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

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In summary, a function like 1-cos(x) or 1-cos3(x) can be represented as an infinite sum called the Taylor series, which can be approximated with a finite number of terms to a desired degree of accuracy. For 1-cos(x), the first non-zero term is x^2/2, and for 1-cos3(x), it would be x^4/24.
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phymatter
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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

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

where [itex]f'(0)[/itex] is the first derivative of the function at 0, and [itex]f''(0)[/itex] 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

[tex]\frac{x^2}{2}[/tex]

Try it out on your other example. What's the first non-zero term this time?
 
  • #3


thanks for the help Rasalhague!
 

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

What is the limit of 1-cos(x) as x approaches 0?

The limit of 1-cos(x) as x approaches 0 is 0.

Why is the limit of 1-cos(x) as x approaches 0 important?

This limit is important because it helps us understand the behavior of the function 1-cos(x) near the point x=0.

Can the limit of 1-cos(x) as x approaches 0 be evaluated using direct substitution?

No, direct substitution does not work for this limit because it would result in an indeterminate form of 1-1, which is 0.

How can we evaluate the limit of 1-cos(x) as x approaches 0?

We can use the limit definition of a derivative, which states that the derivative of a function at a point is equal to the limit of the difference quotient as the change in x approaches 0. In this case, the derivative of 1-cos(x) at x=0 would be equal to the limit of (1-cos(x))/x as x approaches 0.

What is the graph of 1-cos(x) near x=0?

The graph of 1-cos(x) near x=0 is a straight line with a slope of 0 passing through the point (0,0). As x approaches 0, the graph becomes increasingly flat, indicating that the limit of 1-cos(x) as x approaches 0 is indeed 0.

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