One function is greater than other

2 views (last 30 days)
Fatima Majeed
Fatima Majeed on 10 Jun 2024
Answered: Sam Chak on 11 Jun 2024
I would like to determine the range of values for \( z \) where the following inequality holds true:
this is my trying
syms z real
assume(z > exp(1))
% Define the function
f = z - 8.02 * log(z) - (3.359 / 21.233) * log(z) * z;
sol = solve(f > 0, z, 'ReturnConditions', true);
Warning: Unable to find explicit solution. For options, see help.
vpa(sol.conditions)
ans = Empty sym: 0-by-1

Accepted Answer

Matt J
Matt J on 10 Jun 2024
Edited: Matt J on 10 Jun 2024
Let us reorganize the inequality as,
Since log(z)>0 in the given domain of z, this implies which in turn implies that log(z)<1/0.1582, or in turn that z<556.19. Therefore, if the inequality is ever going to be satisfied, it must be satisfied somewhere in the interval [e,556.2]. We can see from plotting that this never occurs,
syms z real
fplot(z, [exp(1),556]); hold on
fplot(log(z).*(0.1582*z+8.02), [exp(1),556.2]); hold on; legend(Location='NW')
but we can also check it with fminbnd,
[zmin,fmin]=fminbnd(@(z) log(z).*(0.1582*z+8.02)-z ,exp(1),556.2)
zmin = 143.8330
fmin = 9.0742

More Answers (2)

Walter Roberson
Walter Roberson on 10 Jun 2024
Q = @(v) sym(v);
syms z real
assume(z > exp(1))
% Define the function
f = z - Q(802)/Q(10)^2 * log(z) == (Q(3359)/Q(10)^3 / (Q(21233)/Q(10)^3)) * log(z) * z;
F = lhs(f) - rhs(f)
F = 
limit(F, z, inf)
ans = 
fplot(F, [exp(1), 1000])
fplot(lhs(f), [exp(1), 200])
hold on
fplot(rhs(f), [exp(1), 200])
legend({'lhs', 'rhs'})
There is no solution. Everywhere in the range, the left hand side is less than the right hand side.
  3 Comments
Walter Roberson
Walter Roberson on 10 Jun 2024
Edited: Walter Roberson on 10 Jun 2024
You have stated
assume(z > exp(1))
so we are justified in starting from exp(1) in the fplot()
Walter Roberson
Walter Roberson on 10 Jun 2024
I do not prove that there are no crossings... but I give evidence.
Q = @(v) sym(v);
syms z real
assume(z > exp(1))
% Define the function
f = z - Q(802)/Q(10)^2 * log(z) == (Q(3359)/Q(10)^3 / (Q(21233)/Q(10)^3)) * log(z) * z;
F = lhs(f) - rhs(f)
F = 
limit(F, z, inf)
ans = 
The limit of the difference is -infinity, so there must be an even number of zero crossings (no zero crossings counts as an even number of them.)
fplot(lhs(f), [exp(1), 1000000])
hold on
fplot(rhs(f), [exp(1), 1000000])
legend({'lhs', 'rhs'})
The two are getting further apart up to at least 1 million.
Additional logic would be needed to establish that there are definitely no crossings.

Sign in to comment.


Sam Chak
Sam Chak on 11 Jun 2024
The search below covers this range .
z = linspace(0, exp(1), 3001);
fun = @(z) z - 8.02*log(z) - (3.359/21.233)*log(z).*z;
plot(z, fun(z)), grid on
sol = fzero(fun, 1)
sol = 1.1506
f1 = z - 8.02*log(z); % LHS of inequality
f2 = (3.359/21.233)*log(z).*z; % RHS of inequality
plot(z, [f1; f2]), grid on, ylim([-0.2 2]), xlabel('z')
txt = sprintf('%.4f', sol);
xline(sol, '--', txt)
legend({'$z - 8.02 \log(z)$', '$\frac{3.359}{21.233} \log(z) z$', ''}, 'interpreter', 'latex', 'fontsize', 14);
Thus, the inequality holds at .

Categories

Find more on Read, Write, and Modify Image in Help Center and File Exchange

Tags

Products

Community Treasure Hunt

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

Start Hunting!