about if statement and solution

Dear sir,

when I run this bellow code, it shows this error,

code:

syms x
a=solve(x^2+4*x+4);
b=a(1)
if b>2
    s=2
else
    s=3
end

error:

Undefined function or method 'gt' for input arguments of type 'sym'.

Error in ==> solution at 4
if b>2

my target is, I have to compare the value of 'x' in the equation with constant. but I got this error.

please help me the solution for my problem..

Answer 1

[Walter Roberson]

If you want numeric answers, work numerically:

a = roots([1, 4, 4]);

solve() does not return numbers: it returns formula, some of which happen to print out like numbers. If you happen to get a solve() result that involves only known functions of constants, then you can use double() on the result to calculate the double-precision approximation of the result.

By the way, you should not assume that the results of solve() are returned in any particular order.

[Dr. Siva Malla]

thanks Walter:

but actually I want a solution of a*x^2+b*sinh((c*x^3+d*x^2+e*x+f)/x)+g=0; hear a, b, c, d, e, f, g are the constants and x is the variable;

I think here x having 2 solutions, I have to compare this 2 solutions with constant. using if statement. so please help me in this portion. here the problem with sinh; i wrote a code to solve x, like

y=a*x^2+b*x*sinh(c*x^3+d*x^2+e*x+f)/x)+g;

f=solve(y);

if f>2

r=3;

else

r=4

end

I wrote this code, but I got error. mentioned above.....later that I changed and I get one of the solution from set of the solution of x. but I got same error. please give me solution for this error. I have to compare this solution of x to constant. it is my target.....

[Walter Roberson]

When f is a vector, the test "if f>2" is the same as "if all(f >2)" . Is that what you want? Or do you want to test "if any(f > 2)" ?

When I tested the formula using some arbitrary constants, I found only one solution, not two: with those particular constants at least, one side of the discontinuity was strictly positive but the other side crossed 0.

[Walter Roberson]

In any case, test double(f) not f itself.

[Dr. Siva Malla]

Sir,

I have to extract first f1, f2....fn in the vector f, means in all values of f, and I have to test any value of f. like if f2>2; because, first I have to take only real values of f and only I should test these real values of f to 2,( any real value of f). this way I want.

here I am giving complete my program please test. in this program, I cant work with if statement. so please rectify this problem and please help me.

program:

L=1.0*1e-4;

a=0.15*1e-4;

W=10*1e-4;

q=1.6*1e-19;

Es=85.845*1e-14;

Nd=5*1e17;

Nc=1.83*1e19;

Nv=4.07*1e19;

Nref=2*1e17;

KT=0.0259;

q1=1;

h1=0.98;

Umin=40;

Umax=950;

r=0.76

x1=1;

vsat=2.07*1e7;

k1=q*Nd*W*a*x1*vsat;

Vbi=0.8-((KT*q1)*log(Nc/Nd));

U0= Umin + ((Umax - Umin)/(1+ ((Nd/Nref)^r)));

G0=(q*Nd*U0*W*a)/L;

Vp=(q*Nd*a^2)/(2*Es);

Ip=G0*Vp/3;

Z=(q*Nd*a^2*U0)/(2*Es*L*vsat);

Ec=vsat*(h1/(1-h1))/U0;

VG=-8

for k=1:8

Vd=1:20;

a1=((Vbi-VG)/Vp)^0.5;

syms x

for i=1:20

b=x^2-a1^2;

c=h1*(1-x);

d=2*(x^3-a1^3)/3;

e=L*Z*((b-d)/c-b);

y=Vp*b+2*Ec*a*x/pi*sinh(pi*(L-e)/(2*a*x))-Vd(i);

f=solve(y);

f

VL1=Vp*(f^2-a1^2);

end

VG=VG+1;

if Vd(i)>VL1

w=2;

else

w=3;

end

end