My question is what is the value i am getting as the output of my code.

syms x
eqn = 0.5959 + 0.0321*(x^2.1)-0.184*(x^8)+0.0143*(x^2.5)-0.2359*(sqrt((1-x^4)/x^4));
solve(eqn == 0,x)

the results are given as

[Varun]

If any one find this question answerable or understandable please do respond or the just give an idea of what are those values whih are so complex by look.

[VBBV]

@Varun You could alternately use fsolve for the expression and solve it for defined initial value limits. Note that your equation/ function involves a term ((sqrt((1-x.^4)./x.^4))) which can potentially cause the result into a indefinite value. Hence, the initial value need to be more specific to solve this type of equation.

x0 = [0.01 1]  % give an initial value excluding 0 !!!

x0 = 1x2

0.0100 1.0000

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eqn = @(x) 0.5959 + 0.0321*(x.^2.1)-0.184*(x.^8)+0.0143*(x.^2.5)-0.2359*(sqrt((1-x.^4)./x.^4));
options = optimoptions('fsolve','Display','iter');                                                                                     % ^
[x] = fsolve(eqn,x0,options)

                                             Norm of      First-order   Trust-region
 Iteration  Func-count     ||f(x)||^2           step       optimality         radius
     0          3         5.56207e+06                        1.11e+09              1
     1          6         1.09835e+06     0.00500015         1.47e+08              1
     2          9              216815      0.0231918         1.93e+07              1
     3         12               42764       0.738655         2.54e+06              1
     4         15             8428.05       0.771889         3.35e+05              1
     5         18             1651.66       0.194943         4.43e+04              1
     6         21             320.676        0.20026         5.87e+03              1
     7         24             60.9821       0.257676              783              1
     8         25             60.9821              1              783              1
     9         28             11.0813           0.25              106           0.25
    10         31             1.75585       0.116714             14.9           0.25
    11         34            0.205127       0.118429             2.17           0.25
    12         37           0.0107402      0.0945807            0.294           0.25
    13         40         7.55946e-05      0.0365412           0.0207           0.25
    14         43         5.37719e-09     0.00365487         0.000172           0.25
    15         46         2.81969e-17    3.13519e-05         1.24e-08           0.25

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

x = 
   0.6015 - 0.0000i  -0.0344 - 1.1446i

real(x)

ans = 1x2

0.6015 -0.0344

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[Varun]

Thank you for your response sir

Answer 1

[Arnav]

Hi @Varun,

The result that you are getting is a closed form way of representing the solution of the equation. Here root(P(x),x,k) represents the kth root of the symbolic polynomial P(x). You can evaluate these roots numerically using the function vpa as follows:

syms x
eqn = 0.5959 + 0.0321*(x^2.1)-0.184*(x^8)+0.0143*(x^2.5)-0.2359*(sqrt((1-x^4)/x^4));
res = solve(eqn == 0,x);
numeric_res = vpa(res)

The numeric result obtained is:

0.6015191969401057095577237699673
- 0.60342554599494554515246506424316 - 0.0032524435318404812187506975163144i
- 0.60342554599494554515246506424316 + 0.0032524435318404812187506975163144i
- 0.0022957627114584795743384488144413 - 0.6148033016363615664866637134719i
- 0.0022957627114584795743384488144413 + 0.6148033016363615664866637134719i
- 0.034350766702267990632625142341669 - 1.1446441922416124914568634765747i
- 0.034350766702267990632625142341669 + 1.1446441922416124914568634765747i
0.045439493798166588383693273719136 - 1.14543346751870257491869864044i
0.045439493798166588383693273719136 + 1.14543346751870257491869864044i
- 1.1691779737703086214274911464114 - 0.029488409879496751528279391787196i
- 1.1691779737703086214274911464114 + 0.029488409879496751528279391787196i
1.1752945738347608763208400050921 - 0.037263119738081124633951256034582i
1.1752945738347608763208400050921 + 0.037263119738081124633951256034582i

You can refer to the below example for more information:

https://www.mathworks.com/help/symbolic/sym.root.html#mw_9bfdcac9-2d0b-4cce-a228-0387033ade2a

[Walter Roberson]

format long g
syms x
eqn = 0.5959 + 0.0321*(x^2.1)-0.184*(x^8)+0.0143*(x^2.5)-0.2359*(sqrt((1-x^4)/x^4));
F = matlabFunction(eqn)

F = function_handle with value:
    @(x)sqrt(-1.0./x.^4.*(x.^4-1.0)).*(-2.359e-1)+x.^(5.0./2.0).*1.43e-2-x.^8.*(2.3e+1./1.25e+2)+x.^(2.1e+1./1.0e+1).*3.21e-2+5.959e-1

numeric_res = fzero(F, [1e-6 1])

numeric_res = 
         0.601519196940106

[Varun]

Thank you for your response

Answer 2

[Walter Roberson]

You are getting garbage values, for a garbage query.

You have used floating point quantities in a symbolic equation. You have used solve() on the equation. solve() is intended to find indefinitely precise solutions. It makes no sense to ask for indefinitely precise solutions to equations involving floating point values, since floating point values are inherently imprecise.