help me --> Taylor series cos(x)

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)

[James Tursa]

HInt: Given a value of x, write a loop to sum up those terms. Since this is an alternating series with monotonically decreasing magnitude terms for x=2, you can quit adding terms when the next term is less than the desired tolerance. Do you know how to write a loop and limit the number of iterations to less than 10?

[Jan]

Please post what you have tried so far.

[justlikethat]

@James Tursa

I don't know how to create loops that limit the number of iterations to less than 10

I'm a very newbie and I don't know how to start.

[justlikethat]

@Jan 처음 코드를 어떻게 시작하면 좋을까요?

[David Hill]

for k=1:9
  %your code
end

[justlikethat]

@David Hill What does "k" mean?

[David Hill]

Sounds like you should go through the MATLAB onramp.

https://www.mathworks.com/learn/tutorials/matlab-onramp.html?s_tid=srchtitle

[justlikethat]

@David Hill

less than 10 --> k = 0:10 (k <= 10)

Isn't it this? not k = 1:9

[승현 표]

저 혹시... 어떤 책을 쓰셨는지 알 수 있을까요

[Ayoub]

Help me please

[Ayoub]

Cos(x)=sum (-1)^(k*2^2k)/(2k!) Sum[k n] K=0

[Jan]

@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.

Answer 1

[Jan]

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.

[justlikethat]

@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]

"[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.

Answer 2

[Mahaveer Singh]

% 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