Answers to Exam 2 Spring 05:
Answers to Exam 2 Fall 03:
Answers to Exam 2 Spring 2002:
You may use a formula sheet: postscript version , and version in pdf format .
n
---
p(x) = > l (x) f
--- i i
i=0
with
n x - x
--- j
l (x) = | | -----
i j=0 x - x
j/=i i j
satisfies
p(x ) = f for i = 0,1,..,n.
i i
Answer:
Observe that l_i(x_j) = 0 for j /= i, because there is one factor in the numerator that is zero. Furthermore, we see that l_i(x_i) = 1 because numerator and denominator are equal to each other. These observations imply p(x_i) = f_i, i = 0,1,2,..,n.
Answers:
0 1
-3 - 1
------- = -4
1 - 0
-3-(-4)
1 -3 ------- = 1
-5 - 1 2-1
------- = -3 1-1
2 - 0 ----- = 0
-2-(-4) 3-2
2 -5 ------- = 1
-5 - 1 3-1
------- = -2
3 - 0
3 -5
The Newton form is
f_0 + f_01*(x-x[0]) + f_012*(x-x[0])*(x-x[1])
+ f_0123*(x-x[0])*(x-x[1])*(x-x[2])
= 1 + (-4)*(x-0) + 1*(x-0)*(x-1) + 0*(x-0)*(x-1)*(x-2)
= 1 - 4*x + x*(x-1)
0 1
p
01
1 -3 p
p 012
12 p
2 -5 p 0123
p 123
23
3 -5
with
(1/2 - 1)*1 - (1/2 - 0)*(-3)
p = ------------------------------ = -1
01 0 - 1
(1/2 - 2)*(-3) - (1/2 - 1)*(-5)
p = --------------------------------- = -2
12 1 - 2
(1/2 - 3)*(-5) - (1/2 - 2)*(-5)
p = --------------------------------- = -5
23 2 - 3
(1/2 - 2)*(-1) - (1/2 - 0)*(-2)
p = --------------------------------- = -5/4
012 0 - 2
(1/2 - 3)*(-2) - (1/2 - 1)*(-5)
p = --------------------------------- = -5/4
123 1 - 3
(1/2 - 3)*(-5/4) - (1/2 - 0)*(-5/4)
p = ------------------------------------- = -5/4
0123 0 - 3
Thus p(0.5) = -5/4.
Notice that, since we are interpolating a second degree polynomial
with four points, we already see the final result appearing twice
in the next to last column.
1 = a*0 + b
-3 = a*1 + b
-5 = a*2 + b
-5 = a*3 + b
or in matrix notation:
[ 0 1 ] [ 1 ]
[ 1 1 ] [ a ] = [ -3 ]
[ 2 1 ] [ b ] [ -5 ]
[ 3 1 ] [ -5 ]
X [ a ] = f
[ b ]
To solve the system in the least squares sense,
we set up the normal equations:
T
X *X = [ 0 1 2 3 ] [ 0 1 ] = [ 14 6 ]
[ 1 1 1 1 ] [ 1 1 ] [ 6 4 ]
[ 2 1 ]
[ 3 1 ]
T
X *f = [ 0 1 2 3 ] [ 1 ] = [ -28 ]
[ 1 1 1 1 ] [ -3 ] [ -12 ]
[ -5 ]
[ -5 ]
T T
The solution to the system X *X [ a ] = X *f
[ b ]
is a = -2 and b = 0.
1 3 2 5 17 7 62 9 10
tan(x) = x + --- x + --- x + ---- x + ----- x + O(x )
3 15 315 2835
Use this Maclaurin expansion to construct a Padé approximation
for tan(x) as a quotient of two quadrics.
Answer:
( 2 1 3 4 ) ( 2 ) ( 2 )
( 0 + x + 0 x + --- x + 0 x ) * ( 1 + b x + b x ) - ( a + a x + a x )
( 3 ) ( 1 2 ) ( 0 1 2 )
expanding product:
2 3
x + b x + b x
1 2
1 3 1 4 1 5
+ --- x + --- b x + --- b x
3 3 1 3 2
setting coefficients to consecutive powers of x to zero:
0
x : 0 - a = 0
0
1
x : 1 - a = 0
1
2
x : b - a = 0
1 2
3 1
x : b + --- = 0
2 3
4 1
x : --- b = 0
3 1
we solve this system:
1
b = 0 b = - --- a = 0 a = 1 a = 0
1 2 3 0 1 2
Thus our approximation is
x
---------------
1 2
1 - --- x
3
2
x + 3 x - 1
r(x) = --------------
2
x - 1
Answers :
2 | 2
x + 3 x - 1 | x - 1
2 |----------
- ( x - 1) 1
-----------------
3 x
2 2
x + 3 x - 1 = 1 ( x - 1 ) + 3 x
3 x 3
r(x) = 1 + ---------- = 1 + ---------
2 2
x - 1 x - 1
--------
x
2 |
x - 1 | x
2 |-------
- x x
---------
- 1
2
x - 1 = x x - 1
3
r(x) = 1 + -----------
1
x - ---
x
Compared to the cost of evaluating the Horner form of numerator and denominator:
2
x + 3 x - 1 (x + 3)*x - 1
-------------- = ---------------
2 (x + 0)*x - 1
x - 1
we see that the evaluation takes two multiplications, one division,
two additions, and two subtractions. Even if we disregard the +0,
the continued-fraction representation is more efficient to evaluate
than the quotient of Horner forms.
Answers:
f(x+h) = f(x) + h*f'(x) + h^2*f''(x) + ...
f(x+h) - f(x)
--------------- = f'(x) + h*f''(x) + ...
h
f(x+h) - f(x-h)
delta f(x,h) = ----------------- is an even function of h
2 h
f(x-h) - f(x+h)
delta f(x,-h) = ----------------- = delta f(x,h)
- 2 h
Because delta f(x,h) is an even function of h, only
even powers of h occur if we expand delta f(x,h) in powers of h.
Thus the power series start at h^2, and delta f(x,h) is a second-order
approximation for the derivative of f at x.
Answers:
f(x+h) - f(x)
Delta f(x,h) = ---------------
h
f(x+2h) - f(x+h) f(x+h) - f(x)
------------------ - ---------------
2 h h
Delta f(x,h) = --------------------------------------
h
f(x+2h) - 2 f(x+h) + f(x)
= ---------------------------
2
h
3 f(x+3h) - f(x+2h) f(x+2h) - f(x+h) f(x+h) - f(x)
Delta f(x,h) = ------------------- - 2 ------------------ + ---------------
3 3 3
h h h
f(x+3h) - 3 f(x+2h) + 3 f(x+h) - f(x)
= ---------------------------------------
3
h
f(x+h) - f(x-h)
delta f(x,h) = -----------------
2 h
f(x+2h) - f(x) f(x) - f(x-2h)
---------------- - ----------------
2 2 h 2 h
delta f(x,h) = -------------------------------------
2 h
f(x+2h) - 2 f(x) + f(x-2h)
= ----------------------------
2
4 h
3 f(x+3h) - f(x+h) f(x+h) - f(x-h) f(x-h) - f(x-3h)
delta f(x,h) = ------------------ - 2 ----------------- + ------------------
3 3 3
8 h 8 h 8 h
f(x+3h) - 3 f(x+h) + 3 f(x-h) - f(x-3h)
= -----------------------------------------
3
8 h
Write pseudo-code for Richardson extrapolation to compute a table with n rows using one of the formulas you derived above.
Answers:
2
Delta f(x,h) = f'(x) + C h + C h + ...
1 2
2 2
Delta f(x,rh) = f'(x) + C rh + C r h + ...
1 2
for some constant C and C .
1 2
To eliminate the first-order term, we multiply the first formula by r
and subtract from it the second formula:
Delta f(x,h)*r - Delta f(x,rh) 2
-------------------------------- = f'(x) + C' h + ...
r - 1 2
for some constant C'.
2
To obtain higher-order approximations by extrapolation, we need
Delta f(x,r^i h) and use r^i instead of r in the formula to obtain
an approximation of order i+1.
Pseudo-code for the algorithm is:
Input: D[i][0] = Delta f(x,r^i*h), for i=0,1,..,n.
Output: D[i][j] contains extrapolated values, j <= i.
for i from 1 to n do
for j from 1 to i do
i
D[i-1][j-1]*r - D[i][j-1]
D[i][j] = ----------------------------
i
r - 1
2 4
delta f(x,h) = f'(x) + C h + C h + ...
1 2
2 2 4 4
delta f(x,rh) = f'(x) + C r h + C r h + ...
1 2
for some constant C and C .
1 2
To eliminate the first-order term, we multiply the first formula by r^2
and subtract from it the second formula:
2
delta f(x,h)*r - delta f(x,rh) 4
-------------------------------- = f'(x) + C' h + ...
2 2
r - 1
for some constant C'.
2
To obtain higher-order approximations by extrapolation, we need
delta f(x,r^(2*i) h) and use r^(2*i) instead of r^2 in the formula
to obtain an approximation of order 2*(i+1).
Pseudo-code for the algorithm is:
Input: d[i][0] = delta f(x,r^i*h), for i=0,1,..,n.
Output: d[i][j] contains extrapolated values, j <= i.
for i from 1 to n do
for j from 1 to i do
2*i
d[i-1][j-1]*r - d[i][j-1]
d[i][j] = ------------------------------
2*i
r - 1
1
/ x
I = | e dx
/
0
Write your answers to the questions below with six decimal places.
Answers:
( 1 ) 1 ( 0 1 ) 1 ( 1/4 1/2 1/4 )
T(h) = T(---) = ---( e + e ) + ---( e + e + e )
( 4 ) 8 ( ) 4 ( )
= 1.72722
First we compute T(1/8) and T(1/16):
( 1 ) 1 ( 0 1 ) 1 ( 1/8 1/4 3/8 1/2 5/8 3/4 7/8 )
T(---) = ----( e + e ) + ---( e + e + e + e + e + e + e )
( 8 ) 16 ( ) 8 ( )
1 ( 1 ) 1 ( 1/8 3/8 5/8 7/8 )
= --- T(---) + ---( e + e + e + e )
2 ( 8 ) 8 ( )
= 1.72052
The relation between two consecutive applications of the composite
trapezoidal rule is especially handy as we get to the next stage:
( 1 ) 1 ( 1 ) 1 ( 1/16 3/16 5/16 7/16 9/16
T(----) = --- T(---) + ----( e + e + e + e + e
( 16 ) 2 ( 8 ) 16 (
11/16 13/16 15/16 )
+ e + e + e )
)
= 1.71884
Now we are ready to extrapolate:
1.72722
1.72052*4 - 1.72722
1.72052 --------------------- = 1.71829
4 - 1
1.71884*4 - 1.72052 1.71828*16 - 1.71829
1.71884 --------------------- = 1.71828 ---------------------- = 1.71828
4 - 1 16 - 1
Theoretically we have a sixth-order approximation, which means that the error is of order O(h^6). With h = 1/16, we compute (1/16)^6, or about 6*10^(-8), which means that we have about seven significant digits correct.
b
/ ( a+b )
| f(x) dx is (b-a) f (-----)
/ ( 2 )
a
Give the formula for the composite midpoint rule that applies
the midpoint rule to n equally sized subintervals of [a,b].
Answer:
n
--- ( ( 1 ) ) b - a
h > f( a + (i + --- ) h ) for h = -------
--- ( ( 2 ) ) n
i=1
b
/ b-a
| f(x) dx = --- ( f(a) + 4*f((a+b)/2) + f(b) )
/ 6
a
Answer:
Simpson's rule uses 3 function evaluations per interval, but when dividing the interval [a,b] into subintervals, the function evaluation at the interior boundaries of the subintervals are shared. Therefore, if we subdivide [a,b] in 3 subintervals, we will use seven function evaluations.
The formula is
b-a (
---*( f(x(0))+4*f(x(1))+2*f(x(2))+4*f(x(3))
18 (
)
+2*f(x(4))+4*f(x(5))+f(x(6)) )
)
with
x(k) = a + k*(b-a)/6, k = 0,1,..,6
3a
/
| f(x) dx
/
a
by the rule w_1 f(a) + w_2 f(2a) + w_3 f(3a).
Answers:
3a
/
| f(x) dx = w f(a) + w f(2a) + w f(3a)
/ 1 2 3
a
2
to hold for f = 1, x, and x . Therefore, we solve the system
3a
/
| 1 dx = 2 a = w + w + w
/ 1 2 3
a
3a
/ 2
| x dx = 4 a = w a + w 2 a + w 3 a
/ 1 2 3
a
3a
/ 2 26 3 2 2 2
| x dx = ---- a = w a + w 4 a + w 9 a
/ 3 1 2 3
a
where a is a parameter. The solution to the linear system is
1 4 1
w = --- a w = --- a w = --- a
1 3 2 3 3 3
The highest algebraic degree of precision we can reach with Gaussian quadrature using three function evaluations is 5 = 2*3 - 1.
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