This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MATH 118, LECTURES 27 & 28: TAYLOR SERIES 1 Taylor Series Suppose we know that the power series summationdisplay n =0 a n ( x c ) n converges on some interval c R < x < c + R to the function f ( x ). That is to say, we have f ( x ) = a + a 1 ( x c ) + a 2 ( x c ) 2 + a 3 ( x c ) 3 + (1) on c R < x < c + R . Thus far, we have been interested in this problem: given a series of the form n =0 a n ( x c ) n , determine the interval and limit of convergence of the series (i.e. find f ( x )). Suppose now that we ask the question in the other direction. Given a function f ( x ), can we find a power series which converges to this function? That is to say, given a function f ( x ), can we determine the values of a n so that the power series with these coefficients converges to f ( x )? Lets revisit equation (1) and see if we can solve for the first value, a . This seems at first glance like an ominous taskthere are, after all, an infinite number of terms on the righthand side, so how can we possibly eliminate all of these terms except for a ? The answer is surprisingly simple, since we see all but the first term vanish when we substitute x = c . This gives f ( c ) = a . So the first term in the power series is a = f ( c )but how do we go about solving for the other terms? We need to isolate a 1 in the same way that a was isolated in the previous derivation. We can do this by differentiating the series to get f prime ( x ) = a 1 + 2 a 2 ( x c ) + 3 a 3 ( x c ) 2 + 4 a 4 ( x c ) 3 + . 1 Again, substituting x = c we get f prime ( c ) = a 1 . A pattern is clearly emergingto get each successive term a n we dif ferentiate and then substitute x = c . We carry this out one more time for clarity, to get f primeprime ( x ) = 2 a 2 + 2 3 a 3 ( x c ) + 3 4( x c ) 2 + 4 5( x c ) 3 + which gives f primeprime ( c ) = 2 a 2 = a 2 = f primeprime ( c ) 2 . We can see that the derivation process leads to a factorial term when solving for a n at each step. It should not be difficult to convince yourself that this process gives rise to the general form a n = f ( n ) ( c ) n ! . This is exactly what we set to out findit is exactly the answer to the question we original posed. Given a function f ( x ), the power series which converges (on some interval) to f ( x ) is given by f ( x ) = summationdisplay n =0 f ( n ) ( c ) n ! ( x c ) n . (2) A series of the form (2) is called a Taylor series . While it may not look like much, it is in fact one of the most versatile and powerful tools in applied mathematical analysis. As we will see in the following weeks, Taylor series can be differentiated, integrated, added, multiplied, and have a variety of other operations performed to iteven when these operations cannot be performed explicitly, or performed easily, to the original function f ( x )! Since a Taylor series consists only of polynomial terms, these operations can be carried out trivially. Many functions of applied interest can, in fact, only becarried out trivially....
View
Full
Document
This note was uploaded on 04/14/2009 for the course ENGINEERIN MATH 118 taught by Professor Soares during the Spring '09 term at Waterloo.
 Spring '09
 Soares

Click to edit the document details