vpasolve gives me only 1 x-value of intersection

I would like to get all 4 points of intersection on the domain -2 < x < 2, but my code is only giving me only 1 value

( exp(x^2) is e^(x^2) right? )

syms x
vpasolve(exp(x^2) == 1+tan(x))

ans = 

0

Answer 1

[Star Strider]

An analytic approach is an option, however it will be necessary first to estimate the approximate zero-crossings in any event.

A numeric approach —

syms x

xv = [-2 2];

figure
fplot(exp(x^2), xv)
hold on
fplot(1+tan(x), xv)
hold off
grid
ylim([-1 16])

clearvars

xv = linspace(-2, 2, 1E+4);
fx = exp(xv.^2) - (1+tan(xv));
L = numel(xv);

ixz = find(diff(sign(fx)))                                          % Indices (Into 'xv') Of Approximate Zero-Crossings

ixz = 1×6

893 1073 5000 7270 8542 8927

for k = 1:numel(ixz)
    idxrng = max(ixz(k)-1,1) : min(ixz(k)+1,L);
    xz(k) = interp1(fx(idxrng),xv(idxrng),0);                       % Use: 'interp1'
    yz(k) = exp(xz(k)^2);
end

xz

xz = 1×6

-1.6428 -1.5715 0.0000 0.9083 1.4169 1.5707

yz

yz = 1×6

14.8623 11.8185 1.0000 2.2818 7.4444 11.7888

figure
plot(xv, exp(xv.^2))
hold on
plot(xv, 1+tan(xv))
plot(xz, yz, 'sr')
hold off
ylim([-1 16])

f = @(x) exp(x.^2) - (1+tan(x));

for k = 1:numel(ixz)
    xf(k) = fzero(f,xv(ixz(k)));                                    % Use: 'fzero'
    yf(k) = exp(xf(k)^2);
end

xf

xf = 1×6

-1.6428 -1.5708 -0.0000 0.9083 1.4169 1.5708

yf

yf = 1×6

14.8623 11.7918 1.0000 2.2818 7.4445 11.7918

figure
plot(xv, exp(xv.^2))
hold on
plot(xv, 1+tan(xv))
plot(xf, yf, 'sr')
hold off
ylim([-1 16])

This approach also detects the intersections (zero-crossings) at so those would have to be eliminated afterwards. I did not try this with vpasolve, so I do not know if it would also detect them. This also requires a relatively high resolution in ‘xv’ in order to detect all the zero-crossings.

.

[Jordi van Selm]

Thank you for your very complete answer. How do I "skip" the intersections for the zero-crossings?

[Star Strider]

My pleasure!

If you are referring to the singularities at it would likely be necessary to use ismembertol (or other appropriate function) to detect them and then use that result to eliminate them —

syms x

xv = [-2 2];

figure
fplot(exp(x^2), xv)
hold on
fplot(1+tan(x), xv)
hold off
grid
ylim([-1 16])

clearvars

xv = linspace(-2, 2, 1E+4);
fx = exp(xv.^2) - (1+tan(xv));
L = numel(xv);

ixz = find(diff(sign(fx)))                                          % Indices (Into 'xv') Of Approximate Zero-Crossings

ixz = 1×6

893 1073 5000 7270 8542 8927

for k = 1:numel(ixz)
    idxrng = max(ixz(k)-1,1) : min(ixz(k)+1,L);
    xz(k) = interp1(fx(idxrng),xv(idxrng),0);                       % Use: 'interp1'
    yz(k) = exp(xz(k)^2);
end

xz

xz = 1×6

-1.6428 -1.5715 0.0000 0.9083 1.4169 1.5707

yz

yz = 1×6

14.8623 11.8185 1.0000 2.2818 7.4444 11.7888

Lv1 = ismembertol(abs(xz), pi/2, 0.01)

Lv1 = 1×6 logical array
   0   1   0   0   0   1

xz = xz(~Lv1)

xz = 1×4

-1.6428 0.0000 0.9083 1.4169

yz = yz(~Lv1)

yz = 1×4

14.8623 1.0000 2.2818 7.4444

figure
plot(xv, exp(xv.^2))
hold on
plot(xv, 1+tan(xv))
plot(xz, yz, 'sr')
hold off
ylim([-1 16])

f = @(x) exp(x.^2) - (1+tan(x));

for k = 1:numel(ixz)
    xf(k) = fzero(f,xv(ixz(k)));                                    % Use: 'fzero'
    yf(k) = exp(xf(k)^2);
end

xf

xf = 1×6

-1.6428 -1.5708 -0.0000 0.9083 1.4169 1.5708

yf

yf = 1×6

14.8623 11.7918 1.0000 2.2818 7.4445 11.7918

Lv2 = ismembertol(abs(xf), pi/2, 0.01)

Lv2 = 1×6 logical array
   0   1   0   0   0   1

xf = xf(~Lv2)

xf = 1×4

-1.6428 -0.0000 0.9083 1.4169

yf = yf(~Lv2)

yf = 1×4

14.8623 1.0000 2.2818 7.4445

figure
plot(xv, exp(xv.^2))
hold on
plot(xv, 1+tan(xv))
plot(xf, yf, 'sr')
hold off
ylim([-1 16])

Desktop computer crashed an hour ago, and also required an app update that took a few minutes to download and install.

.

[Jordi van Selm]

How do I change this code to find an intersection around 1.5*pi ?

[Star Strider]

The functions have to exist in that region, so ‘xv’ must be enlarged to include them. If it does the ‘ixz’ call will locate them and the rest of the code will estimate them.

There do not appear to be any intersections in that region, so it will not find any. The problem appears to be that (1+tan(x)) is not defined there.

syms x

xv = [-1 1]*2*pi

xv = 1×2

-6.2832 6.2832

Check(x) = [exp(x^2); (1+tan(x))]

Check(x) = 

vpaCheck = vpa(Check(1.5*pi*[-1 1]))

vpaCheck = 

figure
fplot(exp(x^2), xv)
hold on
fplot(1+tan(x), xv)
hold off
xline([-1 1]*1.5*pi, '--g')
grid
ylim([-1 2000])

clearvars

xv = [-1 1]*2*pi

xv = 1×2

-6.2832 6.2832

xv = linspace(min(xv), max(xv), 1E+4);
fx = exp(xv.^2) - (1+tan(xv));
L = numel(xv);

ixz = find(diff(sign(fx)))                                          % Indices (Into 'xv') Of Approximate Zero-Crossings

ixz = 1×6

3693 3750 5000 5723 6127 6250

for k = 1:numel(ixz)
    idxrng = max(ixz(k)-1,1) : min(ixz(k)+1,L);
    xz(k) = interp1(fx(idxrng),xv(idxrng),0);                       % Use: 'interp1'
    yz(k) = exp(xz(k)^2);
end

xz

xz = 1×6

-1.6428 -1.5724 0.0000 0.9083 1.4169 1.5698

yz

yz = 1×6

14.8624 11.8505 1.0000 2.2818 7.4444 11.7564

Lv1 = ismembertol(abs(xz), pi/2, 0.01)

Lv1 = 1×6 logical array
   0   1   0   0   0   1

xz = xz(~Lv1)

xz = 1×4

-1.6428 0.0000 0.9083 1.4169

yz = yz(~Lv1)

yz = 1×4

14.8624 1.0000 2.2818 7.4444

figure
plot(xv, exp(xv.^2))
hold on
plot(xv, 1+tan(xv))
plot(xz, yz, 'sr')
hold off
xline([-1 1]*1.5*pi, '--g')
ylim([-1 2000])

f = @(x) exp(x.^2) - (1+tan(x));

for k = 1:numel(ixz)
    xf(k) = fzero(f,xv(ixz(k)));                                    % Use: 'fzero'
    yf(k) = exp(xf(k)^2);
end

xf

xf = 1×6

-1.6428 -1.5708 -0.0000 0.9083 1.4169 1.5708

yf

yf = 1×6

14.8623 11.7918 1.0000 2.2818 7.4445 11.7918

Lv2 = ismembertol(abs(xf), pi/2, 0.01)

Lv2 = 1×6 logical array
   0   1   0   0   0   1

xf = xf(~Lv2)

xf = 1×4

-1.6428 -0.0000 0.9083 1.4169

yf = yf(~Lv2)

yf = 1×4

14.8623 1.0000 2.2818 7.4445

figure
plot(xv, exp(xv.^2))
hold on
plot(xv, 1+tan(xv))
plot(xf, yf, 'sr')
hold off
xline([-1 1]*1.5*pi, '--g')
ylim([-1 2000])

.

[Jordi van Selm]

Thanks for your answer, but it seems that there's an intersection in the "neighbourhood" of 1.5*pi. I think that using vpasolve will do the trick, but I'm unsure how.

[Star Strider]

I doubt that vpasolve or any other function will solve for an undefined value.

With respect to ‘in the "neighbourhood" of 1.5*pi’ I defer to you to define the neighbourhood. In that situation, choosing the value to solve for (e.g. pi*(1.5-eps) or something similar) solves the problem immediately because it defines the result. Then, simply evaluate the functions at that value. I’m not certain how else to define that problem.

[Jordi van Selm]

Thanks again, I was kinda vage about my question. I would like to get the y-value of the intersection near 1.5*pi (which is also the asymptote of tan(x)+1) with the function e^x^2 and not with tan(x)+1 since they don't intersect on the asymptote. So lets say I chose pi*1.5-0.1 (left side of asymptote), what is my y-value when intersecting with e^x^2 ?

[Star Strider]

syms x

xv = [-1 1]*(pi*1.5-0.1);
f1(x) = exp(x^2);
f2(x) = 1+tan(x);

Evaluated = [f1(xv); f2(xv); f1(xv)-f2(xv)]

Evaluated = 

vpaEvaluated = vpa(Evaluated)

vpaEvaluated = 

figure
fplot(exp(x^2), xv)
hold on
fplot(1+tan(x), xv)
hold off
grid
ylim([-1 16])

clearvars

xv = linspace(-(pi*1.5-0.1), (pi*1.5-0.1), 1E+4);
fx = exp(xv.^2) - (1+tan(xv));
L = numel(xv);

ixz = find(diff(sign(fx)))                                          % Indices (Into 'xv') Of Approximate Zero-Crossings

ixz = 1×6

3219 3297 5000 5985 6536 6703

for k = 1:numel(ixz)
    idxrng = max(ixz(k)-1,1) : min(ixz(k)+1,L);
    xz(k) = interp1(fx(idxrng),xv(idxrng),0);                       % Use: 'interp1'
    yz(k) = exp(xz(k)^2);
end

xz

xz = 1×6

-1.6428 -1.5724 0.0000 0.9083 1.4169 1.5705

yz

yz = 1×6

14.8623 11.8514 1.0000 2.2818 7.4444 11.7825

% Lv1 = ismembertol(abs(xz), pi/2, 0.01)
% xz = xz(~Lv1)
% yz = yz(~Lv1)



figure
plot(xv, exp(xv.^2))
hold on
plot(xv, 1+tan(xv))
plot(xz, yz, 'sr')
hold off
ylim([-1 16])

f = @(x) exp(x.^2) - (1+tan(x));

for k = 1:numel(ixz)
    xf(k) = fzero(f,xv(ixz(k)));                                    % Use: 'fzero'
    yf(k) = exp(xf(k)^2);
end

xf

xf = 1×6

-1.6428 -1.5708 0.0000 0.9083 1.4169 1.5708

yf

yf = 1×6

14.8623 11.7918 1.0000 2.2818 7.4445 11.7918

Lv2 = ismembertol(abs(xf), pi/2, 0.01)

Lv2 = 1×6 logical array
   0   1   0   0   0   1

xf = xf(~Lv2)

xf = 1×4

-1.6428 0.0000 0.9083 1.4169

yf = yf(~Lv2)

yf = 1×4

14.8623 1.0000 2.2818 7.4445

figure
plot(xv, exp(xv.^2))
hold on
plot(xv, 1+tan(xv))
plot(xf, yf, 'sr')
hold off
ylim([-1 16])

Looking at ‘(pi*1.5-0.1)’ and ‘vpaEvaluated’, the functions are not equal (and actually differ by more than ) at those values, so we can say with confidence that there is no intersection at that point. It’s not even close.

.

Answer 2

[Walter Roberson]

vpasolve() only ever returns multiple results in the case of where the expressions are all polynomials, and there is the same number of equations as there are variables.

In all cases in which there are more equations than variables, or fewer equations than variables, vpasolve() will error.

In all cases involving a non-linear function of any variable, vpasolve() will try to find a single numeric solution, even if there might be multiple solutions. This includes cases that could be transformed into polynomials by a change of variable: vpasolve() will not even try to find multiple solutions if there is any nonlinear function of one of the variables.

You will need to provide multiple initial guesses to vpasolve()

[Jordi van Selm]

I found this, but I'm having trouble writing code to get all the values:

syms x y 
[sol2x,sol2y] = vpasolve(y-tan(x)-1 == 0,y-exp(x^2) == 0, [-2,2])

sol2x = 

sol2y = 

14.862250796440579193799274998889

[Walter Roberson]

As far as vpasolve() is concerned, you gave it the same initial conditions as if you had specified

[-2, 2;
 -inf, inf]

So you restricted x to be between -2 and +2, and you did not restrict y.

If you wanted to give starting values, then give a column vector such as

[1; 10]

You cannot give starting values and bounds ranges -- only starting values or bounds ranges.