
,
¯
x
2
:= cos
(
3
π
6
)
= cos
(
π
2
)
= 0
and
¯
x
3
:= cos
(
5
π
6
)
=

√
3
2
.
(1P)
Therefore, for the interval
[
a, b
] = [3
,
4]
the interpolating nodes are
x
i
:=
a
+
b
2
+
b

a
2
¯
x
i
=
7
2
+
1
2
¯
x
i
,
for
i
= 1
,
2
,
3
.
(2P)
Their approximations are
x
1
≈
3
.
933
,
x
2
= 3
.
5
and
x
3
≈
3
.
067
.
In total, the required nodes are
(
x
1
, f
(
x
1
)
)
= (3
.
933
,
3
√
3
.
933) = (3
.
933
,
1
.
578528)
,
(
x
2
, f
(
x
2
)
)
= (3
.
5
,
3
√
3
.
5) = (3
.
5
,
1
.
5183)
and
(1P)
(
x
3
, f
(
x
3
)
)
= (3
.
067
,
3
√
3
.
067) = (3
.
067
,
1
.
4529)
.
(b)
The interpolation error formula gives
sup
x
∈
[3
,
4]

f
(
x
)

Q
2
(
x
)

= sup
x
∈
[3
,
4]
n

(
x

x
1
)(
x

x
2
)(
x

x
3
)

3!

f
(3)
(
x
)

o
(1P)
≤
1
2
5
×
3!
×
0
.
019784
≈
0
.
000103
.
(1P)
where we have used that

(
x

x
1
)(
x

x
2
)(
x

x
3
)
 ≤
b

a
2
3
2
2
=
1
2
5
(1P)
and
f
(3)
(
x
) =
1
3
x

2
3
00
=

2
9
x

5
3
0
=
10
27
x

8
3
with
0
≤
f
(3)
(
x
)
≤
f
(3)
(3) = 0
.
019784
for
x
∈
[3
,
4]
(1P)
.