Chapter 2
2.1)
do
x = x - 5
while x > 50
else
if x < 5 then
x = 5
else
x = 7.5
end if
end if
exit
in matlab:
function [ output ] = two( x )
if x >= 10
x = x - 5;
while x > 50
x = x - 5;
end
else
if x < 5
x = 5;
else
x = 7.5;
end
end
output = x;
end
no loops:
function [ output ] = two( x )
if x >= 10
x = x - 5;
if x > 50
x = 50 - (5 -(x - (floor(x / 5) * 5)));
end
else
if x < 5
x = 5;
else
x = 7.5;
end
end
output = x;
end
2.2 Rewrite the following pseudocode using proper indentation
DO
i = i + 1
IF z > 50 EXIT
x = x + 5
IF x > 5 THEN
y = x
ELSE
y = 0
ENDIF
z = x + y
ENDDO
2.3 Develop, debug, and document a program to determine the
roots of a quadratic equation, ax2 + bx + c, in either a high-level
language or a macro language of your choice. Use a subroutine procedure
to compute the roots (either real or complex). Perform test
runs for the cases (a) a = 1, b = 6, c = 2; (b) a = 0, b = −4,
c = 1.6; (c) a = 3, b = 2.5, c = 7.
using this formual:
I wrote the following program in java:
public static void quad(double a, double b, double c){
double tom = (b * b - 4 * a * c);
double real = -b / (2 * a);
if (tom >= 0) {
System.out.println(Math.sqrt(tom) / (2 * a) + real);
System.out.println(-Math.sqrt(tom) / (2 * a) + real);
} else {
System.out.println(real + " + " + (Math.sqrt(-tom) / (2 * a)) + "i");
System.out.println(real + " - " + (Math.sqrt(-tom) / (2 * a)) + "i");
}
}
}
The test cases output is:
-0.3542486889354093
-5.645751311064591
Infinity
NaN
-0.4166666666666667 + 1.4695993407123664i
-0.4166666666666667 - 1.4695993407123664i
matlab:
function [ output ] = quad( a,b,c )
output(1) = (sqrt((b * b - 4 * a * c)) / (2 * a) + (-b / (2 * a)));
output(2) = (-sqrt((b * b - 4 * a * c)) / (2 * a) + (-b / (2 * a)));
end
>> quad(1, 6, 2)
ans =
-0.3542 -5.6458
>> quad(0, -4, 1.6)
ans =
Inf NaN
>> quad(3, 2.5, 7)
ans =
-0.4167 + 1.4696i -0.4167 - 1.4696i
a)
While i < 10
If i is odd
answer = answer - (x^(i*2))/(i * 2)!
Else
answer = answer + (x^(i*2))/(i * 2)!
End If
i = i + 1
print (cos(x) - answer) / cos(x)
End While
b)
2.5) Develop, debug, and document a program for Prob. 2.4 in either a high-level language or a macro language of your choice. Employ the library function for the cosine in your computer to determine the true value. Have the program print out the series approximation and the error at each step. As a test case, employ the program to compute cos(1.25) for up to and including the term x10/10!. Interpret your results.
public class Cos {
public static void main(String[] args) {
double x = 1.25, cosx = Math.cos(x);
double answer = 1.0;
for(int i = 1; i <=5; i++){
if((i & 1) == 1)
answer = answer - Math.pow(x, (i*2))/ fact(i * 2);
else
answer = answer + Math.pow(x, (i*2))/ fact(i * 2);
System.out.println(((cosx - answer)/ cosx));
}
}
public static int fact(int f){
int returnVal = 1;
for(int i = f; i > 1; i--)
returnVal *= i;
return returnVal;
}
}
results:
Approximation: %error
0.21875 30.62655044877899
0.3204752604166667 -1.6341682785373326
0.3151770697699653 0.046077488510394275
0.315324898750063 -8.043688291041844E-4
0.31532233227471407 9.552305259051958E-6
for matlab:
function [ o ] = cos2( x , t)
a = zeros(1,t);
a(1) = 1 - (x^2)/2;
for i = 2:t
if (floor(i/2) * 2) == i
a(i) = a(i-1) + (x^(i*2))/ factorial(i * 2);
else
a(i) = a(i-1) - (x^(i*2))/ factorial(i * 2);
end
end
o = a;
end
function [ o ] = compToCos( c )
l = length(c);
o(1:l) = abs(((cos(1.25) - c(1:l))/ cos(1.25)) * 100);
end
>> temp = cos2(1.25, 5)
temp =
0.2188 0.3205 0.3152 0.3153 0.3153
>> compToCos(temp)
ans =
30.6266 1.6342 0.0461 0.0008 0.0000
for matlab:
function [ o ] = cos2( x , t)
a = zeros(1,t);
a(1) = 1 - (x^2)/2;
for i = 2:t
if (floor(i/2) * 2) == i
a(i) = a(i-1) + (x^(i*2))/ factorial(i * 2);
else
a(i) = a(i-1) - (x^(i*2))/ factorial(i * 2);
end
end
o = a;
end
function [ o ] = compToCos( c )
l = length(c);
o(1:l) = abs(((cos(1.25) - c(1:l))/ cos(1.25)) * 100);
end
>> temp = cos2(1.25, 5)
temp =
0.2188 0.3205 0.3152 0.3153 0.3153
>> compToCos(temp)
ans =
30.6266 1.6342 0.0461 0.0008 0.0000
These results get more accurate the longer it runs and this happens very quickly.
2.6)
input courseNumber, name, WQ, WH, WF
for i = 1 to 5
input quiz
AQ = AQ + quiz
end for
AQ = AQ / 5
for i = 1 to 6
input quiz
AH = AH + quiz
end for
AH = AH / 5
AG = ((WQ * AQ + WH * AH + WF * FE) / (WQ + WH + WF)) * 100
else
AG = ((WQ * AQ + WH * AH) / (WQ + WH)) * 100
end if
print courseNumber, name, AG
b)
function [ o ] = grade( cn, name, WQ, WH, WF, quizes, homeworks, FE )
AQ = sum(quizes) / length(quizes);
AH = sum(homeworks) / length(homeworks);
if isempty(FE)
AG = ((WQ * AQ + WH * AH + WF * FE) / (WQ + WH + WF)) * 100;
else
AG = ((WQ * AQ + WH * AH) / (WQ + WH)) * 100;
end
o(1) = cn;
o(2) = name;
o(3) = AG;
end
>> intrest(100000, .06, 5)
ans =
1.0e+05 *
1.0600 1.1236 1.1910 1.2625 1.3382
function [ o ] = apay( p, i, n )
rv = zeros(1,n);
for t = 1:n
rv(t) = p * ((i * (1 + i)^t) / ((1 + i)^t - 1));
end
o = rv;
end
>> apay(66000, 0.066, 5)
ans =
1.0e+04 *
7.0356 3.6302 2.4966 1.9309 1.5925
function [ o ] = temp( Tm, Tp, t, tp , w)
l = length(t);
T = zeros(1, l);
for i = 1:l
T(i) = Tm + (Tp - Tm) * cos(w * (t(i) - tp));
end
o = T;
end
>> temp( 22.1, 28.3, 0:59, tp, w)
ans =
Columns 1 through 12
16.3593 16.3198 16.2821 16.2460 16.2117 16.1792 16.1484 16.1194 16.0921 16.0667 16.0430 16.0211
Columns 13 through 24
16.0010 15.9827 15.9663 15.9516 15.9388 15.9278 15.9186 15.9112 15.9057 15.9021 15.9002 15.9002
Columns 25 through 36
15.9021 15.9057 15.9112 15.9186 15.9278 15.9388 15.9516 15.9663 15.9827 16.0010 16.0211 16.0430
Columns 37 through 48
16.0667 16.0921 16.1194 16.1484 16.1792 16.2117 16.2460 16.2821 16.3198 16.3593 16.4004 16.4433
Columns 49 through 60
16.4878 16.5340 16.5818 16.6313 16.6824 16.7351 16.7894 16.8452 16.9027 16.9616 17.0221 17.0841
>> temp( 10.7, 22.9, 180:242, tp, w)
ans =
Columns 1 through 12
21.7876 21.8735 21.9562 22.0355 22.1115 22.1841 22.2533 22.3190 22.3813 22.4402 22.4955 22.5474
Columns 13 through 24
22.5958 22.6406 22.6819 22.7197 22.7539 22.7845 22.8115 22.8350 22.8548 22.8711 22.8837 22.8928
Columns 25 through 36
22.8982 22.9000 22.8982 22.8928 22.8837 22.8711 22.8548 22.8350 22.8115 22.7845 22.7539 22.7197
Columns 37 through 48
22.6819 22.6406 22.5958 22.5474 22.4955 22.4402 22.3813 22.3190 22.2533 22.1841 22.1115 22.0355
Columns 49 through 60
21.9562 21.8735 21.7876 21.6983 21.6058 21.5101 21.4111 21.3090 21.2037 21.0953 20.9839 20.8694
Columns 61 through 63
20.7519 20.6314 20.5079
public static LinkedList<BigDecimal> parachutist(double step, double time){
LinkedList<BigDecimal> list = new LinkedList<BigDecimal>();
list.add(new BigDecimal(0));
double g = 9.8, m = 68.1, c = 12.5;
for(double i = (0.0 + step); i <= time; i += step){
double vi = list.getLast().doubleValue();
list.add( new BigDecimal(
vi + (g - ((c/m) * vi)) * step
));
}
return list;
}
matlab:
function [ output ] = eulers( s )
g = 9.8; m=68.1;c=12.5;l = length(s) - 1; v = zeros(1,l + 1);
for i = 1:l
v(2:l+1) = v(1:l) + (g-((c*v(1:l))/ m)) * (s(2) - s(1));
end
output = v;
end
a) the results for t = 10 and step size = 1
t, s v, m/s
0.0 0
1.0 9.800000000000000710542735760100185871124267578125
2.0 17.80117474302496560767394839785993099212646484375
3.0 24.3337050765372708838185644708573818206787109375
4.0 29.6671659655722805837285704910755157470703125
5.0 34.02165092049660444217806798405945301055908203125
6.0 37.57685449602953298153806827031075954437255859375
7.0 40.47948766489341920760125503875315189361572265625
8.0 42.84933207295262747038577799685299396514892578125
9.0 44.784183014040621628737426362931728363037109375
10.0 46.3638851039744253057506284676492214202880859375
in matlab:
>> eulers(0:10)
ans =
0 9.8000 17.8012 24.3337 29.6672 34.0217 37.5769 40.4795 42.8493 44.7842 46.3639
in matlab:
>> eulers(0:.5:10)
ans =
Columns 1 through 12
0 4.9000 9.3503 13.3922 17.0631 20.3971 23.4251 26.1752 28.6729 30.9414 33.0017 34.8729
Columns 13 through 21
36.5724 38.1159 39.5177 40.7909 41.9473 42.9975 43.9513 44.8176 45.6044
2.12 The bubble sort is an inefficient, but easy-to-program, sorting technique. The idea behind the sort is to move down through an array comparing adjacent pairs and swapping the values if they are out of order. For this method to sort the array completely, it may need to pass through it many times. As the passes proceed for an ascending-order sort, the smaller elements in the array appear to rise toward the top like bubbles. Eventually, there will be a pass through the array where no swaps are required. Then, the array is sorted. After the first pass, the largest value in the array drops directly to the bottom. Consequently, the second pass only has to proceed to the second-to-last value, and so on. Develop a program to set up an array of 20 random numbers and sort them in ascending order with the bubble sort (Fig. P2.12).
function [ o ] = sort2( a )
m = length(a)-1;
doswitch = 1;
while doswitch == 1
doswitch = 0;
for i = 1:m
if a(i) > a(i+1)
temp = a(i);
a(i) = a(i+1);
a(i+1) = temp;
doswitch = 1;
end
end
m = m - 1;
end
o = a;
end
>> sort2([4, 69, 666, 82, 111, 420, 9999999, 1,0,21,13,7,7,7,7,5,4,3,1,2])
ans =
Columns 1 through 10
0 1 1 2 3 4 4 5 7 7
Columns 11 through 20
7 7 13 21 69 82 111 420 666 9999999
for i = 1 to 5
input quiz
AQ = AQ + quiz
end for
AQ = AQ / 5
for i = 1 to 6
input quiz
AH = AH + quiz
end for
AH = AH / 5
input FE
if FE existsAG = ((WQ * AQ + WH * AH + WF * FE) / (WQ + WH + WF)) * 100
else
AG = ((WQ * AQ + WH * AH) / (WQ + WH)) * 100
end if
print courseNumber, name, AG
b)
function [ o ] = grade( cn, name, WQ, WH, WF, quizes, homeworks, FE )
AQ = sum(quizes) / length(quizes);
AH = sum(homeworks) / length(homeworks);
if isempty(FE)
AG = ((WQ * AQ + WH * AH + WF * FE) / (WQ + WH + WF)) * 100;
else
AG = ((WQ * AQ + WH * AH) / (WQ + WH)) * 100;
end
o(1) = cn;
o(2) = name;
o(3) = AG;
end
a)
if a > 0
tol = 10^-5
x = a/2
e = 0
while e < tol
y = (x + a/x)/ 2
e = |(y - x) / y|
x = y
end while
squareRoot = x
else
squareRoot = 0
end if
function [ o ] = sqrt2( a )
if a > 0
tol = 10^-5;
x = a/2;
e = 0;
while e < tol
y = (x + a/x)/ 2;
e = abs((y - x) / y);
x = y
end
squareRoot = x;
else
squareRoot = 0;
end
o = squareRoot;
end
function [ o ] = intrest( p, i, n )
rv = zeros(1,n);
for t = 1:n
rv(t) = p * (1 + i)^t;
end
o = rv;
end
ans =
1.0e+05 *
1.0600 1.1236 1.1910 1.2625 1.3382
function [ o ] = apay( p, i, n )
rv = zeros(1,n);
for t = 1:n
rv(t) = p * ((i * (1 + i)^t) / ((1 + i)^t - 1));
end
o = rv;
end
>> apay(66000, 0.066, 5)
ans =
1.0e+04 *
7.0356 3.6302 2.4966 1.9309 1.5925
function [ o ] = temp( Tm, Tp, t, tp , w)
l = length(t);
T = zeros(1, l);
for i = 1:l
T(i) = Tm + (Tp - Tm) * cos(w * (t(i) - tp));
end
o = T;
end
>> temp( 22.1, 28.3, 0:59, tp, w)
ans =
Columns 1 through 12
16.3593 16.3198 16.2821 16.2460 16.2117 16.1792 16.1484 16.1194 16.0921 16.0667 16.0430 16.0211
Columns 13 through 24
16.0010 15.9827 15.9663 15.9516 15.9388 15.9278 15.9186 15.9112 15.9057 15.9021 15.9002 15.9002
Columns 25 through 36
15.9021 15.9057 15.9112 15.9186 15.9278 15.9388 15.9516 15.9663 15.9827 16.0010 16.0211 16.0430
Columns 37 through 48
16.0667 16.0921 16.1194 16.1484 16.1792 16.2117 16.2460 16.2821 16.3198 16.3593 16.4004 16.4433
Columns 49 through 60
16.4878 16.5340 16.5818 16.6313 16.6824 16.7351 16.7894 16.8452 16.9027 16.9616 17.0221 17.0841
>> temp( 10.7, 22.9, 180:242, tp, w)
ans =
Columns 1 through 12
21.7876 21.8735 21.9562 22.0355 22.1115 22.1841 22.2533 22.3190 22.3813 22.4402 22.4955 22.5474
Columns 13 through 24
22.5958 22.6406 22.6819 22.7197 22.7539 22.7845 22.8115 22.8350 22.8548 22.8711 22.8837 22.8928
Columns 25 through 36
22.8982 22.9000 22.8982 22.8928 22.8837 22.8711 22.8548 22.8350 22.8115 22.7845 22.7539 22.7197
Columns 37 through 48
22.6819 22.6406 22.5958 22.5474 22.4955 22.4402 22.3813 22.3190 22.2533 22.1841 22.1115 22.0355
Columns 49 through 60
21.9562 21.8735 21.7876 21.6983 21.6058 21.5101 21.4111 21.3090 21.2037 21.0953 20.9839 20.8694
Columns 61 through 63
20.7519 20.6314 20.5079
public static LinkedList<BigDecimal> parachutist(double step, double time){
LinkedList<BigDecimal> list = new LinkedList<BigDecimal>();
list.add(new BigDecimal(0));
double g = 9.8, m = 68.1, c = 12.5;
for(double i = (0.0 + step); i <= time; i += step){
double vi = list.getLast().doubleValue();
list.add( new BigDecimal(
vi + (g - ((c/m) * vi)) * step
));
}
return list;
}
matlab:
function [ output ] = eulers( s )
g = 9.8; m=68.1;c=12.5;l = length(s) - 1; v = zeros(1,l + 1);
for i = 1:l
v(2:l+1) = v(1:l) + (g-((c*v(1:l))/ m)) * (s(2) - s(1));
end
output = v;
end
t, s v, m/s
0.0 0
1.0 9.800000000000000710542735760100185871124267578125
2.0 17.80117474302496560767394839785993099212646484375
3.0 24.3337050765372708838185644708573818206787109375
4.0 29.6671659655722805837285704910755157470703125
5.0 34.02165092049660444217806798405945301055908203125
6.0 37.57685449602953298153806827031075954437255859375
7.0 40.47948766489341920760125503875315189361572265625
8.0 42.84933207295262747038577799685299396514892578125
9.0 44.784183014040621628737426362931728363037109375
10.0 46.3638851039744253057506284676492214202880859375
in matlab:
>> eulers(0:10)
ans =
0 9.8000 17.8012 24.3337 29.6672 34.0217 37.5769 40.4795 42.8493 44.7842 46.3639
b) results for t = 10 and step = .5:
t, s v, m/s
0.0 0
0.5 4.9000000000000003552713678800500929355621337890625
1.0 9.35029368575624175718985497951507568359375
1.5 13.3921536632015207857193672680296003818511962890625
2.0 17.06306467061694576159425196237862110137939453125
2.5 20.39707121699938596748324926011264324188232421875
3.0 23.42509331529239346991744241677224636077880859375
3.5 26.17521323863193316583419800736010074615478515625
4.0 28.67293595902180669554581982083618640899658203125
4.5 30.941425683781186961596176843158900737762451171875
5.0 33.00172068343416498237274936400353908538818359375
5.5 34.87292840338330535132627119310200214385986328125
6.0 36.57240266885840895838555297814309597015380859375
6.5 38.115904626562297607961227186024188995361328125
7.0 39.51774891560760494257920072413980960845947265625
7.5 40.79093642335286773459301912225782871246337890625
8.0 41.94727485733296390435498324222862720489501953125
8.5 42.997488251483758858739747665822505950927734375
9.0 43.95131642223598333885092870332300662994384765625
9.5 44.8176052968472191651017055846750736236572265625
10.0 45.60438895168869777307918411679565906524658203125
>> eulers(0:.5:10)
ans =
Columns 1 through 12
0 4.9000 9.3503 13.3922 17.0631 20.3971 23.4251 26.1752 28.6729 30.9414 33.0017 34.8729
Columns 13 through 21
36.5724 38.1159 39.5177 40.7909 41.9473 42.9975 43.9513 44.8176 45.6044
function [ o ] = sort2( a )
m = length(a)-1;
doswitch = 1;
while doswitch == 1
doswitch = 0;
for i = 1:m
if a(i) > a(i+1)
temp = a(i);
a(i) = a(i+1);
a(i+1) = temp;
doswitch = 1;
end
end
m = m - 1;
end
o = a;
end
>> sort2([4, 69, 666, 82, 111, 420, 9999999, 1,0,21,13,7,7,7,7,5,4,3,1,2])
ans =
Columns 1 through 10
0 1 1 2 3 4 4 5 7 7
Columns 11 through 20
7 7 13 21 69 82 111 420 666 9999999
posted by Unknown @ 4:57 PM
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