help me --> Taylor series cos(x)

83 views (last 30 days)
justlikethat
justlikethat on 12 May 2021
Commented: 승현 표 on 30 Mar 2022
cos(x) The value of can be represented by the following series.
--> cos(x) = 1 - 1-x^2/2!+x^4/4!-x^6/6! + . . . .
1. Write a mycos function that uses the above series to obtain the value of cos(x).
2. For the difference between the value of cos(2) and mycos(2) provided in Matlab to be 0.001 or less,
Write a code to determine the minimum number of terms of the critical series.
3. Configure the maximum number of iterations to be less than 10.
I'd like to know the matlab code for this problem. Please help me.
My English may be poor and my grammar may be wrong.
function cos(x) = mycos(x,n)
  9 Comments
승현 표
승현 표 on 30 Mar 2022
저 혹시... 어떤 책을 쓰셨는지 알 수 있을까요

Sign in to comment.

Accepted Answer

Jan
Jan on 14 May 2021
Please read the getting started chapters of the documentation and see Matlab's Onramp tutorial.
Then split the question into parts and solve them one by one.
  1. "Write a mycos function"
function y = mycos(x)
end
2. "above series to obtain the value of cos(x)"
function y = mycos(x)
y = 1 - x^2 / factorial(2) + x^4 / factorial(4) - x^6 / factorial(6);
end
This should be expanded in a loop:
function y = mycos(x)
y = 0;
for k = 0:10
y = y + (-1)^k * x^(2*k) / factiorial(2*k);
end
end
But why stop at k==10 oder anyother specific value?
"value of cos(2) and mycos(2) provided in Matlab to be 0.001 or less"
function [y, k] = mycos_2()
realY = cos(2);
y = 0;
k = 0;
while abs(y - realY) > 0.001
y = y + (-1)^k * x^(2*k) / factiorial(2*k);
k = k + 1;
end
end
" Configure the maximum number of iterations to be less than 10."
function [y, k] = mycos_2()
realY = cos(2);
y = 0;
k = 0;
while abs(y - realY) > 0.001 && k <= 10
% ^^^^^^^^^^
y = y + (-1)^k * x^(2*k) / factiorial(2*k);
k = k + 1;
end
end
Fine. But without reading the documentation and to understand how Matlab works, such a solution is completely useless. Do you see it? This wastes your time only.
  2 Comments
Jan
Jan on 19 May 2021
"[y, k]" is the output of the function. So the caller can know, how many iterations have been needed.
It does not matter, if you run a loop from 0 to n-1 and use k as value, or if the loop goes from 1 to n and k-1 is used. Both methods produce the same numbers.

Sign in to comment.

More Answers (1)

Mahaveer Singh
Mahaveer Singh on 19 May 2021
Edited: Mahaveer Singh on 19 May 2021
% n is required length of series.Give initial value of n as your imagination to speed up of %calculation.
function y = mycos(x,n)
y = 0;
for i= 0:2:2*n
y = y + ((-1)^(i/2)) *(x^(i)) / factiorial(i);
end
end
while y-cos(x)>0.001
n=n+1;
y=mycos(x,n);
end

Community Treasure Hunt

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

Start Hunting!