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Unformatted text preview: The Big-O Order Symbol: Another Way to Think of Error To motivate the concept we're about to introduce, consider the following sit- uation: suppose you've shown that the 4 th-order Maclaurin polynomial for a certain function is, say, 1- x 3 + 1 10 x 4 . Suppose also that you go ahead and calculate that the 5 th derivative of this function is zero at zero, so that this same expression is also the 5 th-order Maclaurin polynomial. How would you incorporate this extra piece of information in your work? Well, in the notation we've used in this course we have a fairly e cient way to do this: we could state that f ( x ) ≈ 1- x 3 + 1 10 x 4 + R 5 , ( x ) . Even if we don't feel the need to proceed with nding an upper bound on the magnitude of the error term, the fact that we've written R 5 , shows clearly that we know the x 5 term to be zero. What we are about to introduce is essentially an alternative to that notation, which will be easier to work with and is universally recognized. De nition: Given two functions f and g, we say that f is of order g as x → x and write f ( x ) = O ( g ( x )) as x → x (1) if there exists a constant A (greater than zero) such that | f ( x ) | ≤ A | g ( x ) | on some interval around x (although the point x itself may be excluded from the interval, since the idea is to describe the behaviour of f in the limit as we approach x ). For our purposes the function g will be a power of ( x- x ) , and of course if we're dealing with Maclaurin series then it will simply be a power of x . Note: you might be familiar with a similar concept used in computer science, which deals with behaviour as n → ∞ instead of x → x . Examples: a) Since we know that | x 3 | ≤ | x 2 | for all x [- 1 , 1] , we can state that x 3 = O ( x 2 ) as x → . Since it is also true that | x 3 | ≤ | x | on the same interval, we can also state that x 3 = O ( x ) as x → . Both statements are equally correct (just look at our de nition; in these examples the constant A is just 1. In fact, we can similarly state that x 3 = O ( x 3 ) as x → , and x 3 = O (1) as x → (this last statement just says that x 3 is bounded near zero; | x 3 | ≤ A for some number A when x is small enough)....
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- Spring '08
- Taylor Series, Sin, x3, Hyperbolic function, Euler's formula