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

Question

Varun 2024-8-2，5:21

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评论： Varun 2024-8-2，12:48

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

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VBBV 2024-8-2，12:22

在 MATLAB Online 中打开

@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
<mw-icon class=""></mw-icon>
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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
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>

Varun 2024-8-2，12:48

Thank you for your response sir

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Answer 1

Arnav 2024-8-2，6:16

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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

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Walter Roberson 2024-8-2，6:36

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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 2024-8-2，12:22

Thank you for your response

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Answer 2

Walter Roberson 2024-8-2，6:16

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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.