Lesson 5 A PPROXIMATING T AYLOR SERIES
• In this lecture we investigate a connection between Taylor series and Fourier series: • Taylor series are essentially Fourier series with no negative coefficients • We use this connection to calculate functions inside the unit disk
• Recall the Taylor series representation of the function f : ∞ � c k z k f ( z ) = k =0 • Consider z inside the open unit disk –1 0 1 D = { z ∈ C : | z | < 1 } = • Under what conditions does the Taylor series converge?
������� : If | c k | < Ck d for some d , then f converges in the unit disk D . ����� : Since , the Taylor series converges absolutely, hence converges: ��������� : If for some , then converges in the unit disk for all . ����� : Follows from induction: which satisfies .
������� : If | c k | < Ck d for some d , then f converges in the unit disk D . ����� : Since | z | = r < 1 , the Taylor series converges absolutely, hence converges: � � k d r k < ∞ � � � c k z k � � ≤ C � � k =0 k =0 ��������� : If for some , then converges in the unit disk for all . ����� : Follows from induction: which satisfies .
������� : If | c k | < Ck d for some d , then f converges in the unit disk D . ����� : Since | z | = r < 1 , the Taylor series converges absolutely, hence converges: � � k d r k < ∞ � � � c k z k � � ≤ C � � k =0 k =0 ��������� : If | c k | < Ck d for some d , then f ( λ ) converges in the unit disk D for all λ . ����� : Follows from induction: which satisfies .
������� : If | c k | < Ck d for some d , then f converges in the unit disk D . ����� : Since | z | = r < 1 , the Taylor series converges absolutely, hence converges: � � k d r k < ∞ � � � c k z k � � ≤ C � � k =0 k =0 ��������� : If | c k | < Ck d for some d , then f ( λ ) converges in the unit disk D for all λ . ����� : Follows from induction: � � kc k z k � 1 = � � ( k + 1) c k +1 z k f � ( z ) = k =0 k =0 which satisfies | ( k + 1) c k +1 | < Dk d +1 . �
• Now consider z on the unit circle 0 1 U = { z ∈ C : | z | = 1 } = • Under what conditions does the Taylor series converge on the unit circle?
� � • Let z = � � t • Then the Taylor series becomes ∞ � f ( z ) = f ( � � t ) = c k � � kt k =0 • This is a Fourier series in disguise, just now with only positive terms! • I.e., if are the Fourier series of we have and • Thus convergence of the Taylor series on the unit circle is precisely convergence of the Fourier series, which we already know
• Let z = � � t • Then the Taylor series becomes ∞ � f ( z ) = f ( � � t ) = c k � � kt k =0 • This is a Fourier series in disguise, just now with only positive terms! • I.e., if ˆ f k are the Fourier series of f ( � � t ) we have . . . , ˆ f − 3 = ˆ f − 2 = ˆ f − 1 = 0 and f 0 = c 0 , ˆ ˆ f 1 = c 1 , . . . • Thus convergence of the Taylor series on the unit circle is precisely convergence of the Fourier series, which we already know
Example: exponential function • Recall the definition of the exponential function: ∞ z k � exp( z ) = k ! k =0 • Assume we can calculate exp( z ) on the unit circle • We will calculate its Taylor expansion numerically by using the DFT
• Recall that z denotes the m evenly spaced points on the unit circle 1.0 0.5 z = � � θ = - 1.0 - 0.5 0.5 1.0 - 0.5 - 1.0 • Then the Fourier sample points of f ( � � θ ) are f = f ( � � θ ) = f ( z ) • Recall the discrete Fourier transform F α , β , which maps the m sample points to m coefficients: � � � αθ , f � � � m . . F α , β f = � � . � � � � � βθ , f � m
(Exact) For f ( z ) = exp( z ) , we have: 1. c 0 m = 5 1. c 1 0.5 c 2 0.166667 c 3 0.0416667 c 4 0.00833333 c 5 0.00138889 c 6 0.000198413 c 7 0.0000248016 c 8 2.75573 ¥ 10 - 6 - 0.166642 + 0. ‰ c 9 2.75573 ¥ 10 - 7 - 0.0416639 + 0. ‰ F α , β f = c 10 = 2.50521 ¥ 10 - 8 0.991667 + 0. ‰ c 11 2.08768 ¥ 10 - 9 0.998611 + 0. ‰ c 12 1.6059 ¥ 10 - 10 0.499802 + 0. ‰ c 13 1.14707 ¥ 10 - 11 c 14 7.64716 ¥ 10 - 13 c 15 4.77948 ¥ 10 - 14 c 16 2.81146 ¥ 10 - 15 c 17 1.56192 ¥ 10 - 16 c 18 c 19 8.22064 ¥ 10 - 18 c 20 4.11032 ¥ 10 - 19
For f ( z ) = exp( z ) , we have: (Exact) 1. c 0 1. c 1 0.5 c 2 m = 10 0.166667 c 3 0.0416667 c 4 0.00833333 c 5 0.00138889 0.00833333 + 0. ‰ c 6 0.000198413 0.00138889 + 0. ‰ c 7 0.0000248016 0.000198413 + 0. ‰ c 8 2.75573 ¥ 10 - 6 0.0000248016 + 0. ‰ c 9 2.75573 ¥ 10 - 7 2.75573 ¥ 10 - 6 + 0. ‰ F α , β f = c 10 = 2.50521 ¥ 10 - 8 1. + 0. ‰ c 11 2.08768 ¥ 10 - 9 1. + 0. ‰ c 12 14 1.6059 ¥ 10 - 10 0.5 + 0. ‰ c 13 1.14707 ¥ 10 - 11 0.166667 + 0. ‰ c 14 0.0416667 + 0. ‰ 7.64716 ¥ 10 - 13 c 15 4.77948 ¥ 10 - 14 c 16 2.81146 ¥ 10 - 15 c 17 1.56192 ¥ 10 - 16 c 18 c 19 8.22064 ¥ 10 - 18 c 20 4.11032 ¥ 10 - 19
For f ( z ) = exp( z ) , we have: (Exact) m = 20 1. c 0 1. c 1 0.5 c 2 - 0.0000248016 + 0. ‰ 0.166667 c 3 - 2.75573 ¥ 10 - 6 + 0. ‰ 0.0416667 c 4 - 2.75573 ¥ 10 - 7 + 0. ‰ 0.00833333 c 5 0.00138889 - 2.50521 ¥ 10 - 8 + 0. ‰ c 6 0.000198413 - 2.08768 ¥ 10 - 9 + 0. ‰ c 7 0.0000248016 c 8 - 1.6059 ¥ 10 - 10 + 0. ‰ 2.75573 ¥ 10 - 6 c 9 2.75573 ¥ 10 - 7 - 1.14708 ¥ 10 - 11 + 0. ‰ F α , β f = c 10 = 2.50521 ¥ 10 - 8 1. + 0. ‰ c 11 2.08768 ¥ 10 - 9 1. + 0. ‰ c 12 15 1.6059 ¥ 10 - 10 0.5 + 0. ‰ c 13 1.14707 ¥ 10 - 11 0.166667 + 0. ‰ c 14 7.64716 ¥ 10 - 13 0.0416667 + 0. ‰ c 15 4.77948 ¥ 10 - 14 0.00833333 + 0. ‰ c 16 2.81146 ¥ 10 - 15 0.00138889 + 0. ‰ c 17 0.000198413 + 0. ‰ 1.56192 ¥ 10 - 16 c 18 c 19 8.22064 ¥ 10 - 18 c 20 4.11032 ¥ 10 - 19
For f ( z ) = exp( z ) , we have: (Exact) m = 20 1. c 0 1. c 1 2.75573 ¥ 10 - 7 + 0. ‰ 0.5 c 2 2.50521 ¥ 10 - 8 + 0. ‰ 0.166667 c 3 2.08768 ¥ 10 - 9 + 0. ‰ 0.0416667 c 4 1.60591 ¥ 10 - 10 + 0. ‰ 0.00833333 c 5 1.14708 ¥ 10 - 11 + 0. ‰ 0.00138889 c 6 7.64677 ¥ 10 - 13 + 0. ‰ 0.000198413 c 7 0.0000248016 4.77896 ¥ 10 - 14 + 0. ‰ c 8 2.75573 ¥ 10 - 6 2.75335 ¥ 10 - 15 + 0. ‰ c 9 2.75573 ¥ 10 - 7 1.77636 ¥ 10 - 16 + 0. ‰ F α , β f = c 10 = - 1.11022 ¥ 10 - 17 + 0. ‰ 2.50521 ¥ 10 - 8 c 11 1. + 0. ‰ 2.08768 ¥ 10 - 9 c 12 1. + 0. ‰ 1.6059 ¥ 10 - 10 0.5 + 0. ‰ c 13 1.14707 ¥ 10 - 11 0.166667 + 0. ‰ c 14 0.0416667 + 0. ‰ 7.64716 ¥ 10 - 13 c 15 0.00833333 + 0. ‰ 4.77948 ¥ 10 - 14 c 16 0.00138889 + 0. ‰ 2.81146 ¥ 10 - 15 c 17 0.000198413 + 0. ‰ 1.56192 ¥ 10 - 16 c 18 0.0000248016 + 0. ‰ 2.75573 ¥ 10 - 6 + 0. ‰ c 19 8.22064 ¥ 10 - 18 c 20 4.11032 ¥ 10 - 19
For f ( z ) = exp( z ) , we have: (Exact) m = 40 - 8.13152 ¥ 10 - 21 + 0. ‰ 1. c 0 4.44089 ¥ 10 - 17 + 0. ‰ 1. c 1 4.44089 ¥ 10 - 17 + 0. ‰ 0.5 - 5.55112 ¥ 10 - 18 + 0. ‰ c 2 0.166667 - 3.88578 ¥ 10 - 17 + 0. ‰ c 3 0.0416667 - 5.6205 ¥ 10 - 17 + 0. ‰ c 4 2.55004 ¥ 10 - 17 + 0. ‰ 0.00833333 c 5 3.60822 ¥ 10 - 17 + 0. ‰ 0.00138889 c 6 1. + 0. ‰ 0.000198413 1. + 0. ‰ c 7 0.0000248016 0.5 + 0. ‰ c 8 0.166667 + 0. ‰ 2.75573 ¥ 10 - 6 0.0416667 + 0. ‰ c 9 2.75573 ¥ 10 - 7 0.00833333 + 0. ‰ F α , β f = c 10 0.00138889 + 0. ‰ = 2.50521 ¥ 10 - 8 0.000198413 + 0. ‰ c 11 2.08768 ¥ 10 - 9 0.0000248016 + 0. ‰ c 12 2.75573 ¥ 10 - 6 + 0. ‰ 17 1.6059 ¥ 10 - 10 2.75573 ¥ 10 - 7 + 0. ‰ c 13 1.14707 ¥ 10 - 11 2.50521 ¥ 10 - 8 + 0. ‰ c 14 2.08768 ¥ 10 - 9 + 0. ‰ 7.64716 ¥ 10 - 13 c 15 1.6059 ¥ 10 - 10 + 0. ‰ 4.77948 ¥ 10 - 14 c 16 1.14707 ¥ 10 - 11 + 0. ‰ 2.81146 ¥ 10 - 15 c 17 7.64689 ¥ 10 - 13 + 0. ‰ 4.79126 ¥ 10 - 14 + 0. ‰ 1.56192 ¥ 10 - 16 c 18 2.70894 ¥ 10 - 15 + 0. ‰ c 19 8.22064 ¥ 10 - 18 1.55431 ¥ 10 - 16 + 0. ‰ c 20 4.11032 ¥ 10 - 19 - 4.44089 ¥ 10 - 17 + 0. ‰
D ISCRETE T AYLOR T RANSFORM
For f ( z ) = exp( z ) , we have: (Exact) m = 20 1. c 0 1. c 1 2.75573 ¥ 10 - 7 + 0. ‰ 0.5 c 2 2.50521 ¥ 10 - 8 + 0. ‰ 0.166667 c 3 2.08768 ¥ 10 - 9 + 0. ‰ 0.0416667 c 4 1.60591 ¥ 10 - 10 + 0. ‰ 0.00833333 c 5 1.14708 ¥ 10 - 11 + 0. ‰ 0.00138889 c 6 7.64677 ¥ 10 - 13 + 0. ‰ 0.000198413 c 7 4.77896 ¥ 10 - 14 + 0. ‰ 0.0000248016 c 8 2.75573 ¥ 10 - 6 2.75335 ¥ 10 - 15 + 0. ‰ c 9 1.77636 ¥ 10 - 16 + 0. ‰ 2.75573 ¥ 10 - 7 F α , β f = c 10 = - 1.11022 ¥ 10 - 17 + 0. ‰ 2.50521 ¥ 10 - 8 c 11 1. + 0. ‰ 2.08768 ¥ 10 - 9 c 12 1. + 0. ‰ 19 1.6059 ¥ 10 - 10 0.5 + 0. ‰ c 13 1.14707 ¥ 10 - 11 0.166667 + 0. ‰ c 14 7.64716 ¥ 10 - 13 0.0416667 + 0. ‰ c 15 0.00833333 + 0. ‰ 4.77948 ¥ 10 - 14 c 16 0.00138889 + 0. ‰ 2.81146 ¥ 10 - 15 c 17 0.000198413 + 0. ‰ 1.56192 ¥ 10 - 16 c 18 0.0000248016 + 0. ‰ 2.75573 ¥ 10 - 6 + 0. ‰ c 19 8.22064 ¥ 10 - 18 c 20 4.11032 ¥ 10 - 19
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