How do you find the Maclaurin series of #f(x)=cos(x)# ?

Answer 1
The Maclaurin series of #f(x)=cosx# is #f(x)=sum_{n=0}^infty (-1)^nx^{2n}/{(2n)!}#.

Let's examine a few specifics.

The Maclaurin series for #f(x)# in general can be found by #f(x)=sum_{n=0}^infty {f^{(n)}(0)}/{n!}x^n#
Let us find the Maclaurin series for #f(x)=cosx#.
By taking the derivatives, #f(x)=cosx Rightarrow f(0)=cos(0)=1# #f'(x)=-sinx Rightarrow f'(0)=-sin(0)=0# #f''(x)=-cosx Rightarrow f''(0)=-cos(0)=-1# #f'''(x)=sinx Rightarrow f'''(0)=sin(0)=0# #f^{(4)}(x)=cosx Rightarrow f^{(4)}(0)=cos(0)=1#
Since #f(x)=f^{(4)}(x)#, the cycle of #{1,0,-1,0}# repeats itself.
So, we have the series #f(x)=1-{x^2}/{2!}+x^4/{4!}-cdots=sum_{n=0}^infty(-1)^n x^{2n}/{(2n)!}#
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Answer 2

To find the Maclaurin series of ( f(x) = \cos(x) ), you use the Taylor series expansion formula centered at ( x = 0 ). The Maclaurin series for ( \cos(x) ) is:

[ \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} ]

This series can be derived by substituting ( \cos(x) ) into the Taylor series formula and computing its derivatives at ( x = 0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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