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What are the first 3 nonzero terms in the Taylor series expansion about x = 0 for the function #f(x)=cos(4x)#?

Answer 1

Christopher Agresta

Answer. #1 - 8 x^2 + 32/3 x^4#.

Because

#cos(X) = 1 - X^2/2 + X^4/24 + o_{X->0}(X^4)#,

and because #4x ->0# when #x->0#, you get

#cos(4x) = 1 - (4x)^2/2 + (4x)^4/24 + o_{x->0}((4x)^4)#,

#cos(4x) = 1 - 8 x^2 + 32/3 x^4 + o_{x->0}(x^4)#,

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Answer 2

Aurora Alvarado

The first 3 nonzero terms in the Taylor series expansion about ( x = 0 ) for the function ( f(x) = \cos(4x) ) are:

[ f(x) = \cos(4x) ] [ = \cos(0) + \left( \frac{d}{dx} \cos(4x) \right){x=0}x + \frac{1}{2!}\left( \frac{d^2}{dx^2} \cos(4x) \right){x=0}x^2 + \cdots ]

Where ( \cos(0) = 1 ), the first derivative of ( \cos(4x) ) at ( x = 0 ) is ( -4 \sin(0) = 0 ), and the second derivative is ( -16 \cos(0) = -16 ). Therefore, the expansion becomes:

[ f(x) = 1 + 0 \cdot x - \frac{16}{2}x^2 + \cdots ]

So, the first 3 nonzero terms are ( 1 ), ( 0 ), and ( -8x^2 ).

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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.