sir, i am not able develop a matlab code to solve four transcendental equations using newton raphson method? i had tried till the creation of jacobian matrix but not able to substitute values in the jacobian matrix please help to solve this problem?

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omkari sai krishna 2020-4-19

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编辑： Devineni Aslesha 2020-5-4

采纳的回答： Devineni Aslesha

syms x1 x2 x3 x4 epsi;

g1=cos(x1)+cos(x2)+cos(x3)+cos(x4)-0.9424;

g2=cos(5*x1)+cos(5*x2)+cos(5*x3)+cos(5*x4);

g3=cos(7*x1)+cos(7*x2)+cos(7*x3)+cos(7*x4);

g4=cos(11*x1)+cos(11*x2)+cos(11*x3)+cos(11*x4);

g=[g1;g2;g3;g4];

x=[x1; x2; x3; x4];

epsi=10e-5;

j=jacobian(g,x);

for i=1:4

x(i,1)=input('enter the intial values\n')

end

disp(x)

f=subs(j);

j_inv=inv(f);

disp(-g);

k=subs(-g);

p=j_inv*k;

This is my code i am not able to substitute initial values in 'P' matrix. please help me to solve this.

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Answer 1

Devineni Aslesha 2020-4-24

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In the given code, the values are not substituted in the jacobian matrix because the initial values are assigned to the variable 'x' but the variables used in the equations are 'x1,x2,x3 and x4'. Please find the code below to substitute the initial values.

x1 = x(1,1);
x2 = x(2,1);
x3 = x(3,1);
x4 = x(4,1);

For more information, refer to this file exchange link

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omkari sai krishna 2020-5-3

Thank you mam for your support but still i am not able to solve this problem. I am giving the updated code for four nonlinear equations please give me suggestion where i am doing wrong.

clc;

close all;

clear all;

syms x y z v;

f=cosd(x)+cosd(y)+cosd(z)+cosd(v)-1.884;

fx=diff(f,x);fy=diff(f,y);fz=diff(f,z);fv=diff(f,v);

g=cosd(5*x)+cosd(5*y)+cosd(5*z)+cosd(5*v);

gx=diff(g,x);gy=diff(g,y);gz=diff(g,z);gv=diff(g,v);

h=cosd(7*x)+cosd(7*y)+cosd(7*z)+cosd(7*v);

hx=diff(h,x);hy=diff(h,y);hz=diff(h,z);hv=diff(h,v);

l=cosd(11*x)+cosd(11*y)+cosd(11*z)+cosd(11*v);

lx=diff(l,x);ly=diff(l,y);lz=diff(l,z);lv=diff(l,v);

n=input('Enter the number of decimal places:');

epsilon = 5*10^-(n+1)

x0 = input('Enter the x intial approximation:');

y0 = input('Enter the y intial approximation:');

z0 = input('Enter the z intial approximation:');

v0 = input('Enter the v initial value:');

xk = x0;yk = y0;zk = z0;vk = v0;

for i=1:100

fk=vpa(subs(f,[x y z v],[xk yk zk vk]));

gk=vpa(subs(g,[x y z v],[xk yk zk vk]));

hk=vpa(subs(h,[x y z v],[xk yk zk vk]));

lk=vpa(subs(l,[x y z v],[xk yk zk vk]));

fxk=vpa(subs(fx,[x y z v],[xk yk zk vk]));

fyk=vpa(subs(fy,[x y z v],[xk yk zk vk]));

fzk=vpa(subs(fz,[x y z v],[xk yk zk vk]));

fvk=vpa(subs(fv,[x y z v],[xk yk zk vk]));

gxk=vpa(subs(gx,[x y z v],[xk yk zk vk]));

gyk=vpa(subs(gy,[x y z v],[xk yk zk vk]));

gzk=vpa(subs(gz,[x y z v],[xk yk zk vk]));

gvk=vpa(subs(gv,[x y z v],[xk yk zk vk]));

hxk=vpa(subs(hx,[x y z v],[xk yk zk vk]));

hyk=vpa(subs(hy,[x y z v],[xk yk zk vk]));

hzk=vpa(subs(hz,[x y z v],[xk yk zk vk]));

hvk=vpa(subs(hv,[x y z v],[xk yk zk vk]));

lxk=vpa(subs(lx,[x y z v],[xk yk zk vk]));

lyk=vpa(subs(ly,[x y z v],[xk yk zk vk]));

lzk=vpa(subs(lz,[x y z v],[xk yk zk vk]));

lvk=vpa(subs(lv,[x y z v],[xk yk zk vk]));

xyzvk = [xk yk zk vk]'; %The old values of x,y,z

J = [fxk fyk fzk fvk;gxk gyk gzk gvk;hxk hyk hzk hvk;lxk lyk lzk lvk];

Fk = [fk gk hk vk]'; % Matrix of function values

xyzvkplus1 = (xyzvk - inv(J)*Fk);

if abs(xyzvkplus1 - xyzvk) < epsilon

break;

else

display('no solution');

end

xyzvans = xyzvkplus1-rem(xyzvkplus1,10^-n);

fprintf('The Root Matrix is :\n');

display(xyzvans);

for this code i am getting soltion more than 90 degrees.

Devineni Aslesha 2020-5-4

编辑：Devineni Aslesha 2020-5-4

In the above code, you are computing the answer 'xyzvans' even if there is no solution available. Use the return command after display('no solution'); to stop running this script when there is no solution. In this way, the answer will be computed only when the error value is less than epsilon.

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