How to solve non_linear equation

1 次查看（过去 30 天）
Fatima Majeed
Fatima Majeed 2024-6-12
评论： Sam Chak 2024-6-13
I want to solve this eqution

请先登录，再回答此问题。

采纳的回答

Star Strider
Star Strider 2024-6-12
Ran in:
Solve it symbolically —
syms z
Eqn = z^3 == log(z)*(482036/0.18525)^5
Eqn = 
Z = solve(Eqn)
Z = 
Z = vpa(Z)
Z = 
format longG
Zd = double(Z)
Zd =
-76151096277.1022 + 123912303259.595i 145274860824.135 + 0i 1 + 0i -76151096277.1022 - 123912303259.595i
.
  4 个评论
显示 2更早的评论
Sam Chak
Sam Chak 2024-6-13
编辑：Sam Chak 2024-6-13
Ran in:
@Star Strider, @Torsten, Wolfram Alpha also returned the perfect "1" as one of the solutions. But we all know that . Maybe that's merely an approximation because ?
syms z
f = (z^3)/(482036/0.18525)^5;
limit(f, z, 1)
ans = 
double(ans)
ans = 8.3829e-33
Plot:
z = linspace(0.9, 1.1, 20001);
y1 = z.^3;
y2 = log(z)*(482036/0.18525)^5;
plot(z, [y1; y2]), grid on, ylim([0 2])
Sam Chak
Sam Chak 2024-6-13
Ran in:
I guess both MATLAB and Wolfram Alpha analytically computed the solution:
c = (4820360/0.018525)^7; % constant
sol = exp(-lambertw(-3/c)/3)
sol = 1
However, I mathematically believe that this is just an approximation with the real solution very close to being 1.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Numeric Solvers 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by