Symbolic Toolbox Solving for Zeros
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Hey! I've tried a few things but cant seem to get this script to solve for the x values of a functions values. Any ideas?
%% function
function [zeros] = dPlotInfo(func,range)
%
%This function plots a function and its 1st and 2nd derivatives
%It also returns any zeros
%
figure(1);
fplot(func,range,1);
hold on;
fplot(diff(func,1),range);
fplot(diff(func,2),range);
syms x;
zeros = fzero(func == 0 ,range);
%AHHHHH
end
%% test script used for the function
range = [-5 5];
syms x;
f = sin(3*x);
z = dPlotInfo(f, range);
disp(z);
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Answers (2)
Paul
on 15 Dec 2022
2 Comments
Paul
on 15 Dec 2022
Here is one way to use solve, taken nearly verbatim from its doc page
syms x real
f(x) = sin(3*x);
Solve the equation, parameterically if needed. If the solution(s) isn't parametertized, then it can be be checked manually to determine if it's in the desired range, which isn't the case here.
[solx,parameters,conditions] = solve(f(x),'ReturnConditions',true)
Now solve for the parameter that forces the solution to lie within the desired range
assume(conditions)
restrictions = [solx > -5 , solx < 5];
solp = solve(restrictions,parameters)
And sub those parameters back into the solution:
valx = subs(solx,parameters,solp)
Star Strider
on 15 Dec 2022
One approach —
range = [-5 5];
syms x;
f(x) = sin(3*x);
z = dPlotInfo(f, range);
disp(z);
function [zeros] = dPlotInfo(func,range)
%
%This function plots a function and its 1st and 2nd derivatives
%It also returns any zeros
%
digits(5) % Set Precision On Returned Symbolic Numbers (Convenience)
figure(1);
hfp = fplot(func,range,1);
hold on;
fplot(diff(func,1),range);
fplot(diff(func,2),range);
syms x;
zix = find(diff(sign(hfp.YData))); % Approximate Indices Of Zero-Crossings
for k = 1:numel(zix)
zeros(k) = vpasolve(func, x, hfp.XData(zix(k))); % Use The X-Values For Each Approximate Index Value As A Starting Value
end
%AHHHHH
end
.
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