plot the equation graph

how to plot graph V vs x

equation is

V*sqrt(1-x)=m*pi+atan((x)/(1-x))+atan((a+x)/(1-x))

taking V=1:10

m=0

a=1;

then plot V vs x

[Wan Ji]

It's your homework, not our duty to help you. Your words are impolite and made me uncomfortable. I delelted the code that I wrote. Looking back to the questions you've asked, you are lacking in basic respect and gratefulness to others who have helped you. Hope you behave yourself better

[shiv gaur]

sorry for inconviniennce but graph is not showing exact value on page 63 fig 3.14 for a=10 the answer is not showing as shown in fig equation 3,44 sorry again for again

[shiv gaur]

Answer 1

[Wan Ji]

Hi,

It is not neccessary to solve this equation

x = 0:0.00001:1;
f = @(x,m,a)(m*pi+atan((x)/(1-x))+atan((a+x)/(1-x)))./sqrt(1-x);
m = [0,1,2];
a = [0,10,1e15];
V1 = f(x,m(1),a(1));
V2 = f(x,m(1),a(2));
V3 = f(x,m(1),a(3));
plot(V1,x)
hold on 
plot(V2,x)
plot(V3,x)
V1 = f(x,m(2),a(1));
V2 = f(x,m(2),a(2));
V3 = f(x,m(2),a(3));
plot(V1,x)
hold on 
plot(V2,x)
plot(V3,x)

V1 = f(x,m(3),a(1));
V2 = f(x,m(3),a(2));
V3 = f(x,m(3),a(3));
plot(V1,x)
hold on 
plot(V2,x)
h = plot(V3,x);
axis([0, 15, 0, 1.1])
grid on

[shiv gaur]

graph is fine and matching with fig but sir you have assume the value of x this equation is transcedental equation so you have to required some other method so that find the root of the equation satisfy the equation by varying the value of V suppose 1:10 so 10 value of x can find so these 10 value of V and 10 value of x gives the graph so sir we an use numerical method

[shiv gaur]

plot n vs x n vs y

for n=1:10

[t,r]=ode45(@fn,[0 140],[0.01+n 0.01 ])

plot (n,r(:,1))

end

function dr=fn(t,r)

x=r(1); y=r(2);

dr=zeros(2,1)

dr(1)=x+y;

dr(2)=x-y;

end

[shiv gaur]

using array index

[Wan Ji]

Is this the right way to access x when V = 0:10?

the main function

function main
mu = 0;
a = 1;
V_arr = 0:1:15;
[V_arr, X_arr] = getX(V_arr,mu(1),a(1));
plot(V_arr, X_arr,'rs-','markerfacecolor','r','markersize',10);
axis([0 15 -0.5 1])
end

function [V_arr, xnorm] = getX(V_arr,mu,a)

f = @(V,x,m,a)V*sqrt(1-x)-(m*pi+atan((x)/(1-x))+atan((a+x)/(1-x)));
m = mu;
a = a;
x0=-0.999;
x1=0;
x2=0.999;
TOL=1e-14;
Nmax=1000;
% V_arr = 0:1:15;
X_arr = zeros(size(V_arr));
for i = 1:1:numel(V_arr)
V = V_arr(i);
X_arr(i) = muller(@(x)f(V,x,m,a), x0, x1, x2, TOL, Nmax);
% X_arr(i) = real(fsolve(@(x)f(V,x,m,a), 0.8));
end

% xnorm = (X_arr-min(X_arr))/(max(X_arr)-min(X_arr));
xnorm = X_arr;
end

Muller method

function X = muller(f, x0, x1, x2, TOL, Nmax)
for n=1:Nmax
    h1=x1-x0;
    h2=x2-x1;
    d1=(f(x1)-f(x0))/h1;
    d2=(f(x2)-f(x1))/h2;
    a=(d1-d2)/(h2+h1);
    b=d2+h2*a;
    c=f(x2);
    t=x2-(2*c*sign(b))/(abs(b)+sqrt(b^2-4*a*c));
    s=f(t);
    if abs(s)<TOL
        break
    else
        x0=x1;
        x1=x2;
        x2=t;
    end
end
X = real(t);
end

[shiv gaur]

sir muller method giving right value test program for x^2+1 but it is showing only one root

[shiv gaur]

sir muller method giving right value test program for x^2+1 but it is showing only one root

[Wan Ji]

you can change x0,x1 and x2 to get more roots if there are

[shiv gaur]

Mj=[cos (kj hj) i*sin(kj hj) /kj ; i*kj*sin(kj hj) cos (kj hj) ];

kj=sqrt(nj2*k02-x2);

Take n1=1.521;n2=2.66;k0=1;h1=1.5e-6;h2=1e-6

We take M=M1*M2;

Where M=( m11 m12;m21 m22) as usual in

Matrix;

How to multiply this matrix for M find the multiplication of matrix and find the value of

x;

and solve the eq

ns=1.512;

nc=1.2-i7;

gs=x2-ns2k02;

gc= x2-ns2k02;

f(x)=igsm11+gcm22)-m21+gsgcm12;

and solve the eq and value of x