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How to solve a function with uknown parameter with bisection method

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Nikolas
Nikolas 2022-10-29
评论： Torsten 2022-10-29
y(x) = (x^3 + p^2x^2 - 10) sin (x) for x=13.61015 y= 13257 p= unknown 5.14<p<11.47

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回答（2 个）

David Hill
David Hill 2022-10-29
Ran in:
x=13.61015;y=13257;
f=@(p)(x^3+p.^(2*x^2)-10)*sin(x)-y;
p=fzero(f,1)
p = 1.0259
P=1.01:.00001:1.03;
plot(P,f(P))%plot shows zero crossing cannot be between 5.14 and 11.47
Torsten
Torsten 2022-10-29
编辑：Torsten 2022-10-29
Ran in:
syms x y p
eqn = y - (x^3 + p^2*x^2 - 10)*sin(x) == 0;
eqn = subs(eqn,[x y],[13.61015,13257]);
p = vpasolve(eqn,p);
p = p(p>5.14 & p<11.47)
p = 
8.3215048459659113138901560210847

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