Numerical Analysis Programs Using Fortran 90 - University of ...

Kurdistan Region-Iraq
Sulaimani University
College of Science
Physics Department
Numerical Analysis Programs
Using Fortran 90
(2009)
Prepared by
Dr. Omed Gh. Abdullah
1

Problem: Write a program to find the roots of the equation
x
y = e − 3x
using Bisection method in the interval [ 1,2]
, within
3
the tolerance 10 − .
f(x)=exp(x)-3*x
tol=.001
a=1
b=2
10 c=(a+b)/2
print*,a,b,c
if (f(c)*f(a))20,30,30
20 b=c
goto 40
30 a=c
40 if (abs (a-b).lt.tol) goto 50
goto 10
50 print*,'The Root is',c
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
1.000000 2.000000 1.500000
1.500000 2.000000 1.750000
1.500000 1.750000 1.625000
1.500000 1.625000 1.562500
1.500000 1.562500 1.531250
1.500000 1.531250 1.515625
1.500000 1.515625 1.507813
1.507813 1.515625 1.511719
1.511719 1.515625 1.513672
1.511719 1.513672 1.512695
The Root is 1.512695
2

Problem: Write a program to find the roots of the equation
f ( x)
= x sin( x)
− 1 using False position method in the interval
[ 1,2] , within the tolerance
4
10 − .
f(x)=x*sin(x)-1
tol=.0001
a=1
b=2
10 c=(a*f(b)-b*f(a))/(f(b)-f(a))
print*,a,b,c
if (f(a)*f(c))20,30,30
20 b=c
goto 40
30 a=c
40 if (abs (a-b).lt.tol) goto 50
goto 10
50 print*,'The Root is',c
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
1.000000 2.000000 1.162241
1.000000 1.162241 1.114254
1.000000 1.114254 1.114157
The Root is 1.114157
3

Problem: Write a program to find the roots of the equation
x
f ( x)
= 3x
+ sin x − e using Secant method in the interval [ 0,1]
,
4
within the tolerance 10 − .
f(x)=3*x+sin(x)-exp(x)
tol=.00001
x1=0
x2=1
10 x3=(x1*f(x2)-x2*f(x1))/(f(x2)-f(x1))
print*,x1,x2,x3
x1=x2
x2=x3
if (abs (x1-x2).lt.tol) goto 20
goto 10
20 print*,'The Root is',x3
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
0.000000 1.000000 4.709896E-01
1.000000 4.709896E-01 3.075085E-01
4.709896E-01 3.075085E-01 3.626132E-01
3.075085E-01 3.626132E-01 3.604615E-01
3.626132E-01 3.604615E-01 3.604217E-01
3.604615E-01 3.604217E-01 3.604217E-01
The Root is 3.604217E-01
4

Problem: Write a program to find the roots of the equation
2
f ( x)
= x − 2x
− 3 by iterative method, within the tolerance
4
10 − .
read*,xo
tol=0.0001
5 x=sqrt(2*xo+3)
print*,xo,x
if(abs(x-xo)

Problem: Write a program using the Newton-Raphson procedure to
3 2
determine the roots of the equation f ( x)
= x − x − 2x
+ 1 ,
6
within the tolerance 10 − .
f(x)=x**3-x**2-2*x+1
df(x)=3*x**2-2*x-2
tol=1e-6
xo=2
k=0
10 k=k+1
x=xo-f(xo)/df(xo)
if(abs(x-xo).lt.tol) goto 20
print *,k,x
xo=x
goto 10
20 print *,'The Root is',x
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
1 1.833333
2 1.802935
3 1.801939
4 1.801938
The Root is 1.801938
If xo=0
The Root is 4.450419E-01
If xo=-1
The Root is -1.246980
6

Problem: Write a program to determine the roots of the equation
4
3
2
x −1.99x
−1.76x
+ 5.22x
− 2.23 = 0 that are close to x = 1. 5
by Newton-Raphson method.
f(x)=x**4-1.99*x**3-1.76*x**2+5.22*x-2.23
df(x)=4*x**3-5.97*x**2-3.52*x+5.22
ddf(x)=12*x**2-11.94*x-3.52
tol=0.00001
x=1.5
5 rat=df(x)/ddf(x)
x=x-rat
if(abs(rat)-tol)10,10,5
10 eps=sqrt(-2*f(x)/ddf(x))
sign=1
do i=1,2
k=0
y=x+sign*eps
15 k=k+1
print *,y
rat=f(y)/df(y)
y=y-rat
if (abs(rat)-tol)20,15,15
20 sign=-1
print*,'++++++++++++++++++++++++++++++++'
enddo
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
1.569134
1.565973
1.565888
++++++++++++++++++++++++++
1.428165
1.424016
1.424112
++++++++++++++++++++++++++
7

Problem: Write a program to determine the roots of the equation
4
3
2
x −1.99x
−1.76x
+ 5.22x
− 2.23 = 0 that are close to x = 1. 5
by Newton-Raphson method.
f(x)=x**4-1.99*x**3-1.76*x**2+5.22*x-2.23
df(x)=4*x**3-5.97*x**2-3.52*x+5.22
ddf(x)=12*x**2-11.94*x-3.52
tol=0.00001
xo=1.5
10 x=xo-df(xo)/ddf(xo)
if(abs(x-xo).lt.tol) goto 20
xo=x
goto 10
20 eps=sqrt(-2*f(x)/ddf(x))
y1=x+eps
30 y=y1-f(y1)/df(y1)
if(abs(y-y1).lt.tol) goto 40
print *,y
y1=y
goto 30
40 print *,'The First Root is',y
print*,'************************'
y1=x-eps
50 y=y1-f(y1)/df(y1)
if(abs(y-y1).lt.tol) goto 60
print *,y
y1=y
goto 50
60 print *,'The Second Root is',y
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
1.565973
1.565888
The First Root is 1.565888
************************************
1.424016
1.424112
The Second Root is 1.424112
++++++++++++++++++++++++++
8

Problem: Write a program to determine the roots of the equation
3 2
8x − 20x
− 26x
+ 33 = 0 that are close to x = 0. 8 using Birge-
Vieta method.
dimension a(0:10),b(0:10),c(0:10)
a(0)=8
a(1)=-20
a(2)=-26
a(3)=33
m=3
xo=0.8
tol=1e-6
5 k=k+1
b(0)=a(0); c(0)=b(0)
do j=1,m
b(j)=a(j)+xo*b(j-1)
c(j)=b(j)+xo*c(j-1)
enddo
x=xo-b(m)/c(m-1)
print *,k,x
if(abs(x-xo)

Problem: Write a program to determine the roots of the equation
3 2
8x − 20x
− 26x
+ 33 = 0 that are close to x = 0. 8 using Birge-
Vieta method.
dimension a(0:10),b(0:10),c(0:10)
data (a(i),i=0,3)/8,-20,-26,33/
m=3
k=0
xo=0.8
tol=1e-6
5 k=k+1
b(0)=a(0); c(0)=b(0)
do j=1,m
b(j)=a(j)+xo*b(j-1)
c(j)=b(j)+xo*c(j-1)
enddo
x=xo-b(m)/c(m-1)
print *,k,x
if(abs(x-xo)

Problem: Write a program to determine the roots of the system of
2
2
equations x + xy −10
= 0 ; y + 3xy
− 57 = 0
using the fixed point iteration. Initiate the computation with
guesses of x = 1. 5 and y = 3. 5 .
xo=1.5
yo=3.5
tol=0.0001
5 x=sqrt(10-xo*yo)
y=sqrt((57-yo)/(3*x))
print*,x,y
if((abs(x-xo)

Problem: Write a program to solve the following system of equations,
by using Newton-Raphson Method
2
2
x − y − 0.2 = 0 ; y − x − 0.3 = 0
Initiate the computation with guesses of x = 1. 2 and y = 1. 2 .
f(x,y)=x**2-y-.2
g(x,y)=y**2-x-.3
fx(x,y)=2*x
fy(x,y)=-1
gx(x,y)=-1
gy(x,y)=2*y
xo=1.2
yo=1.2
tol=0.0001
5 xj=fx(xo,yo)*gy(xo,yo)-fy(xo,yo)*gx(xo,yo)
hh=(g(xo,yo)*fy(xo,yo)-f(xo,yo)*gy(xo,yo))/xj
hk=(f(xo,yo)*gx(xo,yo)-g(xo,yo)*fx(xo,yo))/xj
x=xo+hh
y=yo+hk
print*,x,y
if((abs(x-xo)

Problem: Write a program to find the interpolated value for x =1. 15,
using Newton forward method, for these tabulated data.
X 1 1.2 1.4 1.6 1.8 2.0
f 2.317 2.425 2.522 2.609 2.689 2.762
program Newton_Forward_Iterpolation
dimension x(20),y(20),d(20,20)
n=6
xx=1.15
data (x(i), i=1,6) /1,1.2,1.4,1.6,1.8,2/
data (y(i), i=1,6)/2.317,2.425,2.522,2.609,2.689,2.762/
print*,(x(i), i=1,6)
print*,(y(i), i=1,6)
s=y(1)
do i=1,n-1
do j=1,n-i
y(j)=y(j+1)-y(j)
d(i,j)=y(j)
enddo
enddo
r=(xx-x(1))/(x(2)-x(1))
do i=1,n-1
p=1
do j=0,i-1
p=p*(r-j)
enddo
f=1
do j=1,i; f=f*j; enddo
s=s+d(i,1)*p/f
enddo
print*,'Interpolation of(',xx,')=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Interpolation of (1.15)= 2.398955
13

Problem: Write a program to find the interpolated value for x =1. 15,
using Newton forward method, for these tabulated data.
X 1 1.2 1.4 1.6 1.8 2.0
f 2.317 2.425 2.522 2.609 2.689 2.762
dimension f(0:20),diff(0:20)
xo=1
x=1.15
h=0.2
n=5
data (f(i), i=0,5)/2.317,2.425,2.522,2.609,2.689,2.762/
r=(x-xo)/h
do i=0,n-1
diff(i)=f(i+1)-f(i)
enddo
coeff=r
yx=f(0)+coeff*diff(0)
do i=2,n
coeff=coeff*(r-i+1)/(i)
do j=0,n-i
diff(j)=diff(j+1)-diff(j)
enddo
yx=yx+coeff*diff(0)
enddo
print*,yx
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Interpolation of (1.15)= 2.398955
14

Problem: Write a program to find the interpolated value for x = 0. 55,
using Newton forward method, for these tabulated data.
X 0.5 0.7 0.9 1.1 1.3 1.5
Y 0.47943 0.64422 0.78333 0.89121 0.96356 0.99749
dimension y(20),diff(20)
x0=0.5
x=0.55
h=0.2
n=6
m=4
data (y(i), i=1,6) /0.47943,0.64422,0.78333,0.89121,0.96356,0.99749/
j=int((x-x0)/h-0.5)
r=(x-x0)/h-j
if ((j.ge.0).and.(j.lt.n-m)) goto 10
print*,'Not enough function values'
goto 20
10 do i=1,m-1
diff(i)=y(j+i+1)-y(j+i)
enddo
coeff=r
yx=y(1)+coeff*diff(1)
do i=2,m-1
coeff=coeff*(r-i+1)/i
do j=1,m-i
diff(j)=diff(j+1)-diff(j)
enddo
yx=yx+coeff*diff(1)
enddo
print*,'Interpolation of(',xx,')=',yx
20 end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Interpolation of (0.55)= 5.227315 E-01
15

Problem: Write a program to find the interpolated value for x =1. 93,
using Newton backward method, for these tabulated data.
X 1 1.2 1.4 1.6 1.8 2.0
f 2.317 2.425 2.522 2.609 2.689 2.762
program Newton_Backward_Iterpolation
dimension x(20),y(20),d(20,20)
n=6
xx=1.93
data (x(i), i=1,6) /1,1.2,1.4,1.6,1.8,2/
data (y(i), i=1,6)/2.317,2.425,2.522,2.609,2.689,2.762/
print*,(x(i), i=1,6)
print*,(y(i), i=1,6)
s=y(n)
do i=1,n-1
do j=1,n-i
y(j)=y(j+1)-y(j)
d(i,j)=y(j)
enddo
enddo
r=(xx-x(n))/(x(2)-x(1))
do i=1,n-1
p=1
do j=0,i-1
p=p*(r+j)
enddo
f=1
do j=1,i; f=f*j; enddo
s=s+d(i,n-i)*p/f
enddo
print*,'Interpolation of(',xx,')=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Interpolation of(1.93)=2.737522
16

Problem: Write a program to find the interpolated value for x =1. 47 ,
using Newton backward method, for these tabulated data.
x 0.5 0.7 0.9 1.1 1.3 1.5
y 0.47943 0.64422 0.78333 0.89121 0.96356 0.99749
program Newton_Backward_Iterpolation
dimension x(20),y(20),d(20,20)
n=6
xx=1.47
data (x(i), i=1,6)/.5,.7,.9,1.1,1.3,1.5/
data (y(i), i=1,6) /0.47943,0.64422,0.78333,0.89121,0.96356,0.99749/
print*,(x(i), i=1,6)
print*,(y(i), i=1,6)
s=y(n)
do i=1,n-1
do j=1,n-i
y(j)=y(j+1)-y(j)
d(i,j)=y(j)
enddo
enddo
r=(xx-x(n))/(x(2)-x(1))
do i=1,n-1
p=1
do j=0,i-1
p=p*(r+j)
enddo
f=1
do j=1,i; f=f*j; enddo
s=s+d(i,n-i)*p/f
enddo
print*,'Interpolation of(',xx,')=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Interpolation of(1.47)= 9.9492 E-01
17

Problem: Write a program to find the interpolated value for x =1. 05 ,
using Stirling Interpolation method, for these tabulated data.
X 0.5 0.7 0.9 1.1 1.3 1.5
y 0.48 0.64 0.78 0.89 0.96 0.99
program Stirling_Formula_Iterpolation
dimension x(20),y(20),d(20,20)
n=6
xx=1.05
data (x(i), i=1,6) /.5,.7,.9,1.1,1.3,1.5/
data (y(i), i=1,6)/.48,.64,.78,.89,.96,.99/
print*,(x(i), i=1,6)
print*,(y(i), i=1,6)
s=y(4)
do i=1,n-1
do j=1,n-i
y(j)=y(j+1)-y(j)
d(i,j)=y(j)
print*,d(i,j)
enddo
enddo
r=(xx-x(4))/(x(2)-x(1))
p1=r*(d(1,3)+d(1,4))/2
p2=r**2/2*d(2,3)
p3=r*(r**2-1)/6*(d(3,2)+d(3,3))/2
p4=r**2*(r**2-1)/24*d(4,2)
s=s+p1+p2+p3+p4
print*,'Interpolation of(',xx,')=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Interpolation of(1.05)=8.660302 E-01
18

Problem: Write a program to find the interpolated value for x =1. 05 ,
using Bessel’s Interpolation method, for these tabulated data.
X 0.5 0.7 0.9 1.1 1.3 1.5
y 0.48 0.64 0.78 0.89 0.96 0.99
program Bessels_Formula_Iterpolation
dimension x(20),y(20),d(20,20)
n=6
xx=1.05
data (x(i), i=1,6) /.5,.7,.9,1.1,1.3,1.5/
data (y(i), i=1,6)/.48,.64,.78,.89,.96,.99/
print*,(x(i), i=1,6)
print*,(y(i), i=1,6)
s=(y(3)+y(4))/2
do i=1,n-1
do j=1,n-i
y(j)=y(j+1)-y(j)
d(i,j)=y(j)
print*,d(i,j)
enddo
enddo
r=(xx-x(3))/(x(2)-x(1))
p1=(r-.5)*d(1,3)
p2=r*(r-1)/2*(d(2,2)+d(2,3))/2
p3=r*(r-1)*(r-.5)/6*d(3,2)
p4=r*(r-1)*(r+1)*(r-2)/24*(d(4,1)+d(4,2))/2
s=s+p1+p2+p3+p4
print*,'Interpolation of(',xx,')=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Interpolation of(1.05)=8.659448 E-01
19

Problem: Write a program to find the interpolated value for x = 0. 7 ,
using divided difference formula, for these tabulated data.
X 0 0.5 1 1.4 2 3.2 5.5
F(x) -1.25 -3.5 6 2.32 1.5 1.27 5.6
program Divided_Differences_Iterpolation
dimension x(20),y(20),d(20,20)
n=7
xx=.7
data (x(i), i=1,7) /0,.5,1,1.4,2,3.2,5.5/
data (y(i), i=1,7)/-1.25,-3.5,6,2.32,1.5,1.27,5.6/
s=y(1)
do i=1,n-1
do j=1,n-i
y(j)=(y(j+1)-y(j))/(x(j+i)-x(j))
d(i,j)=y(j)
print*,d(i,j)
enddo
enddo
do i=1,n-1
p=1
do j=1,i
p=p*(xx-x(j))
enddo
s=s+d(i,1)*p
enddo
print*,'Interpolation of(',xx,')=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Interpolation of (0.7) =2.290929
20

Problem: Write a program to find the interpolated value for x =1. 5,
using divided difference formula, for these tabulated data.
X 1 2 3 4 5
F(x) 0 1 4 6 10
program Divided_Differences_Iterpolation
dimension x(20),y(20),d(20,20)
n=5
xx=1.5
data (x(i), i=1,5)/1,2,3,4,5/
data (y(i), i=1,5)/0,1,4,6,10/
s=y(1)
do i=1,n-1
do j=1,n-i
y(j)=(y(j+1)-y(j))/(x(j+i)-x(j))
d(i,j)=y(j)
enddo
enddo
do i=1,n-1
p=1
do j=1,i
p=p*(xx-x(j))
enddo
s=s+d(i,1)*p
enddo
print*,'Interpolation of(',xx,')=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Interpolation of (1.5) = -0.171875
21

Problem: Write a program to find the interpolated value for x = 3, using
Lagrangian Polynomial, from the following data.
X 3.2 2.7 1 4.8
F(x) 22 17.8 14.2 38.3
program Lagrange_Iterpolation
dimension x(20),y(20)
n=4
xx=3
data (x(i), i=1,4)/3.2,2.7,1,4.8/
data (y(i), i=1,4)/22,17.8,14.2,38.3/
s=0
do i=1,n
p=1
do j=1,n
if (i.eq.j) goto 5
p=p*(xx-x(j))/(x(i)-x(j))
5 enddo
s=s+p*y(i)
enddo
print*,'Interpolation of(',xx,')=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Interpolation of (3) = 20.21196
22

Problem:
Write a program to determine the parameters a
1 & a
2 so that
( x)
= a + a x , fits the following data in least squares sense.
f
1 2
X 0 0.3 0.6 0.9 1.2 1.5 1.8 2.1
y 1 2.7 4.3 6 7.5 9 10.6 12
program List_Square_Fitting
dimension x(20),y(20),f(20)
n=8
data (x(i), i=1,8)/0,.3,.6,.9,1.2,1.5,1.8,2.1/
data (y(i), i=1,8)/1,2.7,4.3,6,7.5,9,10.6,12/
sx=0; sxx=0;sy=0;sxy=0
do i=1,n
sx=sx+x(i)
sy=sy+y(i)
sxx=sxx+x(i)**2
sxy=sxy+x(i)*y(i)
enddo
print *,'sx=',sx
print *,'sxx=',sxx
print *,'sy=',sy
print *,'sxy=',sxy
d=n*sxx-sx**2
a1=(sxx*sy-sx*sxy)/d
a2=(n*sxy-sx*sy)/d
print*,'a1=',a1
print*,'a2=',a2
s=0
do i=1,n
f(i)=a1+a2*x(i)
s=s+(y(i)-f(i))**2
enddo
print*,'Standerd Deviation=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
sx=8.4 sxx=12.6 sy=53.1 sxy=75.57
a1=1.133333
a2=5.242064
Standard deviation = 6.726178 E -02
23

Problem:
Write a program to determine the parameters A & B so that
f ( x)
= A ln( x)
+ B , fits the following data in least squares sense.
X 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
y 0.27 0.72 1.48 2.66 4.48 7.26 11.43 17.64 26.78
program List_Square_Fitting
dimension x(20),y(20),f(20)
n=9
data (x(i), i=1,9)/.1,.2,.3,.4,.5,.6,.7,.8,.9/
data (y(i), i=1,9)/.27,.72,1.48,2.66,4.48,7.26,11.43,17.64,26.78/
sx=0; sxx=0;sy=0;sxy=0
do i=1,n
sx=sx+log(x(i))
sy=sy+y(i)
sxx=sxx+(log(x(i)))**2
sxy=sxy+log(x(i))*y(i)
enddo
print *,'sx=',sx
print *,'sxx=',sxx
print *,'sy=',sy
print *,'sxy=',sxy
d=n*sxx-sx**2
a1=(sxx*sy-sx*sxy)/d
a2=(n*sxy-sx*sy)/d
a=a2; b=a1
print*,'A=',a
print*,'B=',b
s=0
do i=1,n
f(i)=a*log(x(i))+b
s=s+(y(i)-f(i))**2
enddo
print*,'Standerd Deviation=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
A=9.74113 B=16.66255
S=260.5141
24

Problem:
Write a program to determine the parameters A & C so that
f =
A x
( x)
C e , fits the following data in least squares sense.
X 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
y 0.27 0.72 1.48 2.66 4.48 7.26 11.43 17.64 26.78
program List_Square_Fitting
dimension x(20),y(20),f(20)
n=9
data (x(i), i=1,9)/.1,.2,.3,.4,.5,.6,.7,.8,.9/
data (y(i), i=1,9)/.27,.72,1.48,2.66,4.48,7.26,11.43,17.64,26.78/
sx=0; sxx=0;sy=0;sxy=0
do i=1,n
sx=sx+x(i)
sy=sy+log(y(i))
sxx=sxx+(x(i))**2
sxy=sxy+x(i)*log(y(i))
enddo
print *,'sx=',sx
print *,'sxx=',sxx
print *,'sy=',sy
print *,'sxy=',sxy
d=n*sxx-sx**2
a1=(sxx*sy-sx*sxy)/d
a2=(n*sxy-sx*sy)/d
a=a2; c=exp(a1)
print*,'A=',a
print*,'C=',c
s=0
do i=1,n
f(i)=c*exp(a*x(i))
s=s+(y(i)-f(i))**2
enddo
print*,'Standerd Deviation=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
A= 5.512738
C= .2359106
S=52.53117
25

Problem:
Write a program to determine the parameters A & C so that
A
f ( x)
= C x , fits the following data in least squares sense.
X 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
y 0.27 0.72 1.48 2.66 4.48 7.26 11.43 17.64 26.78
program List_Square_Fitting
dimension x(20),y(20),f(20)
n=9
data (x(i), i=1,9)/.1,.2,.3,.4,.5,.6,.7,.8,.9/
data (y(i), i=1,9)/.27,.72,1.48,2.66,4.48,7.26,11.43,17.64,26.78/
sx=0; sxx=0;sy=0;sxy=0
do i=1,n
sx=sx+log(x(i))
sy=sy+log(y(i))
sxx=sxx+(log(x(i)))**2
sxy=sxy+log(x(i))*log(y(i))
enddo
print *,'sx=',sx
print *,'sxx=',sxx
print *,'sy=',sy
print *,'sxy=',sxy
d=n*sxx-sx**2
a1=(sxx*sy-sx*sxy)/d
a2=(n*sxy-sx*sy)/d
a=a2; c=exp(a1)
print*,'A=',a
print*,'C=',c
s=0
do i=1,n
f(i)=c*x(i)**a
s=s+(y(i)-f(i))**2
enddo
print*,'Standerd Deviation=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
A= 2.092605
C= 23.42711
S= 75.07242
26

Problem:
Write a program to determine the parameters C & D so that
D x
f ( x)
= C x e , fits the following data in least squares sense.
X 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
y 0.27 0.72 1.48 2.66 4.48 7.26 11.4 17.6 26.7
3 4 8
program List_Square_Fitting
dimension x(20),y(20),f(20)
n=9
data (x(i), i=1,9)/.1,.2,.3,.4,.5,.6,.7,.8,.9/
data (y(i), i=1,9)/.27,.72,1.48,2.66,4.48,7.26,11.43,17.64,26.78/
sx=0; sxx=0;sy=0;sxy=0
do i=1,n
sx=sx+x(i)
sy=sy+log(y(i)/x(i))
sxx=sxx+x(i)**2
sxy=sxy+x(i)*log(y(i)/x(i))
enddo
print *,'sx=',sx
print *,'sxx=',sxx
print *,'sy=',sy
print *,'sxy=',sxy
d=n*sxx-sx**2
a1=(sxx*sy-sx*sxy)/d
a2=(n*sxy-sx*sy)/d
d=a2; c=exp(a1)
print*,'D=',d
print*,'C=',c
s=0
do i=1,n
f(i)=c*x(i)*exp(d*x(i))
s=s+(y(i)-f(i))**2
enddo
print*,'Standerd Deviation=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
C= 1.993407
D= 3.004763
S= 1.117279 E-01
27

Problem:
Write a program to determine the parameters A & B so that
A
f ( x)
= + B , fits the following data in least squares sense.
x
X 0.5 2 3.5 4.1 5 6.2 7.5 9.2
y 3.3 2 1.4 1.2 1 0.8 0.64 0.5
program List_Square_Fitting
dimension x(20),y(20),f(20)
n=8
data (x(i), i=1,8)/.5,2,3.5,4.1,5,6.2,7.5,9.2/
data (y(i), i=1,8)/3.3,2,1.4,1.2,1,.8,.64,.5/
sx=0; sxx=0;sy=0;sxy=0
do i=1,n
sx=sx+1/x(i)
sy=sy+y(i)
sxx=sxx+(1/x(i))**2
sxy=sxy+y(i)*(1/x(i))
enddo
print *,'sx=',sx
print *,'sxx=',sxx
print *,'sy=',sy
print *,'sxy=',sxy
d=n*sxx-sx**2
a1=(sxx*sy-sx*sxy)/d
a2=(n*sxy-sx*sy)/d
a=a2; b=a1
print*,'A=',a
print*,'B=',b
s=0
do i=1,n
f(i)=a/x(i)+b
s=s+(y(i)-f(i))**2
enddo
print*,'Standerd Deviation=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
A= 1.353131
B= .7405201
S=.7070401
28

Problem:
Write a program to determine the parameters D & C so that
D
f ( x)
= , fits the following data in least squares sense.
x + C
X 0.5 2 3.5 4.1 5 6.2 7.5 9.2
y 3.3 2 1.4 1.2 1 0.8 0.64 0.5
program List_Square_Fitting
dimension x(20),y(20),f(20)
n=8
data (x(i), i=1,8)/.5,2,3.5,4.1,5,6.2,7.5,9.2/
data (y(i), i=1,8)/3.3,2,1.4,1.2,1,.8,.64,.5/
sx=0; sxx=0;sy=0;sxy=0
do i=1,n
sx=sx+x(i)*y(i)
sy=sy+y(i)
sxx=sxx+(x(i)*y(i))**2
sxy=sxy+(x(i)*y(i))*y(i)
enddo
print *,'sx=',sx
print *,'sxx=',sxx
print *,'sy=',sy
print *,'sxy=',sxy
d=n*sxx-sx**2
a1=(sxx*sy-sx*sxy)/d
a2=(n*sxy-sx*sy)/d
c=-1/a2; d=a1*c
print*,'C=',c
print*,'D=',d
s=0
do i=1,n
f(i)=d/(x(i)+c)
s=s+(y(i)-f(i))**2
enddo
print*,'Standerd Deviation=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
C= 1.369226
D= 6.209051
S=5.723841 E-02
29

Problem:
Write a program to determine the parameters A & B so that
1
f ( x)
= , fits the following data in least squares sense.
A x + B
X 0.5 2 3.5 4.1 5 6.2 7.5 9.2
y 3.3 2 1.4 1.2 1 0.8 0.64 0.5
program List_Square_Fitting
dimension x(20),y(20),f(20)
n=8
data (x(i), i=1,8)/.5,2,3.5,4.1,5,6.2,7.5,9.2/
data (y(i), i=1,8)/3.3,2,1.4,1.2,1,.8,.64,.5/
sx=0; sxx=0;sy=0;sxy=0
do i=1,n
sx=sx+x(i)
sy=sy+1/y(i)
sxx=sxx+x(i)**2
sxy=sxy+x(i)*1/y(i)
enddo
print *,'sx=',sx
print *,'sxx=',sxx
print *,'sy=',sy
print *,'sxy=',sxy
d=n*sxx-sx**2
a1=(sxx*sy-sx*sxy)/d
a2=(n*sxy-sx*sy)/d
a=a2; b=a1
print*,'A=',a
print*,'B=',b
s=0
do i=1,n
f(i)=1/(a*x(i)+b)
s=s+(y(i)-f(i))**2
enddo
print*,'Standerd Deviation=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
A= 1.953443 E-01
B= 9.250808 E-02
S= 3.864387
30

Problem:
Write a program to determine the parameters A & B so that
x
f ( x)
= , fits the following data in least squares sense.
A x + B
X 0.5 2 3.5 4.1 5 6.2 7.5 9.2
y 3.3 2 1.4 1.2 1 0.8 0.64 0.5
program List_Square_Fitting
dimension x(20),y(20),f(20)
n=8
data (x(i), i=1,8)/.5,2,3.5,4.1,5,6.2,7.5,9.2/
data (y(i), i=1,8)/3.3,2,1.4,1.2,1,.8,.64,.5/
sx=0; sxx=0;sy=0;sxy=0
do i=1,n
sx=sx+1/x(i)
sy=sy+1/y(i)
sxx=sxx+(1/x(i))**2
sxy=sxy+1/(x(i)*y(i))
enddo
print *,'sx=',sx
print *,'sxx=',sxx
print *,'sy=',sy
print *,'sxy=',sxy
d=n*sxx-sx**2
a1=(sxx*sy-sx*sxy)/d
a2=(n*sxy-sx*sy)/d
a=a1; b=a2
print*,'A=',a
print*,'B=',b
s=0
do i=1,n
f(i)=x(i)/(a*x(i)+b)
s=s+(y(i)-f(i))**2
enddo
print*,'Standerd Deviation=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
A= 1.279117
B= -5.697291 E-01
S= 16.410690
31

Problem:
Write a program to determine the parameters A & B so that
1
f ( x)
= , fits the following data in least squares sense.
A x + B
( ) 2
X 0.5 2 3.5 4.1 5 6.2 7.5 9.2
y 3.3 2 1.4 1.2 1 0.8 0.64 0.5
program List_Square_Fitting
dimension x(20),y(20),f(20)
n=8
data (x(i), i=1,8)/.5,2,3.5,4.1,5,6.2,7.5,9.2/
data (y(i), i=1,8)/3.3,2,1.4,1.2,1,.8,.64,.5/
sx=0; sxx=0;sy=0;sxy=0
do i=1,n
sx=sx+x(i)
sy=sy+(y(i))**(-.5)
sxx=sxx+x(i)**2
sxy=sxy+x(i)*(y(i))**(-.5)
enddo
print *,'sx=',sx
print *,'sxx=',sxx
print *,'sy=',sy
print *,'sxy=',sxy
d=n*sxx-sx**2
a1=(sxx*sy-sx*sxy)/d
a2=(n*sxy-sx*sy)/d
a=a2; b=a1
print*,'A=',a
print*,'B=',b
s=0
do i=1,n
f(i)=1/(a*x(i)+b)**2
s=s+(y(i)-f(i))**2
enddo
print*,'Standerd Deviation=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
A= 9.919496 E-02
B= 5.035566 E-01
S= 2.275830 E-03
32

Problem: Write a Fortran program to find the value of the integral
1 dt
∫
using Trapezoidal rule with n = 6
0 2
2
( t + 1) (3t
+ 4)
program Trapezoidal_Rule
f(t)=1/((t**2+1)*(3*t**2+4))**.5
a=0
b=1
n=6
h=(b-a)/n
s=0
do i=1,n-1
x=a+h*i
s=s+f(x)
enddo
t=h*(f(a)/2+s+f(b)/2)
print*,'Integration by Trapezoidal Rule=',t
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Integration by Trapezoidal rule = 4.016085 E-01
33

Problem: Write a Fortran program to find the value of the integral
π
2
dx
∫ 2
sin x + cos
0
dimension y(20)
f(x)=1/((sin(x))**2+1./4*(cos(x))**2)
pi=22./7
a=0; b=pi/2; n=10
h=(b-a)/n
do i=1,n+1
x=a+h*(i-1)
y(i)=f(x)
print*,y(i)
enddo
t=0
do i=1,n
t=t+h*(y(i)+y(i+1))/2
enddo
print*,'Integration by Trapezoidal rule =',t
end
1
4
2
x
using Trapezoidal rule with n = 10
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Integration by Trapezoidal rule = 3.142227
34

Problem: Write a Fortran program to find the value of the integral
1 dt
∫
using Simpson’s 1/3 rule with n = 6
0 2
2
( t + 1) (3t
+ 4)
! program Simpson's 1/3 Rule
f(t)=1/((t**2+1)*(3*t**2+4))**.5
a=0
b=1
n=6
h=(b-a)/n
s1=0;s2=0
do i=1,n-1,2
x=a+h*i
s1=s1+f(x)
enddo
do i=2,n-1,2
x=a+h*i
s2=s2+f(x)
enddo
s=(h/3)*(f(a)+4*s1+2*s2+f(b))
print*,'Integration by Simpsons 1/3 Rule=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Integration by Simpson's 1/3 rule= 4.021834 E-01
35

Problem: Write a Fortran program to find the value of the integral
1 dt
∫
using Simpson’s 1/3 rule with n = 6
0 2
2
( t + 1) (3t
+ 4)
! program Simpson's 1/3 Rule
dimension y(20)
f(t)=1/((t**2+1)*(3*t**2+4))**.5
a=0
b=1
n=6
h=(b-a)/n
s=0
do i=1,n+1
x=a+h*(i-1)
y(i)=f(x)
enddo
do i=1,n,2
s=s+(h/3)*(y(i)+4*y(i+1)+y(i+2))
enddo
print*,'Integration by Simpsons 1/3 Rule=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Integration by Simpson's 1/3 rule= 4.021834 E-01
36

Problem: Write a Fortran program to find the value of the integral
1 dt
∫
using Simpson’s 3/8 rule with n = 6
0 2
2
( t + 1) (3t
+ 4)
! program Simpson's 3/8 Rule
dimension y(0:10)
f(t)=1/((t**2+1)*(3*t**2+4))**.5
a=0
b=1
n=6
h=(b-a)/n
s=0
do i=0,n
x=a+h*i
y(i)=f(x)
enddo
do i=1,(n/2-1)
s=s+y(3*i-3)+3*(y(3*i-2)+y(3*i-1))+y(3*i)
enddo
s=(3*h/8)*s
print*,'Integration by Simpsons 3/8 Rule=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Integration by Simpson’s 3/8 rule= 4.021832 E-01
37

Problem: Write a Fortran program to find the value of the integral
1 dt
∫
using Simpson’s 3/8 rule with n = 6
0 2
2
( t + 1) (3t
+ 4)
! program Simpson's 3/8 Rule
dimension y(0:10)
f(t)=1/((t**2+1)*(3*t**2+4))**.5
a=0
b=1
n=6
h=(b-a)/n
s=0
do i=1,n+1
x=a+h*(i-1)
y(i)=f(x)
enddo
do i=1,n,3
s=s+(3*h/8)*(y(i)+3*y(i+1)+3*y(i+2)+y(i+3))
enddo
print*,'Integration by Simpsons 3/8 Rule=',s
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Integration by Simpson’s 3/8 rule= 4.021832 E-01
38

Problem: Write a Fortran program to find the value of the integral
∫ 0.
8
0
e
t 2 dt
with error of order
using Trapezium rule and Romberg integration
8
h .
dimension y(20),trap(20),romb(20,20)
f(t)=exp(t**2)
a=0; b=0.8; m=4
do k=1,m
n=2**k
h=(b-a)/n
do i=1,n+1
x=a+h*(i-1)
y(i)=f(x)
enddo
t=0
do i=1,n
t=t+h*(y(i)+y(i+1))/2
enddo
trap(k)=t
print*,t
enddo
print*,'---------------------------'
fact=2**2
do k=1,m-1
fact=fact**k
do j=1,m-k
trap(j)=(fact*trap(j+1)-trap(j))/(fact-1)
romb(k,j)=trap(j)
print*,romb(k,j)
enddo
print*,'---------------------------'
enddo
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
39

1.048701
1.019178
1.011646
1.009753
--------------------------------------
1.009338
1.009135
1.009122
--------------------------------------
1.009122
1.009121
--------------------------------------
1.009121
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
40

Problem: Write a Fortran program to find the value of the integral
∫ 0.
8
0
e
t 2 dt
with error of order
dimension trap(20,20)
f(t)=exp(t**2)
a=0; b=0.8; n=2; tol=1e-8
h=(b-a)/n
sum=(f(a)+f(b))/2
do i=1,n-1
sum=sum+f(a+i*h)
enddo
trap(1,1)=h*sum
print*,trap(1,1)
j=1
5 do k=0,n-1
sum=sum+f(a+k*h+h/2)
enddo
n=2*n
j=j+1
factor=1
h=h/2
trap(1,j)=h*sum
print*,trap(1,j)
using Trapezium rule and Romberg integration
8
h .
do i=1,j-1
factor=2**2*factor
trap(i+1,j)=(factor*trap(i,j)-trap(i,j-1))/(factor-1)
enddo
if (abs(trap(j,j)-trap(j-1,j)).lt.tol) goto 15
goto 5
15 print*,'+++++++++++++++++++++++++++'
do k=1,j
do i=1,k
print*,trap(i,k)
enddo
print*,'---------------------------'
enddo
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
41

1.048701
1.019178
1.011646
1.009753
+++++++++++++++++++++
1.048701
--------------------------------------
1.019178
1.009338
--------------------------------------
1.011646
1.009135
1.009122
--------------------------------------
1.009753
1.009122
1.009121
1.009121
--------------------------------------
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
42

Problem: Write a Fortran program to find the value of the integral
∫ 0.
8
0
e
t 2 dt
with error of order
using Simpson’s 1/3 rule and Romberg integration
8
h .
dimension y(20),simps(20),romb(20,20)
f(t)=exp(t**2)
a=0; b=0.8; m=4
do k=1,m
n=2**k
h=(b-a)/n
do i=1,n+1
x=a+h*(i-1)
y(i)=f(x)
enddo
s=0
do i=1,n,2
s=s+(h/3)*(y(i)+4*y(i+1)+y(i+2))
enddo
simps(k)=s
print*,s
enddo
print*,'---------------------------'
fact=2**2
do k=1,m-1
fact=fact**k
do j=1,m-k
simps(j)=(fact*simps(j+1)-simps(j))/(fact-1)
romb(k,j)=simps(j)
print*,romb(k,j)
enddo
print*,'---------------------------'
enddo
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
43

1.012070
1.009338
1.009135
1.009122
--------------------------------------
1.008427
1.009067
1.009117
--------------------------------------
1.009110
1.009121
--------------------------------------
1.009121
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
44

Problem:
Write a Fortran program to find first, second, third, and fourth
3
derivation of the function f ( x)
= ln( x ) at x = 1 and h = 0. 01
! Derivation using Different Formula
f(x) = log(x**3)
x=1
h=0.01
! df: Forward, db: Backward, dc: Central
df=(f(x+h)-f(x))/h
db=(f(x)-f(x-h))/h
dc=(f(x+h)-f(x-h))/(2*h)
print*,'Forward derivation is', df
print*,'Backward derivation is', db
print*,'Central derivation is', dc
print*,'-----------------------------------------'
! d3p: Three Point, d5p: Five Point
d3p=(f(x+h)-f(x-h))/(2*h)
d5p=(-1* f(x+2*h)+8*f(x+h)-8*f(x-h)+f(x-2*h))/(12*h)
print*,'Three Point derivation is', d3p
print*,'Five Point derivation is', d5p
print*,'-----------------------------------------'
! d2: Second derivation, d3: Third Derivation, d4: Fourth Derivation"
d2=(f(x+h)-2*f(x)+f(x-h))/h**2
d3=(f(x+2*h)-2*f(x+h)+2*f(x-h)-f(x-2*h))/(2*h**3)
d4=(f(x+2*h)-4*f(x+h)+6*f(x)-4*f(x-h)+f(x-2*h))/h**4
print*,'Second derivation is', d2
print*,'Third derivation is ', d3
print*,'Fourth derivation is ', d4
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
45

Forward derivation is 2.985099
Backward derivation is 3.015098
Central derivation is 3.000097
-------------------------------------------------------------------
Three Point derivation is 3.000097
Five Point derivation is 2.999997
-------------------------------------------------------------------
Second derivation is -3.000144
Third derivation is 6.001784
Fourth derivation is -18.005940
46

Problem: Write a Fortran program to find first derivation from the
following data using derivation of Newton Forward
polynomial.
x 1 2 3 4 5 6 7
y -4 -1 10 35 80 151 254
! Derivative from Newton Forward
dimension x(10),y(10),d(10,10)
n=7
data (x(i), i=1,7)/1,2,3,4,5,6,7/
data (y(i), i=1,7)/-4,-1,10,35,80,151,254/
!print*,(x(i), i=1,7)
!print*,(y(i), i=1,7)
do i=1,n-1
do j=1,n-i
y(j)=y(j+1)-y(j)
d(i,j)=y(j)
enddo
enddo
s=0
do i=1,n-1
s=s+d(i,1)/i*(-1)**(i+1)
enddo
fd=s/(x(2)-x(1))
print*,'First derivation is', fd
end
3 2
y = x − 2 x + 2 x − 5
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
The First Derivation= 1.000000
47

Problem: Write a Fortran program to find first derivation from the
following data at x = 0. 1 using derivation of Lagrange
polynomial.
x 0.1 0.2 0.3 0.4
y 0.01 0.04 0.09 0.16
! Derivation using Lagrange Formula
dimension x(40),y(40)
n=4
xx=0.1
data (x(i), i=1,4)/.1,.2,.3,.4/
data (y(i), i=1,4)/.01,.04,.09,.16/
d=0
do i=1,n
s=0
do j=1,n
if (j.eq.i) goto 10
p=1
do k=1,n
if ((k.eq.j).or.(k.eq.i)) goto 20
p=p*(xx-x(k))/(x(i)-x(k))
20 enddo
s=s+p/(x(i)-x(j))
10 enddo
d=d+s*y(i)
enddo
print*,'The First Derivation=',d
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
The First Derivation= 2.000000E-01
48

Problem: Write a Fortran program to find first derivation of the function
2 3
f ( x)
= ln(5x
) + 8x
−8
at x =1. 1 using derivation of Lagrange polynomial.
! Derivation using Lagrange Formula
f(x)=log(5*x**2)+8*x**3-8
n=4
xo=1.1
h=.1
d=0
do i=1,n
s=0
do j=1,n
if (j.eq.i) goto 10
p=1
do k=1,n
if ((k.eq.j).or.(k.eq.i)) goto 20
p=p*(1-k)/(i-k)
20 enddo
s=s+p/((i-j)*h)
10 enddo
d=d+s*f(xo+h*(i-1))
enddo
print*,'The First Derivation=',d
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
The First Derivation= 30.856790
49

Problem: Write a Fortran program to find first derivation of the function
2 3
f ( x)
= ln(5x
) + 8x
−8
at x =1. 1 using derivation of Lagrange polynomial.
! Derivation using Lagrange Formula
f(x)=log(5*x**2)+8*x**3-8
n=4
xo=1.1
h=.1
d=0
do i=1,n
s=0
do j=1,n
if (j.eq.i) goto 10
p=1
do k=1,n
if ((k.eq.j).or.(k.eq.i)) goto 20
p=p*(-h*(k-1))/(h*(i-1)-h*(k-1))
20 enddo
s=s+p/(h*(i-1)-h*(j-1))
10 enddo
d=d+s*f(xo+h*(i-1))
enddo
print*,'The First Derivation=',d
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
The First Derivation= 30.856790
50

Problem: Write a Fortran program to find second derivation of the
function:
2 3
f ( x)
= ln(5x
) + 8x
−8
at x =1. 1 using derivation of Lagrange polynomial.
! Second Derivation using Lagrange Formula
f(x)=log(5*x**2)+8*x**3-8
n=4
xo=1.1
h=.1
d=0
do i=1,n
s=0
p=1
do j=1,n
if (j.eq.i) goto 10
s=s+(-h*(j-1))
p=p*(i-j)*h
10 enddo
d=d+2*s/p*f(xo+h*(i-1))
enddo
print*,'The Second Derivation=',d
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
The Second Derivation= 51.200130
51

Problem: Write a program in Fortran to solve the differential equation
′
−
2 x
y = e − 2 y using Euler method over [ 0,2]
, with y ( 0) = 0. 1,
let h = 0. 1 (i.e. n = 20).
! Euler's Method
dimension x(0:30), y(0:30)
a=0
b=2
n=20
y(0)=0.1
x(0)=a
h=(b-a)/n
do i=0,n-1
x(i+1)=x(i)+h
y(i+1)=y(i)+h*(exp(-2*x(i))-2*y(i))
enddo
do i=0,n
print*,x(i),y(i)
enddo
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
52

0 1.000000 E-01
0.1 1.800000 E-01
0.2 2.258731 E-01
0.3 2.477305 E-01
0.4 2.530655 E-01
0.5 2.473853 E-01
0.6 2.346962 E-01
0.7 2.178764 E-01
0.8 1.989608 E-01
0.9 1.793583 E-01
1 1.600165 E-01
1.1 1.415467 E-01
1.2 1.243177 E-01
1.3 1.085260 E-01
1.4 9.424812E-02
1.5 8.147950 E-01
1.6 7.016230 E-01
1.7 6.020606 E-02
1.8 5.150217 E-02
1.9 4.393411E-02
2 3.738436 E-02
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
−2
x
Exact solution for
y′ = e − 2 y
is:
1 −2
x − 2 x
y = e − x e ⇒ y ( 2) = 0. 03846284
10
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
53

Problem: Write a program in Fortran to solve the differential equation
′
−
2 x
y = e − 2 y using Runge-Kutta method over [ 0,2]
, with
y ( 0) = 0.1, let n = 20.
! Runge Kutta Method
real k1,k2,k3,k4
dimension x(0:25),y(0:25)
a=0
b=2
n=20
x(0)=a
y(0)=0.1
h=(b-a)/n
do i=0,n-1
xo=x(i)
yo=y(i)
k1=h*(exp(-2*xo)-2*yo)
xo=x(i)+h/2
yo=y(i)+k1/2
k2=h*(exp(-2*xo)-2*yo)
xo=x(i)+h/2
yo=y(i)+k2/2
k3=h*(exp(-2*xo)-2*yo)
xo=x(i)+h
yo=y(i)+k3
k4=h*(exp(-2*xo)-2*yo)
y(i+1)=y(i)+1./6*(k1+2*k2+2*k3+k4)
x(i+1)=x(i)+h
enddo
do i=0,n
print*,x(i),y(i)
enddo
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
54

0 1.000000 E-01
0.1 1.637440 E-01
0.2 2.010928 E-01
0.3 2.195209 E-01
0.4 2.246607 E-01
0.5 2.207241 E-01
0.6 2.108327 E-01
0.7 1.972747 E-01
0.8 1.817045 E-01
0.9 1.652969 E-01
1 1.488672 E-01
1.1 1.329626 E-01
1.2 1.179324 E-01
1.3 1.039823 E-01
1.4 9.121463 E-02
1.5 7.965902 E-02
1.6 6.929560 E-02
1.7 6.007185 E-02
1.8 5.191512 E-02
1.9 4.474165 E-02
2 3.846299 E-02
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
−2
x
Exact solution for y′ = e − 2 y is:
1 −2
x − 2 x
y = e − x e ⇒ y ( 2) = 0. 03846284
10
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
55

Problem:
Write a program in Fortran to solve the differential equation y'
= −x
using Taylor series method for values x
o
= 0. 0, x = 0. 1
1 , x = 0. 2,
2 x = 0.3 , x = 0. 4,
and 5
3 4
x = 0. 5 , with the initial condition y ( 0) = 1 .
y
! Taylor Series Method
dimension x(0:25),y(0:25)
x(0)=0
y(0)=1
h=.1
n=5
do i=0,n-1
d1=-1*x(i)*y(i)
d2=(x(i)**2-1)*y(i)
d3=(-1*x(i)**3+3*x(i))*y(i)
d4=(x(i)**4-6*x(i)**2+3)*y(i)
y(i+1)=y(i)+h*d1+h**2/2*d2+h**3/6*d3+h**4/24
x(i+1)=x(i)+h
enddo
do i=0,n
print*,x(i),y(i)
enddo
end
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
0 1.000000
0.1 9.950042 E-01
0.2 9.801826 E-01
0.3 9.559749 E-01
0.4 9.230893 E-01
0.5 8.824676 E-01
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
2
x
−
2
0.125
Exact solution is: y = e ⇒ y (0.5) = e
− = 0. 08824969
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
56