If we know that as x -> to 0 , 1-cos(x)

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The discussion focuses on the Taylor series expansion of the function 1 - cos(x) as x approaches 0, establishing that it is equivalent to x²/2. It further explores the Taylor series representation of 1 - cos³(x), emphasizing the importance of derivatives at 0 in determining the series' terms. The first non-zero term for 1 - cos³(x) can be derived similarly, using the function's derivatives. This method allows for accurate approximations of functions near x = 0.

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if we know that as x -> to 0 , 1-cos(x) is equivalent to x2/2 ,
how can we find what 1-cox3(x) is equivalent to ?
 
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A function such as 1-cos(x) or 1-cos3(x) can be represented as a power series: an infinite sum of the form

f(0) + \frac{f'(0)x^1}{1} + \frac{f''(0)x^2}{2 * 1} + \frac{f'''(0)x^3}{3*2*1} + ...

where f'(0) is the first derivative of the function at 0, and f''(0) the second derivative at 0, and so on. I used the * to mean "times". This sum is called the Taylor series representation of the function "expanded at 0". (A Taylor series expanded at 0 is also called a MacLaurin series.)

You can approximate the function with some finite number of terms in this series. For any given number of terms, and any given degree of accuracy (any error you're willing to accept), you can always make your approximation accurate to within that amount of error by chosing a small enough value of x (a value of x close enough to 0).

For your first example, the first non-zero term in the sum is

\frac{x^2}{2}

Try it out on your other example. What's the first non-zero term this time?
 


thanks for the help Rasalhague!
 

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