Also
ı dont want to use syms subprogram
value must be numerical
Solve equation for one unknown with one known parameter
3 次查看（过去 30 天）
显示 更早的评论
clear all
clc
a=0.5
x=linspace(-2*a,2*a,100)
eqn=@(x,y,a)(1/2*a^2)*[x^2./(2-(1-(x./(2*a))^2-(y./(2*a))^2)+sqrt((1-(x./(2*a))^2-(y./(2*a))^2)+(y./a)^2))-y^2./((1-(x./(2*a))^2-(y./(2*a))^2)+sqrt((1-(x./(2*a))^2-(y./(2*a))^2)+(y./a)^2))]-1==0;
solve(eqn,y)
Answer is
Unrecognized function or variable 'y'.
Error in naca (line 6)
solve(eqn,y)
how can I solve y for x matrix value
2 个评论
Star Strider
2021-5-18
The actual equations are:
I = imread('https://www.mathworks.com/matlabcentral/answers/uploaded_files/621828/image.png');
figure
imshow(I)
.
采纳的回答
Star Strider
2021-5-18
Try something like this —
a=0.5;
% x=linspace(-2*a,2*a,100)
x=linspace(-2*a,2*a,20);
eqn=@(x,y,a)(1/2*a^2)*[x^2./(2-(1-(x./(2*a))^2-(y./(2*a))^2)+sqrt((1-(x./(2*a))^2-(y./(2*a))^2)+(y./a)^2))-y^2./((1-(x./(2*a))^2-(y./(2*a))^2)+sqrt((1-(x./(2*a))^2-(y./(2*a))^2)+(y./a)^2))]-1;
for k = 1:numel(x)
yv(k,:) = fsolve(@(y)eqn(x(k),y,a), 10);
end
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
Results = table(x', yv)
Results = 20×2 table
Var1 yv
_________ ______
-1 1.9833
-0.89474 2.1056
-0.78947 2.2028
-0.68421 2.2812
-0.57895 2.3444
-0.47368 2.3946
-0.36842 2.4335
-0.26316 2.4619
-0.15789 2.4805
-0.052632 2.4897
0.052632 2.4897
0.15789 2.4805
0.26316 2.4619
0.36842 2.4335
0.47368 2.3946
0.57895 2.3444
It might be necessary to use the uniquetol function to eliminate duplicate (or near-duplicate) values of ‘yv’ and thier associated ‘x’ values.
.
9 个评论
onur karakurt
2021-5-18
I think it is not correct
because first and last value of y must be zero.
Actually ı want to solve these equation.
(x/(2*a*cos(theta)))^2-(y/(2*a*sin(theta)))=1
(2*(sin(theta))^2)=p+sqrt(p^2+(y/a)^2)
p=1-(x/2*a)^2-(y/2*a)^2
Here I simplified these function to f(x,y) shown below. May be I make a mistake while simplifying equations. Here first and last y value must be zero
(1/2*a^2)*[x^2./(2-(1-(x./(2*a))^2-(y./(2*a))^2)+sqrt((1-(x./(2*a))^2-(y./(2*a))^2)+(y./a)^2))-y^2./((1-(x./(2*a))^2-(y./(2*a))^2)+sqrt((1-(x./(2*a))^2-(y./(2*a))^2)+(y./a)^2))]=1
Star Strider
2021-5-18
I’m lost.
What variables are known, what variables are to be solved for?
The fsolve fucntion can solve systems such as these, however more needs to be known about them.
onur karakurt
2021-5-18
x and a are known, y unknown.
Now ı trying to find the roots of y because it is nonlinear function.
when I fits x and a to any constant to find equation answers against the y values from 0 to 1 with 100 pieces. I saw infinite roots of y, I guess. But it must be any solution for this.
clear all
clc
a=0.5;
x=0.5;
y=linspace(0,1,100);
for i=1:size(y)
y=y(i,:);
z=(1e15)*((x./(2*a*cos(asin(sqrt(((1-(x./(2*a)).^2-(y./(2*a)).^2)+sqrt((1-(x./(2*a)).^2-(y./(2*a)).^2).^2+(y./a).^2))./2))))).^2-(y./(2*a*sin(asin(sqrt(((1-(x./(2*a)).^2-(y./(2*a)).^2)+sqrt((1-(x./(2*a)).^2-(y./(2*a)).^2).^2+(y./a).^2))./2))))).^2-1);
end
plot(y,z)
onur karakurt
2021-5-18
can you fid the roots of y for each x value while z=zero, a=0.5 and x=linspace(-2*a,2*a,100)
equation is this
z=(1e15)*((x./(2*a*cos(asin(sqrt(((1-(x./(2*a)).^2-(y./(2*a)).^2)+sqrt((1-(x./(2*a)).^2-(y./(2*a)).^2).^2+(y./a).^2))./2))))).^2-(y./(2*a*sin(asin(sqrt(((1-(x./(2*a)).^2-(y./(2*a)).^2)+sqrt((1-(x./(2*a)).^2-(y./(2*a)).^2).^2+(y./a).^2))./2))))).^2-1);
Star Strider
2021-5-19
Finding all the roots is going to be a challenge.
First, using fsolve (I also experimented with fzero) is not very revealing —
zfcn = @(x,y,a) (1e15)*((x./(2*a*cos(asin(sqrt(((1-(x./(2*a)).^2-(y./(2*a)).^2)+sqrt((1-(x./(2*a)).^2-(y./(2*a)).^2).^2+(y./a).^2))./2))))).^2-(y./(2*a*sin(asin(sqrt(((1-(x./(2*a)).^2-(y./(2*a)).^2)+sqrt((1-(x./(2*a)).^2-(y./(2*a)).^2).^2+(y./a).^2))./2))))).^2-1);
a=0.5;
x=linspace(-2*a,2*a,100);
S = zeros(numel(x),3);
for k = 1:numel(x)
[y,fv] = fsolve(@(y)zfcn(x(k),y,a), x(k));
S(k,:) = [x(k), y, fv];
end
Equation solved at initial point.
fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
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.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.
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.
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.
Equation solved at initial point.
fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
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.
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.
Equation solved at initial point.
fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
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.
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.
Equation solved at initial point.
fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
Equation solved at initial point.
fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
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.
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.
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.
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.
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.
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.
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.
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.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
Equation solved at initial point.
fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
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.
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.
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.
No solution found.
fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.
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.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
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.
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.
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.
No solution found.
fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.
No solution found.
fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
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.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
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.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.
No solution found.
fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.
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.
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.
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.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
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.
No solution found.
fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.
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.
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.
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.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
Equation solved at initial point.
fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
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.
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.
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.
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.
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.
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.
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.
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.
Equation solved at initial point.
fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
Equation solved at initial point.
fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
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.
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.
Equation solved at initial point.
fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
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.
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.
Equation solved at initial point.
fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
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.
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.
No solution found.
fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
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.
No solution found.
fsolve stopped because the relative size of the current step is less than the
value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the value of the function tolerance.
Equation solved at initial point.
fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
Out = array2table(S,'VariableNames',{'x','y','FunctionValue'})
Out = 100×3 table
x y FunctionValue
________ ________ _____________
-1 -1 0
-0.9798 -0.9798 -0.22204
-0.9596 -0.9596 0
-0.93939 -0.93939 0.22204
-0.91919 -0.91919 0.22204
-0.89899 -0.89899 0
-0.87879 -0.87879 0
-0.85859 -0.85859 0
-0.83838 -0.83838 0
-0.81818 -0.81818 0
-0.79798 -0.79798 0
-0.77778 -0.77778 0
-0.75758 -0.75758 0
-0.73737 -0.73737 0
-0.71717 -0.71717 0
-0.69697 -0.69697 0
Second, a relatively easy way to determine the roots is to plot them using the contour function, drawing contours at the roots of ‘z’ —
y = linspace(-2*pi, 2*pi);
[Xm,Ym] = ndgrid(x,y);
Zm = zfcn(Xm,Ym,a);
figure
[M,h] = contour(Xm,Ym,Zm, [0 0]); % Draw Contours Only At z=0
grid
title('Contour Map of (x,y) Coordinates Where z=0')
xlabel('x')
ylabel('y')
% axis('equal')
Choose whatever limits you want for ‘y’. Here, I chose the interval 
.
. So there are myriad symmetrical roots —
format shortE
M
M = 2×10722
0 -1.0000e+00 -9.7980e-01 -9.5960e-01 -9.5296e-01 -9.5960e-01 -9.6839e-01 -9.7980e-01 -9.9197e-01 -1.0000e+00 0 -1.0000e+00 -9.7980e-01 -9.6466e-01 -9.5960e-01 -9.5339e-01 -9.3939e-01 -9.2945e-01 -9.1919e-01 -9.1044e-01 -9.1528e-01 -8.9899e-01 -8.8995e-01 -8.8131e-01 -8.7879e-01 -8.7605e-01 -8.5859e-01 -8.4236e-01 -8.3838e-01 -8.1818e-01
9.0000e+00 -6.2233e+00 -6.2400e+00 -6.1954e+00 -6.1563e+00 -6.0802e+00 -6.0293e+00 -5.9283e+00 -6.0293e+00 -6.0905e+00 6.0000e+01 -5.8454e+00 -5.8875e+00 -5.7755e+00 -5.7244e+00 -5.6485e+00 -5.5255e+00 -5.6485e+00 -5.7244e+00 -5.6485e+00 -5.5216e+00 -5.4117e+00 -5.5216e+00 -5.6485e+00 -5.7041e+00 -5.6485e+00 -5.5861e+00 -5.5216e+00 -5.4043e+00 -5.4184e+00
format short
See the documentation for contour and specifically for M to understand how to retrieve the values that contour creates.
This is the only way I can think of to solve this.
.
更多回答（0 个）
另请参阅
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!发生错误
由于页面发生更改，无法完成操作。请重新加载页面以查看其更新后的状态。
Translated by 
您也可以从以下列表中选择网站：
如何获得最佳网站性能
选择中国网站（中文或英文）以获得最佳网站性能。其他 MathWorks 国家/地区网站并未针对您所在位置的访问进行优化。
美洲
- América Latina (Español)
- Canada (English)
- United States (English)
欧洲
- Belgium (English)
- Denmark (English)
- Deutschland (Deutsch)
- España (Español)
- Finland (English)
- France (Français)
- Ireland (English)
- Italia (Italiano)
- Luxembourg (English)
- Netherlands (English)
- Norway (English)
- Österreich (Deutsch)
- Portugal (English)
- Sweden (English)
- Switzerland
- United Kingdom (English)
亚太
- Australia (English)
- India (English)
- New Zealand (English)
- 中国
- 日本Japanese (日本語)
- 한국Korean (한국어)