Clear Filters
Clear Filters

Can someone help me check if I did this correctly?

3 views (last 30 days)
Here's the problem:
And the script:
syms r a0;
% Part (a) - Probability Density Function p(r) and its integral from 0 to infinity
p_r = 4 / a0^3 * r^2 * exp(-2 * r / a0);
integral_p = int(p_r, r, 0, Inf);
disp("Part (a) - Integral of p(r) from 0 to infinity:");
Part (a) - Integral of p(r) from 0 to infinity:
disp(integral_p);
% Part (b) - Limit as r approaches infinity
limit_r_infinity = limit(p_r * r^2 / exp(2 * r / a0), r, Inf);
disp("Part (b) - Limit as r approaches infinity:");
Part (b) - Limit as r approaches infinity:
disp(limit_r_infinity);
% Part (c) - Visualization of p(r) and its maximum
a0 = 5.59e-11;
p_r = @(r) 4 / a0^3 * r.^2 .* exp(-2 * r / a0);
r_vals = linspace(0, 2*a0, 1000);
p_values = p_r(r_vals);
[max_value, max_index] = max(p_values);
max_r = r_vals(max_index);
plot(r_vals, p_values);
hold on;
plot(max_r, max_value, 'ro');
text(max_r, max_value, sprintf('(%.2e, %.2e)', max_r, max_value), 'VerticalAlignment', 'bottom');
title('Probability Density Function p(r)');
xlabel('r');
ylabel('p(r)');
legend('p(r)', 'Maximum');
hold off;
% Part (d) - Calculation of P(4a0)
P_4a0 = integral(p_r, 0, 4 * a0);
disp("Part (d) - Integral P(4a0):");
Part (d) - Integral P(4a0):
disp(P_4a0);
0.9862
% Part (e) - Expected value mu
syms r a0;
% Define the integrand for the expected value mu
integrand_mu = r^3 * exp(-2 * r / a0);
% Calculate the expected value mu
mu = 4 / a0^3 * int(integrand_mu, r, 0, Inf);
disp("Part (e) - Expected value mu:");
Part (e) - Expected value mu:
disp(mu);

Accepted Answer

Torsten
Torsten on 1 Dec 2023
Replace
syms r a0
by
syms r a0 positive
to get the limits resolved.

More Answers (0)

Categories

Find more on Programming in Help Center and File Exchange

Tags

Community Treasure Hunt

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

Start Hunting!