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

조회 수: 5 (최근 30일)
Varun 2024년 8월 2일
댓글: Varun 2024년 8월 2일
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 2024년 8월 2일
@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 2024년 8월 2일
Thank you for your response sir

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### 채택된 답변

Arnav 2024년 8월 2일
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
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Walter Roberson 2024년 8월 2일
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 2024년 8월 2일

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### 추가 답변 (1개)

Walter Roberson 2024년 8월 2일
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.

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