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### AnalyticContinuation

Course: ECE 6382, Fall 2008
School: U. Houston
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Word Count: 607

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6382 Fall ECE 2008 Analytic Continuation D. R. Wilton ECE Dept. Analytic Continuation of Functions It seems clear how to extend linear combinations of the elementary functions 1, x 1 , x 2 , x 3 , x 4 ,x 5 , ,x n , x 1 2 , Pn ( x ) (rational function) Qm ( z ) e x , ln x, sin x, cos x,sinh x, cosh x, tan x, cot x, tanh x, coth x in such a way that the new function is analytic; i.e., x z. We define as...

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6382 Fall ECE 2008 Analytic Continuation D. R. Wilton ECE Dept. Analytic Continuation of Functions It seems clear how to extend linear combinations of the elementary functions 1, x 1 , x 2 , x 3 , x 4 ,x 5 , ,x n , x 1 2 , Pn ( x ) (rational function) Qm ( z ) e x , ln x, sin x, cos x,sinh x, cosh x, tan x, cot x, tanh x, coth x in such a way that the new function is analytic; i.e., x z. We define as analytic continuation this process of continuing the function off the real axis and into the complex plane such that the resulting function is analytic. More generally, we'll use the term for extending the "representation" of a function in one region of the complex plane to another where the original "representation" may not have been valid. A "representation" of a function means an explicit rule for computing it, such as a formula involving elementary functions, a series or product form, etc. Many representations may be possible, and they may be valid only in certain regions of the complex plane. E.g., consider the Bessel function, J n ( x ) ? I.e., how do we a) define J n ( z ) so that it is computable b) in some region and c) how do we know it is analytic? One approach is to start with, say, a Taylor series valid in some region of the complex plane within its radius of convergence. Analytically Continuing the Geometric Series two alternative representations 1 f (z) = = 1- z n =0 zn , z <1 y -1 2 Expand about z = - 1 . Since both series are 2 valid there, find coefficients of new series by differentiating the original series : m =0 1 x bm ( z + 1 ) 2 m , z+1 < 2 3 2 where used the series representation 1 dm bm = m! dz m n z n =0 z =- 1 2 1 = n ( n - 1)( n - 2 ) m! n=m n - m + 1) z n-m ( z =- 1 2 1 n! 1 n-m = (- ) m! n=m n ( - m ) ! 2 Coefficients of the new series --- with extended region of convergence --- are determined from the coefficients of the original series, though that series did not even converge in the extended region. I.e., the information to extend the convergence region is contained in the coefficients of the original series! Analytically Continuing the Geometric Series, cont'd f ( z) = where 1 dm n bm = z m! dz m n=0 1 = n ( n - 1)( n - 2 ) m! n=m m=0 bm ( z + 1 ) 2 m , z+1 < 2 3 2 ( n - m + 1) z 1 + [1 - z ]m=-11 z 2 n-m z =- 1 2 dm n (Note m z = 0 ,n = 0 ,1, dz z =- 1 2 ,m - 1 ) n-m 1 1 n! = ( - 1 ) = m! (1)( 2 )( 3) m! n=m ( n - m ) ! 2 m +1 ( m) 2 = 3 m +1 m 2 3 f ( z) = ( z + 1 ) , z+1 < 2 2 2 m=0 3 As long as we take only a finite number --- say, N + 1 --- of terms in the new series, we could always expand out each polynomial term and collect coefficients to obtain a finite power series approximation : f ( z) 2 3 1 + 2 z + 1 + 2 2 z + 1 2 + 2) (3) ( 2) 3( 2 3 ( 2) (z + 1) 3 2 N N = 2 1 + 1 + 3 (3) N + ( 1 ) +[ 3 ] z +[ ] z2 +[ ] z N Note, e.g., the constant term tends to 1 = 1 as N , the same as in the orginal series. 1- 1 3 The same result holds for all the coefficients of powers of z . Such rearranging of the order of summation of terms in an infinite series is generally invalid, as it is here; clearly, however, the new series is simply a rearrangement of the order of summation of terms of the original series! Analytic Continuation 1 Continuing f ( z ) = from 1- z its power series representation in the region z < 1 into the entire complex plane y -1 2 1 x If the singularities are isolated, we can continue any function into the entire complex plane via a sequence of continuations using Taylor and / or Laurent series !
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