help me --> Taylor series cos(x)
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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)
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Jan
le 14 Mai 2021
Modifié(e) : Jan
le 8 Déc 2022
Then split the question into parts and solve them one by one.
- "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) / factorial(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) / factorial(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 commentaires
Jan
le 19 Mai 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.
Plus de réponses (1)
Mahaveer Singh
le 19 Mai 2021
Modifié(e) : Mahaveer Singh
le 19 Mai 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
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