help me --> Taylor series cos(x)

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justlikethat
justlikethat 2021-5-12
编辑： Jan 2022-12-8
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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Ayoub
Ayoub 2022-12-7
Cos(x)=sum (-1)^(k*2^2k)/(2k!) Sum[k n] K=0
Jan
Jan 2022-12-8
@Ayoub: Please open a new thread for a new question. It is not clear, what you are asking for or how we can help you.

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Jan
Jan 2021-5-14
编辑：Jan 2022-12-8
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) / 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 个评论
justlikethat
justlikethat 2021-5-14
编辑：justlikethat 2021-5-15
@Jan
I'm trying to run this code. I changed mycos_2() to mycos(x)
Is this right?
And I don't understand this part --> [y, k]
why did you write it like this?
lastly, I made ' ( y = y + (-1)^k * x^(2*k) / factiorial(2*k);) ' like this.
--> y = (-1)^(k-1)*x^((k-1)*2/factorial((k-1)*2);
What's the difference?
Jan
Jan 2021-5-19
"[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.

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Mahaveer Singh
Mahaveer Singh 2021-5-19
编辑：Mahaveer Singh 2021-5-19
% 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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