midterm2_21c_topics

midterm2_21c_topics - Calculus: Math 21C, Fall 2010 Summary...

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Calculus : Math 21C, Fall 2010 Summary of basic results for Midterm 2. 1. Power series Convergence of power series. A power series centered at x = a is a series of the form n =0 c n ( x a ) n . (1) Every power series has a radius of convergence 0 R ≤ ∞ such that the series converges absolutely if | x a | < R and diverges if | x a | > R . The series may or may not converge at the endpoints of the interval of convergence where | x a | = R . Ratio test. The power series (1) converges absolutely for all x such that lim n →∞ ± ± ± ± c n +1 ( x a ) n +1 c n ( x a ) n ± ± ± ± < 1 if the limit exists. The radius of convergence R is given by 1 R = lim n →∞ ± ± ± ± c n +1 c n ± ± ± ± if the limit exists (with the natural conventions for R = 0 and R = ). Differentiation and integration of power series. If a function f ( x ) = n =0 c n ( x a ) n is the sum of a power series with nonzero radius of convergence R > 0, then f ( x ) has derivatives of all orders inside the interval of convergence | x a | < R . Its derivative f is given by differentiating the power series of f term-by-term, f ( x ) = n =0 nc n ( x a ) n 1 , and this power series has the same radius of convergence as the power series for f . Power series for higher-order derivatives of f are obtained by the 1
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repeated application of this result. Similarly, the integral of f is given by term-by-term integration of the power series of f f ( x ) dx = n =0 c n n + 1 ( x a ) n +1 + C, and this power series has the same radius of convergence as the power series for f . Multiplication of power series. If f ( x ), g ( x ) have power series expansions f ( x ) = n =0 a n ( x a ) n , g ( x ) = n =0 b n ( x a ) n , which both converge in | x a | < R , then h ( x ) = f ( x ) g ( x ) has the power series expansion h ( x ) = n =0 c n ( x a ) n , c n = n k =0 a k b n k , which converges in | x a | < R . Writing out the first few terms explicitly, we have [ a 0 + a 1 ( x a ) + a 2 ( x a ) 2 + a 3 ( x a ) 3 + . . . ] · [ b 0 + b 1 ( x a ) + b 2 ( x a ) 2 + b 3 ( x a ) 3 + . . . ] = a 0 b 0 + ( a 0 b 1 + a 1 b 0 ) ( x a ) + ( a 0 b 2 + a 1 b 1 + a 2 b 0 ) ( x a ) 2 + ( a 0 b 3 + a 1 b 2 + a 2 b 1 + a 3 b 0 ) ( x a ) 3 + . . . . Taylor series.
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midterm2_21c_topics - Calculus: Math 21C, Fall 2010 Summary...

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