Can someone please check my code?

2 vues (au cours des 30 derniers jours)
Eduardo Gallegos
Eduardo Gallegos le 16 Déc 2022
Réponse apportée : Nikhil le 19 Déc 2022
Here are the instructions : The purpose of this assignment is to demonstrate these MATLAB skills:
a. Code a function
b. Code a for loop that sums the terms of a series.
c. Use a conditional statement to break the loop
d. Display the sum after the for loop ends.
Part 1A: Code a Matlab function file called fcn1.m
This loop will sum the terms of a sequence.
The function’s inputs are:
b = a constant used in the calculations.
loop_limit = the value to break for loop.
The function’s output is:
total = the sum of the terms.
Part 1B: Inside this function:
Initialize total = 0;
Initialize term = zeros(1, 100);
Code a for loop with k = 1 to 100 that does the following:
a. Computes term(k) = pi/(12+ k^b);
b Adds term(k) to total. (This will be the sum of all the terms.)
c. Breaks the loop when the absolute value of term(k) is less than loop_limit
Part 2A : Create a Matlab test file that sets these parameters and calls your function:
b = 3;
loop_limit = 0.001
Part 2B:
In the test file, after the call to your function, use the disp() and num2str() functions to display the sum of the series.
Here is the function
function [total] = fcn1(x)
b = 3;
loop_limit = 0.001
total = 0;
term = zeros(1, 100);
for k = 1:100
term(k) = pi/(12+ k^b);
total = total + term(k);
if (term(k) < loop_limit)
break
end
end
end
The calling function
clc; close all; clear all;
y = fcn1(1)
disp(num2str(y))
Thank you so much.
  1 commentaire
Eduardo Gallegos
Eduardo Gallegos le 16 Déc 2022
I made an adjustment to the code.
function
function [total] = fcn1(b,loop_limit)
total = 0;
term = zeros(1, 100);
for k = 1:100
term(k) = pi/(12+ k^b);
total = total + term(k);
if (term(k) < loop_limit)
break
end
end
end
Calling Function
clc; close all;
y = fcn1(3,0.001);
disp(num2str(y))

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Réponses (1)

Nikhil
Nikhil le 19 Déc 2022
Hey Eduardo, you have covered all the required aspects in your code. It looks fine.

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