numerical analysis laboratory

8/2/2019 Numerical Analysis Laboratory

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MADE BY ROLL NUMBER

1)GOURANGA SARKAR 1) 0008103010072) SANJOY CHANDRA MANDAL 2) 000810301013

3) BAPPA SAHA 3) 000810301018

4) KUNAL SINHA RAY 4) 000810301025


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C0NTENTS

SET 1 PROBLEM NO.3

SET 2 PROBLEM NO.7

SET 3 PROBLEM NO.3

SET 4 PROBLEM NO.3

SET 5 PROBLEM NO.3

SET 5 PROBLEM NO.9

SET 6 PROBLEM NO.3

SET 6 PROBLEM NO.9

SET 7 PROBLEM NO.1

GROUP NO.3

CLASS: B.CH.E 2nd

YEAR SECTION: A - 1


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SET 1 PROBLEM NO.3

TITLE: WRITE A PROGRAM TO FIND SUM OF THE FOLLOWING INFINITE SERIES

CORRECT TO THE FIFTH DECIMAL PLACES.

1+(1/2!)+(1/3!)+(1/4!)+(1/5!)+.

ALGORITHM:

Step 1: start algorithm.

Step 2: declare integer type variables a ,x ,float term=1,float sum=0. Initialize x with zero.

Step 3: clear the output screen.

Step 4: print the value of a.

Step 5: repeat step6 and step7 when a>0.

Step 6: term=term/x.

Step 7: sum=sum+term,x++, a--.

Step 8: print the sum.

Step 9: in the output and go to the next.

Step 10: end the algorithm.


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PROGRAM:

// program to calculate sum of series //

# include

# include

# include

# define eps 1.0e-05

void main(void)

{

int a;

int x=1;

float term=1;

float sum=0;

clrscr();

printf("\n\n\n enter the value of term at which term sum will be needed a:");

scanf("%d" ,&a);

} while(a>0)

{

term=term/x;

sum=sum+term;

x++;

a--;

}

printf("\n\n\n sum of series upto following term is =%f",sum);

getch();

}


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SAMPLE INPUT: Enter the value of a

1) 2

2) 3

3) 4

4) 5

5) 6

SAMPLE OUTPUT: Sum of series of corresponding value of a.

1) 1.5000000

2) 1.6666667

3) 1.7083334

4) 1.7166667

5) 1.7180556

.

REMARKS: From this program we can easily calcute the sum of this series for upto various

terms.


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SET 2 PROGRAM NO.7

TITLE:CALCULATE THE SPCIFIC VOLUME OF CHLORINE AT T=460K AND

P=15atm USING VANDER-WAALS EQUATION OF STATE USING BISECTION

TECHNIQUE.

P={RT/(v-b)}{ a/(v*v)}

a=27(R*R)*(T_c*T_c)/64P_c, b=RT_c/8P_c.

TAKE THE IDEAL GAS VOLUME AS THE INITIAL GUESS,P_c=76atm, T_c=417K

ALGORITHM:

Step 1: define the equestion f(x)=0.

Step 2: read the diserid accuracy say,epsilon(e) & maxit.

Step 3: do

{ Enter the initial approximation say, a & b.

} while(f(a)*f(b)>0);

Step 4: k < - 0.

Step 5: do

{ x < - (a+b)/2

if(f(a)*f(b)e & k < maxit);

Step 6: if (k


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PROGRAM:

//program for bisection method//

#include

#include

#include

main()

{

float f(float);

float x0,x1,x2,f0,f1,f2,R=0.0821,T=417,P=76;

int i=1,j;

printf("\n the volume of ideal gas is =%f",(R*T)/8*P);

printf("Give the guess values x0,x1 (x0


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printf("%d %f %f %f %f %f %f\n",i,x0,x1,x2,f0,f1,f2);

if((f2/f0)>0)

{

x0=x2;

f0=f2;

}

else

{

x1=x2;

f1=f2;

}

i++;

}

}

printf("\n\nThe root of the eqn converges to %f.",x2);

}

float f(float x)

{

float f;

f=(pow(x,3)-0.257*x*x+0.43*x-0.024);

return(f);


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SAMPLE INPUT : the volume of ideal gas =325.239136

the gauss value is X0 , X1 ( X0 < X1 )

-0.5

1

Upto which decimal point you want to the root?

0.0001

SAMPLE OUTPUT : 1 -0.500000 1.000000 0.250000 -0.428250 1.149000

2 -0.500000 0.250000 -0.125000 -0.428250 0.000000

22 -0.500000 -0.499999 -0.500000 -0.428250 0.000000

23 -0.500000 -0.500000 -0.500000 -0.428250 0.000000

REMARKS : the equestion converges to "-0.500000"


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SET 3 PROBLEM NO.3

TITLE: SOLVE THE FOLLOWING EQUESTION USING GAUSS-SIEDEL TECHNIQUE.

10*X1 X2 + 2*X3 =4

X1 + 10*X2 X3 =3

2*X1 + 3*X2 + 20*X3 =7

ALGORITHM:

Step 1: start the algorithm.

Step 2: sum < - 0.0

Step 3: read the desired accuracy say epsilone (e) & maxit.

Step 4: read the number of equestions.

Step 5: read the right handside constant ( b[][]).

Step 6: read the co-efficient matrix row wise ( a[][]).

Step 7: for i < - 1 to n.

X[i] < - 0.0

Repeat

Step 8: for i < - 1 to n.

for j < - 1 to n.

for k < - 1 to maxit

if( j!=1)

sum=sum + a[i][j]*X[j]

Previous x[i] < - X[i]

X[i] < - ( b[i] (sum) )/ a[i] [j]

Step 9: write i , x[i]

Step 10: fabs ( x[i]- prex[i])


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PROGRAM:

// program for gauss-siedel method //

#include

#include

#include

#define n 3

void main ( )

{

int i,j,p,k,maxit;

float x[n],a[n][n],sum=0,px[n],e=pow(e,-0.5),b[n];clrscr( );

printf("\n enter the number of equations say n:");

scanf("%d",&p);

printf("\enter the maximum iteration number say maxit:");

scanf("%d", & maxit);

printf("\n enter the right hand side constant say b:");

for(i=0;i


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for(i=0;i


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SAMPLE INPUT : enter the number of equestion ,say n : 3

enter the maximum iteration number say maxit :50

enter the right hand side constant, say b :

b[ 0 ] = 4

b[ 0 ] = 3

b[ 0 ] = 7

enter the coefficient matrix (raw wise),say a[ ][ ] :

a[ 0 ][ 0 ] = 10

a[ 0 ][ 1 ] = -1

a[ 0 ][ 2 ] = 2

a[ 1 ][ 0 ] = 1

a[ 1 ][ 1 ] = 10

a[ 1 ][ 2 ] = -1

a[ 2 ][ 0 ] = 2

a[ 2 ][ 1 ] = 3

a[ 2 ][ 2 ] = 20

CORRESPONDING SAMPLE OUTPUT :

the solution x[ 0 ] = 0.400000



REMARKS: this approximate value is very close to the actual mathematical value.


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SET 4 PROBLEM NO.3

TITLE: USE LAGRANGIAN FORMULA FOR POLYNOMIAL APPROXIMATION AND

SIMPSOMS RULE TO EVALUTE THE INTEGRAL .

ALGORITHM:

Step 1: start the algorithm.

Step 2: define function f(x).

Step 3: read the values a(lower limit) , b(upper limit) & n(no. of strips).

Step 4: h < - (b-a)/n

Step 5: ends < - f(a) + f(b)

Step 6: k < - 1sum 1< - 0.0

sum 2< - 0.0

do

{

If (k%2==0)

sum 1+=f(a+k*h)

else if (k%2==1)

sum 2+=f(a+k*h)

k< - k+1}while (K


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PROGRAM:

// program for lagrangian formula //

#include

#include

#include

void main()

{

float x[50],y[50],sum=0.0,product,z;

int n,i,j,k,X,Y;

clrscr();

printf("\n enter the total terms:");

scanf("%d",&x);

printf("\n \t\t--:enter the table:--");

printf("\n X:");

scanf("%d",&X);

printf("\n Y:");

scanf("%d",&Y);

for(i=0;i


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for(k=0;k


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SAMPLE INPUT:

enter the total terms :

--:enter the table

x:

y:

enter the point at which you want to inter polate:

SAMPLE OUTPUT:

sum=

interpolation at (x=)=


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PROGRAM:

// program for simson's rule//

#include

#include

#include

#define f(x) ((((x-1)*(x-2)*(x-4)*(x-5)*(x-6))/((0-1)*(0-2)*(0-4)*(0-5)*(0-6))*1)+(((x-0)*(x-

2)*(x-4)*(x-5)*(x-6))/((1-0)*(1-2)*(1-4)*(1-5)*(1-6))*15)+(((x-0)*(x-1)*(x-4)*(x-5)*(x-

6))/((2-0)*(2-1)*(2-4)*(2-5)*(2-6))*14)+(((x-0)*(x-2)*(x-5)*(x-1)*(x-6))/((4-0)*(4-1)*(4-

5)*(4-6)*(4-2))*5)+(((x-0)*(x-2)*(x-4)*(x-6)*(x-1))/((5-0)*(5-1)*(5-2)*(5-4)*(5-

6))*6)+(((x-0)*(x-2)*(x-1)*(x-4)*(x-6))/((6-0)*(6-1)*(6-2)*(6-4)*(6-5))*19))

void main()

{

float a,b,h,sum1,sum2,ends,i;

int j,n;

clrscr();

printf("\n enter the limits of integration:");

printf("\n lower limit (a)=\t");

scanf("%f",&a);printf("\n uppar limit(b)=\t");

scanf("%f",&b);

printf("\n enter the no. of subinterval:\n by which the interval[a,b]to be

devided :\n");

scanf("%d",&n);

if(n%2==0)

{

h=(b-a)/(float)n;

ends=f(a)+f(b);sum1=0.0;

sum2=0.0;

for(j=1;j


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else

sum2+=f(a+j*h);

}

i=(h/3)*(ends+(4*sum2)+(2*sum1));

printf("\n required integral(i)=%f",i);}

else

{

printf("\n no need for calculations");

}

getch();

}


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SAMPLE INPUT:

Enter the limits of integration :

Lower limit (a) = 0

Upper limit (b) = 6

Enter the no.of subinterval by which the interval [a,b] to be devided: 2

SAMPLE OUTPUT:

Required integral (i) =38.700001

REMARKS: the numerical value is approximately equal to the mathematical

value.


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SET 5 PROGRAM NO.3

TITLE: THE EMPIRICAL EQUESTION FOR THE WORLDS POPULATION ,y AS A

FUNCTION OF TIME IS GIVEN BY(1)dy/dt=4.25713*pow(10,-12)*pow(y,2010101) ,y0= pow(10,9)

THE BASE YEAR IS 1840 AD. OBTAIN THE WORLD POPULATION AT

DIFFERENT YEARS HENCEFORTH.

(2)Dy/dt= -200[y F(t)] + dF(t)/dt ; y0=10

F(t)=10 (10 + t)*pow(e,-t)

THE ANALYTIC SOLUTION IS

y= 10- (10 + t)*pow(e,-t) + 10*exp( -200t)

USING RANGA-KUTTA(4th

ORDER).

ALGORITHM:

Step 1: Difine f(t,y) [=RHS of ODE dy/dx= f(t,y)

Step 2: Read t0,y0,h [initial value of t,y & step size h no of t]

Step 3: Read tp [value of t at which y required]

Step 4: n=( tp-t)/h + 0.5 [n= number of step size]

Step 5: For I < - 1 to n

Step 6: K1 < - f(t0,y0)

Step 7: K2 < - f(t0 +1/2*h, y0 + 1/2*K1*h)

Step 8: K3 < - f(t0 +1/2*h, y0 + 1/2*K2*h)

Step 9: K4 < - f(t0 + 1/2*h,y0 +1/2* K3*h)

Step 10: t0 < - t0 + h

Step 11: K < - h/6*(K1 + 2*K2 + 2*K3 + 1/2*K3*h)

Step 12: yi < - yi + K

Step 13: Write I,ti,yi

Step 14: next i /(repeat)



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PROGRAM (1);

//program for range kutta//

#include

#include

#include

main()

{

int i,n;

float p,t,y,xp,h,m11,m12,m21,m22,m13,m23,m14,m24;

float func(float,float,float);

printf("enter the base year and pop.\n");

scanf("%f%f",&t,&y);

printf("\n enter the year at which pop. req.:");

scanf("%f",&xp);

printf("\n enter the step size,h:");

scanf("%f",&h);

n=(int)((xp-t)/h+0.5);

for(i=1;i


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m23=func(p+0.5*h,t+0.5*m12,y+0.5*m22);

m14=func(p+h,t+m13,y+m23);

m24=func(p+h,t+m13,y+m23);

t=t+(m11+2.0*m12+2.0*m13+m14)/6.0;

y=y+(m21+2.0*m22+2.0*m23+m24)/6.0;

printf("%5d%15.6f%15.6f\n",i,t,y);

}

printf("\n value of pop.at req. year=%f is %f\n ",t,y);

}

float func(float p,float t,float y)

{

float f;

f=(4.25713*pow(10,-12)*pow(y,2.010101));

return(f);

}


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SAMPLE INPUT :

ENTER THE BASE YEAR AND POPULATION -

1840

0.2

ENTER THE YEARAT WHICH POPULATION REQUIRED : 1841

ENTER THE STEP SIZE , h = 0.02


1 1840.000000 0.200000

REMARKS :

VALUE OF POPULATION AT REQUIRED YEAR = 1840 IS 0.200000


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PROGRAM (2):

//program for range kutta//

#include

#include

#include

main()

{

int i,n;

float t,y,xp,h,m1,m2,m3,m4;

float func(float,float);

printf("enter the base year and pop.\n");

scanf("%f%f",&t,&y);

printf("\n enter the year at which pop. req.:");

scanf("%f",&xp);


scanf("%f",&h);

n=(int)((xp-t)/h+0.5);

for(i=1;i


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m4=func(t+h,y+m3*h);

t=t+h;

y=y+(m1+2.0*m2+2.0*m3+m4)*h/6.0;

printf("%5d%15.6f%15.6f\n",i,t,y);

}

printf("\n value of pop.at req. year=%f is %f\n ",t,y);

}

float func(float t,float y)

{

float f;

f=(4.25713*pow(10,-12)*pow(y,2.010101));

return(f);

}


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SAMPLE INPUT :ENTER THE BASE YEAR AND POPULATION -

1840

0.2

ENTER THE YEARAT WHICH POPULATION REQUIRED : 1841

ENTER THE STEP SIZE , h = 0.02


1 1840.000000 0.200000

REMARKS :

VALUE OF POPULATION AT REQUIRED YEAR = 1840 IS 0.200000


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SET 6 PROBLEM NO.3

TITLE: (1)COMPUTE THE NUMERICAL SOLUTION OVER THE INTERVEL [0,2]. USING

4TH

ORDER RANGA KUTTA METHOD.

dy1/dt= -0.2*y1 + 0.2*y2

dy2/dt= 10*y1 (60+0.125*t)*y2 + 0.124*t ,[y0]=[0,0]T

(2)COMPUTE THE NUMERICAL SOLUTION, USING 4TH

ORDER RANGA KUTTA.

dy1/dt= 1.3*(y2-y1) + 1.04*x*10^4ky2

dy2/dt= 1.88*x*10^3*[y4-(1+k)]

dy3/dt= 1752 + 266.1y2 - 269.3*y3

dy4/dt= 0.1 + 320y2 321*y4

k= 6*x*10^-3exp[20.7 - 15*x*10^3/y1]

[y0]= [759.167,0,600,0.1]^T

ALGORITHM:

Step 1: Difine f(t,y) [=RHS of ODE dy/dx= f(t,y) Step 2: Read t0,y0,h [initial value of t,y & step size h no of t]

Step 3: Read tp [value of t at which y required]

Step 4: n=( tp-t)/h + 0.5 [n= number of step size]

Step 5: For I < - 1 to n

Step 6: K1 < - f(t0,y0)

Step 7: K2 < - f(t0 +1/2*h, y0 + 1/2*K1*h)

Step 8: K3 < - f(t0 +1/2*h, y0 + 1/2*K2*h)

Step 9: K4 < - f(t0 + 1/2*h,y0 +1/2* K3*h)

Step 10: t0 < - t0 + h Step 11: K < - h/6*(K1 + 2*K2 + 2*K3 + 1/2*K3*h)

Step 12: yi < - yi + K

Step 13: Write I,ti,yi

Step 14: next i /(repeat)



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PROGRAM (1):

//program for 4th order ranga kutta//

#include

#include

#include

main()

{

int i,n;

float t,y1,y2,xp,h,m11,m12,m21,m22,m13,m23,m14,m24;

float func1(float,float,float);

float func2(float,float,float);

printf("enter the initial values of t,y1,y2 \n");

scanf("%f%f%f",&t,&y1,&y2);

printf("\n enter the time at which y1 and y2 are req.:");

scanf("%f",&xp);


scanf("%f",&h);

n=(int)((xp-t)/h+0.5);

for(i=1;i


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m13=h*func1(t+0.5*h,y1+0.5*m12,y2+0.5*m22);

m23=h*func2(t+0.5*h,y1+0.5*m12,y2+0.5*m22);

m14=h*func1(t+h,y1+m13,y2+m23);

m24=h*func2(t+h,y1+m13,y2+m23);

t=t+h;

y1=y1+(m11+2.0*m12+2.0*m13+m14)/6.0;

y2=y2+(m21+2.0*m22+2.0*m23+m24)/6.0;

printf("%5d%15.6f%15.6f%15.6f\n",i,t,y1,y2);

}

printf("\n values of y1 and y2 at t=%f is %f and %f",t,y1,y2);

}

float func1(float t,float y1,float y2)

{

float f1;

f1=(-0.2*y1+0.2*y2);

return(f1);

}

float func2(float t,float y1, float y2)

{

float f2;

f2=(10.0*y1-(60.0+0.125*t)*y2+0.124*t);

return(f2);

}


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SAMPLE INPUT :

ENTER THE INITIAL VALUES OF t , y1 , y2

0

0

0ENTER THE TIME AT WHICH Y1 AND Y2 ARE REQUIRED :2

ENTER THE STEP SIZE ,h :0.001


VALUES OF y1 AND y2 AT t = 2.000037 IS0.000728

AND 0.004201


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PROGRAM (2):

//program for 4th order multidimentional range kutta//

#include

#include

#include

main()

{

int i,n;

float

t,y1,y2,y3,y4,xp,h,m11,m12,m21,m22,m13,m23,m14,m24,m31,m32,m33,m34,m41,m42,m43,

m44;

float func1(float,float,float,float,float);




printf("enter the initial values of t,y1,y2,y3,y4 \n");

scanf("%f%f%f%f%f",&t,&y1,&y2,&y3,&y4);

printf("\n enter the time at which y1,y2,y3 and y4 are req.:");

scanf("%f",&xp);


scanf("%f",&h);

n=(int)((xp-t)/h+0.5);


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for(i=1;i


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printf("%5d%15.6f%15.6f%15.6f%15.6f%15.6f\n",i,t,y1,y2,y3,y4);

}

printf("\n values of y1,y2,y3 and y4 at t=%f are %f,%f,%f and %f",t,y1,y2,y3,y4);

}

float func1(float t,float y1,float y2,float y3,float y4)

{

float f1;

f1=1.3*(y2-y1)+1.04*pow(10,4)*(6*pow(10,-4)*pow(2.73,(20.7-

(15*pow(10,3))/y1)))*y2;

return(f1);

}

float func2(float t,float y1, float y2,float y3,float y4)

{

float f2;

f2=1.88*pow(10,3)*(y4-(1.0+(6*pow(10,-4)*pow(2.73,(20.7-(15*pow(10,3))/y1)))));

return(f2);

}


{

float f3;

f3=(1752+266.1*y2-269.3*y3);

return(f3);

}


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{

float f4;

f4=(0.1+320*y3-321.0*y4);

return(f4);

}


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SAMPLE INPUT :

ENTER THE INITIAL VALUES OF t , y1 , y2

0

00

ENTER THE TIME AT WHICH Y1 AND Y2 ARE REQUIRED :2

ENTER THE STEP SIZE ,h :0.001


VALUES OF y1 AND y2 AT t = 2.000037IS 0.000728

AND 0.004201


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SET 7 PROBLEM NO.1

TITLE:MULTIDIMENSIONAL NEWTON RAPSONS METHOD.

#include

#include

#include

#define f(x,y) (x*x+x*y+y*y-7)

#define g(x,y) (pow(x,3)+pow(y,3)-9)

main()

{

int i,j,k,n;

double x_0=1.5,y_0=0.5,det_J,b[2][1],a[2][1],x[2][1];

b[0][0]=x_0;

b[1][0]=y_0;

x[0][0]=x_0;

x[1][0]=y_0;

for(k=0;k


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x[0][0]=b[0][0]-(a[0][0]/det_J);

x[1][0]=b[1][0]-(a[1][0]/det_J);

printf("\nFor the %d iteration current value: %f %f",k+1,x[0][0],x[1][0]);

b[0][0]=x[0][0];

b[1][0]=x[1][0];

}

getch();

return 0;

}


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SAMPLE OUTPUT:

for the 1 iteration current value:2.267544 0.925439






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