Maclaurin series representation of cos(2x^3)

[hjaleta]

Hello!

I came across an problem in Calculus. Exercise 2 in chapter 9.6. It goes:

Find the Maclaurin representation of cos(2x³)

In the method used in the book, they used the series expansion for cos(x)= 1 - x²/2! + x⁴/4! + x⁶/6! ...
and then substituted x with 2x³, resulting in the series 1 - (2x³)²/2! + (2x³)⁴/4! + (2x³)⁶/6! ...

However, the second derivative of cos(2x³) is equal to zero, if x=0, which is the definition of a maclaurin polynomial.

f'=-6x²sin(2x³), f'' = -12xcos(2x³) - 36x⁴sin(2x³)

How can the term (2x³)²/2! then exist in the expansion?

 

[tkhunny]

Hello!

I came across an problem in Calculus. Exercise 2 in chapter 9.6. It goes:

Find the Maclaurin representation of cos(2x³)

In the method used in the book, they used the series expansion for cos(x)= 1 - x²/2! + x⁴/4! + x⁶/6! ...
and then substituted x with 2x³, resulting in the series 1 - (2x³)²/2! + (2x³)⁴/4! + (2x³)⁶/6! ...

However, the second derivative of cos(2x³) is equal to zero, if x=0, which is the definition of a maclaurin polynomial.

f'=-6x²sin(2x³), f'' = -12xcos(2x³) - 36x⁴sin(2x³)

How can the term (2x³)²/2! then exist in the expansion?

Thought Question: Is there a term-by-term equivalence between the result of the more convenient method given by your book and what you would expect if you calculate the individual derivatives?

 

[Steven G]

Hello!

I came across an problem in Calculus. Exercise 2 in chapter 9.6. It goes:

Find the Maclaurin representation of cos(2x³)

In the method used in the book, they used the series expansion for cos(x)= 1 - x²/2! + x⁴/4! + x⁶/6! ...
and then substituted x with 2x³, resulting in the series 1 - (2x³)²/2! + (2x³)⁴/4! + (2x³)⁶/6! ...

However, the second derivative of cos(2x³) is equal to zero, if x=0, which is the definition of a maclaurin polynomial.

f'=-6x²sin(2x³), f'' = -12xcos(2x³) - 36x⁴sin(2x³) This is not true

How can the term (2x³)²/2! then exist in the expansion?

f'' \(\displaystyle \neq\) -12xcos(2x³) - 36x⁴sin(2x³)