SOME PROBLEMS FOR THE FINAL EXAM PREPARATION
FINAL FALL, 2001, INSTRUCTOR: LUDMIL ZIKATANOV
Problem:
a) Find a polynomial q3 (x) such that q3 (2) = 1 and
Z 2
q3 (x)xk dx = 0 for k = 0, 1, 2
−2
R2
R1
R2
(Hint: −2 x dx = −2 x3 dx = −2 x5 dx = 0). Also watch out the interval is not [−1, 1],
but this is very similar to what we have done in class.
Solution: We know that q3 (x) should have the form
q3 (x) = ax3 + bx2 + cx + d.
We use the hint, the equations:
Z 2
q3 (x)xk dx = 0 for k = 0, 1, 2
−2
and also q3 (2) = 1 to find a, b and c and d.
16
k=0
4d + b = 0.
3
64
16
d + b = 0.
k=2
3
5
16
64
a + c = 0.
k=1
5
3
q3 (2) = 1 8a + 4b + 2c + d = 1.
5
, c = − 43 . Finally
First two equations give b = d = 0 and third and forth a = 16
x
(5x2 − 12
q3 (x) =
16
b) Given a function f on [−1, 1]. Let p2 (x) be the Lagrange interpolation polynomial using the zeros of the polynomial q3 as above. We have the following Gaussian
Quadrature Rule:
Z
2
Z
2
10
f
f (x) dx ≈
p2 (x) dx =
9
−2
−2
Verify that A0 =
r r 10
3
3
−2
+ A1 f 0 + f 2
.
5
9
5
16
9 .
Solution:
Since when f (x) = 1 the formulae has to be exact we get that A1 + 20
9 = 4, i.e.
.
A1 = 16
9
Another solution is obtained via the definition.
R1
c) Using the above Gaussian quadrature rule to approximately evaluate −1 x6 dx
(with a proper change of variable)
1
2
FINAL FALL, 2001, INSTRUCTOR: LUDMIL ZIKATANOV
Z 1
Z
1 2
t
Solution: Change the variables t = 2x and get
x6 dx =
f ( ) dt ≈
2
2
−1
−2
r !6
r !6
5
6
8
5 27
3
3
5
−
=
.
+ ·0+
=2
9
5
9
9
5
9 125
25
Problem:
Find the formula with which to replace ??? below so that the
resulting function s(x) is a natural cubic spline on the interval [−3, 1] with knots
at −3, −2, −1, 0, and 1.

x ∈ [−3, −2],
1 − 2x + 9x2 + x3 ,



9 + 10x + 15x2 + 2x3 , x ∈ [−2, −1],
s(x) =
x ∈ [−1, 0],
6 + x + 6x2 − x3 ,



???,
x ∈ [0, 1].
Solution: We can write the question marks as

1 − 2x + 9x2 + x3 ,



9 + 10x + 15x2 + 2x3 ,
s(x) =
6 + x + 6x2 − x3 ,



ax3 + bx2 + cx + d,
a third degree polynomial
x ∈ [−3, −2],
x ∈ [−2, −1],
x ∈ [−1, 0],
x ∈ [0, 1].
We only need to determine a, b, c, d so that our function satisfies the definition of
a spline. In the case we consider this means that, we have to check the continuity
of the function, first and second derivative at x = 0. Also we will need S 00 (1) = 0
(remember, we are looking for natural cubic spline, i.e. S 00 (−3) = 0 = S 00 (1) = 0).
To find a, b, c, d so that S, S 0 and S 00 are continuous at x = 0 we need to satisfy the
equations:
i.e.
d = 6.
S+ (0) = S− (0)
0
0
(0) = S−
(0)
i.e.
c = 1.
S+
00
00
(0) = S−
(0)
i.e.
b = 6.
S+
6a + 12 = 0 a = −2.
S 00 (1) = 0
Problem: Find the LU decomposition of the matrix


4 2 0 0
2 3 1 0


0 2 4 1 .
0 0 2 3
Solution: You can solve this problem by using the procedure we had used in
class, or by using the procedure described in the book. You can also determine L
and U in the following way:

 


0
d1 u1 0
4 2 0 0
1 0 0 0
l1 1 0 0  0 d2 u2 0  2 3 1 0

 


 0 l2 1 0  0 0 d3 u3  = 0 2 4 1 .
0 0
0 d4
0 0 l3 1
0 0 2 3
Now just multiply the corresponding row and column to get the corresponding
entry in the matrix on the right. In what follows m × n means that we multiply
m-th row of L by the n-th row of U :
1 × 1 d1 = 4.
1 × 2 u1 = 2
2 × 1 l1 d1 = 2, l1 = 1/2.
SOME PROBLEMS FOR THE FINAL EXAM PREPARATION
2×2
2×3
3×2
3×3
3×4
4×3
4×4
3
l1 u1 + d2 = 3, d2 = 2.
u2 = 1
2l2 = 2, l2 = 1
l2 u2 + d3 = 4, d3 = 3
u3 = 1
l3 d3 = 2, l3 = 2/3.
l3 u3 + d4 = 3, d4 = 7/3.
Problem:
Find both Newton’s form and Lagrange’s form of the polynomial
of lowest degree passing through the points (−2, 9), (−1, 0), (1, 0), (2, 15). (You
don’t need to simplify.)
-2 -1 1 2
Solution We first write the interpolation table:
9 0 0 15
To determine the Lagrange form we have to write down the cardinal functions
`i (x), i = 0, 1, 2, 3. Then the polynomial will be
L(x) = 9`0 (x) + 0`1 (x) + 0`2 (x) + 15`3 (x) = 9`0 (x) + 15`3 (x).
As you can see we only need `0 (x) and `3 (x). We have
(x + 1)(x − 1)(x − 2)
(x + 1)(x − 1)(x − 2)
=−
(−2 + 1)(−2 − 1)(−2 − 2)
12
(x + 2)(x + 1)(x − 1)
(x + 2)(x + 1)(x − 1)
=
.
`3 (x) =
(2 + 2)(2 + 1)(2 − 1)
12
`0 (x) =
The polynomial then is
L(x) = −9
(x + 2)(x + 1)(x − 1)
(x + 1)(x − 1)(x − 2)
+ 15
.
12
12
The Newtons form of the polynomial will be (n = 4)
pn (x) = f [x0 ] + f [x0 , x1 ](x − x0 ) + . . . + f [x0 , x1 , . . . , xn ]
n−1
Y
(x − xk ).
k=1
We construct the divided difference table
x
−2
−1
1
2
f [·] f [·, ·] f [·, ·, ·] f [·, ·, ·, ·]
9
9
−3
0
0
2
0
5
15
15
Problem:
Suppose that f (x) is a smooth function. Give an estimate (as a
power of h) on the difference between f [−h, 0, h] and 12 f 00 (0), where f [−h, 0, h] is
the second divided difference of f . Try to justify your answer.
Solution: The second divided difference is:
f [−h, 0, h] =
f (−h) − 2f (0) + f (h)
.
2h2
4
FINAL FALL, 2001, INSTRUCTOR: LUDMIL ZIKATANOV
We also know that
f (x − h) − 2f (x) + f (x + h)
+ O(h2 ).
h2
(We have done this in class several times, one can use Taylor formula to derive
it). Take x = 0. A short answer to this problem is: The difference is estimated by
h2 . A longer answer (with justification) would use the derivation of:
f 00 (x) =
f (x − h) − 2f (x) + f (x + h)
+ O(h2 ).
h2
by Taylor formula, which follows:
f 00 (x) =
f (x − h) =
2f (x) =
f (x − h) =
f (x) − hf 0 (x) +
2f (x)
f (x) + hf 0 (x) +
h2 00
2 f (x)
−
h3 00
6 f (x)
+ O(h4 ).
h2 00
2 f (x)
+
h3 00
6 f (x)
+ O(h4 ).
Then do “first equation”+“third equation”- “second equation” and get the expression for f 00 .
Problem: Find the LU decomposition of the matrix


4 −2 6
−2 2 −2
6 −2 15
Solution: Again there are several ways to determine this LU decomposition.
A straightforward one is to determine lij , di and uij by multiplying the matrices
below:

 


d1 u12 u13
4 −2 6
1
0 0
l21 1 0  0 d2 u23  = −2 2 −2 .
0
0
d3
l31 l32 1
6 −2 15
Follow the following rule when you try to write the corresponding equations for
lij , uij , di :
First determine d1 . Then l21 followed by u12 . Then d2 . Now proceed with the
third row in L and third column in U , i.e. First determine l31 , then l32 then u31
and u32 . Finally get d3 . This precisely follows the procedure we derived in class.
The answer is:




1
0 0
4 −2 6
L = −1/2 1 0 , U = 0 1 1 .
3/2 1 1
0 0 5
Problem: Write down the formulas for the Newton and secant method applied
to the function
f (x) = exp(cos x) − x3 . (You don’t need to simplify.)
To solve this problem one needs to know Newton’s and secant formulae. Newton
formula approximates the root, by solving linear equation. This linear equation
is is nothing else but the equation of the tangent line passing through (xk , f (xk ),
which looks in the following way:
y = f (xk ) + f 0 (xk )(x − xk ).
Therefore xk+1 will be the solution of the above equation, namely, you put y = 0
and solve for x. This gives the Newton’s formula:
xk+1 = xk −
f (xk )
,
f 0 (xk )
x0 given initial guess.
SOME PROBLEMS FOR THE FINAL EXAM PREPARATION
5
The secant formula is obtained from this one by replacing f 0 (xk ) by a divided
difference (one needs two points to get the divided difference, xk and xk−1 . Here
is the secant method formula:
Given x0 and x1 the sequence of the approximations to the root of f (x) = 0 is
obtained via:
f (xk )(xk − xk−1
.
xk+1 = xk −
f (xk ) − f (xk−1 )
To solve this problem you need just to substitute in these two formulae.
Problem: Given the system of linear equation with a tridiagonal matrix:
 
 


x1
b1
α1 β1
0
  x2   b2 
 γ1 α2
β2
 
 


  ..   .. 

.
.
.
..
..
..
  .  =  . .

 
 



γn−2 αn−1 βn−1   xn−1   bn−1 
0
γn−1
αn
xn
bn
Derive general formulae for the solution.
Solution: This is in the book. You can proceed as follows: Perform couple of
Gaussian-elimination steps and then observe the pattern and write the formulae for
xi .
1
Problem: Derive a secant iteration formula, to compute 3 3 . Then set x0 = 1
and x1 = 2 and compute one step of the secant iteration.
Solution: This is a simple application of the secant method for finding the root
of the equation
x3 − 3 = 0.
Problem:
R4
(1) Write down the composite trapezoid rule and use it to approximate 0 2x dx
using two and four equal-length intervals.
(2) Apply the Romberg extrapolation technique to get a better approximation
R4
to 0 2x dx.
Solution: For part (a), the solution is straightforward by using composite trapezoidal rule with n = 1 and n = 2 (formula (1) page 211). You have to compute R(1, 0) and R(2, 0). For part (b) use formula (2) page 211 to get R(2, 1) =
R(2, 0) + 13 (R(2, 0) − R(1, 0).
Problem:
(1) Find a polynomial L2 (x) of degree two such that
Z 1
L2 (x)xk dx = 0, k = 0, 1.
L2 (1) = 1;
0
(note that the interval is now [0, 1]).
(2) Let x0 and x1 be the roots of L2 (x). Derive an integration formula of the
form
Z 1
f (x)dx ≈ A0 f (x0 ) + A1 f (x1 )
0
such that it is accurate for polynomials of as high a degree as possible; what
is the highest degree of polynomials for which the formula is exact ?
6
FINAL FALL, 2001, INSTRUCTOR: LUDMIL ZIKATANOV
Solution: This problem is similar to the first one (but simpler). Try to solve it
by following the procedure from the solution of the first problem. Pay attention to
the fact that the interval is not [−1, 1] but all the integration should be performed
on [0, 1]. It is also possible to solve this problem by using what you know for the
interval [−1, 1]. A simple change of variables will yield the values of x0 and x1 .
Then the polynomial will be c(x − x0 )(x − x1 ), where the constant is determined
by the condition L2 (1) = 1.
Problem: Write a formulae to compute to solve a n × n backward tridiagonal
system of linear equations:
 

 

a1 d1
b1
x1

a2 d2 c2 


x2 
 

  b2 
 

a
d3 c3
  b3 
3
x
 

3
 

.

···
··· ···
..  =  .. 
 




  .   . 

an−2 dn−2 cn−2

 

 xn−1  bn−1 
an−1 dn−1 cn−1
xn
bn
dn
cn
Solution: This problem is not complicated at all. You start by permuting the rows
of the matrix (the n-th row becomes 1-st, n − 1-st becomes second, etc. This will
give you a tridiagonal matrix. Get the formulae for tridiagonal matrix and write
the solution.
Problem:
(1) Find the LU factorization for the matrix


6 −1 8
A = 2 −3 3
2 −1 2
(2) Use the above LU factorization to solve the equation
Ax = b with b = (6, 5, 28)t.
Solution: The LU decomposition is:



1
0 0
6
L = 1/3 1 0 , U = 0
1/3 1/4 1
0

−1
8
−8/3 1/3  .
0
−3/4
The solution of the corresponding system is x = (1, 2, 3)t . You can find this
solution by first solving
Ly = b,
and find that y = (28, −13/3, −9/4). Then you obtain x such that U x = y.
Problem:
Q0 (−1) = 0
Find a quadratic spline interpolating the table and such that
t −1 0 1
y 13 7 9
SOME PROBLEMS FOR THE FINAL EXAM PREPARATION
7
Solution: Proceed by definition. You need to find a function Q(x) which is
piece-wise quadratic (namely a quadratic polynomial Q1 (x) on [−1, 0] and another
quadratic polynomial Q2 (x) on [0, 1]) such that they satisfy the interpolation and
Q1 (−1) = 13,
Q1 (0) = Q2 (0) = 7,
and the smoothness condition
0(0)
Q2 (1) = 9.
0(0)
Q1 = Q2 .
Both Q1 and Q2 are quadratic polynomials as stated above, and therefore Q(x)
should have the form:
a1 x2 + b1 x + c1 , x ∈ [−1, 0]
Q(x) =
a2 x2 + b2 x + c2 , x ∈ (0, 1]
Let us use the interpolation and smoothness conditions and try to determine ai , bi
and ci . We have
a1 − b1 + c1 = 13
c 1 = c2 = 7
a 2 + b 2 + c2 = 9
b1 = b2 , from the condition for the derivatives.
−2a1 + b1 = 0, from Q0 (x) = 0.
Let us set b1 = b2 = b. Then we have the following equations:
a1 − b = 6,
a2 + b = 2,
b = 2a1 .
This gives the solution:
a1 = −6, b1 = b2 = b = −12, c1 = 7
c2 = 7.
a2 = 14,
Finally the spline is
Q(x) =
−6x2 − 12x + 7, x ∈ [−1, 0]
14x2 − 12x + 7. x ∈ (0, 1]
Problem:
Determine a, b and c and d so that
3
0≤x<3
x ,
S(x) =
a + b(x − 3) + c(x − 3)2 + d(x − 3)3 , 3 ≤ x ≤ 4
is a natural cubic spline.
This we have solved in class with 2 instead of 3. Here one more coefficient d is
unknown, but you have one more condition, namely the spline has to be natural,
i.e. S 00 (4) = 0.