sine approximation using a loop

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Yogesh Bhambhwani
Yogesh Bhambhwani on 30 Sep 2020
Edited: Walter Roberson on 1 Oct 2020
  2 Comments
Yogesh Bhambhwani
Yogesh Bhambhwani on 30 Sep 2020
%% Problem 3.2 Template
%{
test cases:
x = 0.2; N = 7; -> s = 0.198669330795061
x = 1.85; N = 4; -> s = 0.960597005280723
%}
function [s] = prob3_2(x,N)
% function [s] = prob3_2(x,N)
% takes a scalar angle measure x (in radians) and estimates sin(x) using
% the first N terms of an alternating series
format long
%edit below this line ---------------------
x = pi/2;
error = 1;
n = 1;
count = 0;
while error >= 1*(10^-3);
count = count+1;
terms(count) = (-1)^(count+1)*(x^n)/factorial(n);
s = sum(terms);
n=n+2;
error = abs((sin(x)-s)/sin(x))*100;
end
disp(s)
%edit above this line ----------------------
return

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Answers (1)

Walter Roberson
Walter Roberson on 1 Oct 2020
Edited: Walter Roberson on 1 Oct 2020
x = pi/2;
That is overriding the user input about the value to find the sine of.
function [s] = prob3_2(x,N)
You never use the user input N
n = 1;
you have a variable named n which is confusing considering the input named N
while error >= 1*(10^-3);
You are proceeding until your error is in a certain range, but the problem requirement says
% takes a scalar angle measure x (in radians) and estimates sin(x) using
% the first N terms of an alternating series
so you are not intended to go until error is a particular range: you are to go until you have used a particular number of terms.
error = abs((sin(x)-s)/sin(x))*100;
Is error a percentage, or is an absolute value?
What about the cases where sin(x) = 0 ? You would be dividing by 0, which is not going to give you a useful error measure.

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