ENGG2013 Unit 21 Power Series Apr, 2011.. Charles Kao Vice-chancellor of CUHK from 1987 to 1996. Nobel prize laureate in 2009. kshum2 K. C. Kao and G.

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ENGG2013 Unit 21 Power Series Apr, 2011.

Charles Kao Vice-chancellor of CUHK from 1987 to Nobel prize laureate in kshum2 K. C. Kao and G. A. Hockham, "Dielectric-fibre surface waveguides for optical frequencies," Proc. IEE, vol. 133, no. 7, pp.1151–1158, 1966.Dielectric-fibre surface waveguides for optical frequencies “It is foreseeable that glasses with a bulk loss of about 20 dB/km at around 0.6 micrometer will be obtained, as the iron impurity concentration may be reduced to 1 part per million.”

Special functions From the first paragraph of Prof. Kao’s paper (after abstract), we see J n = nth-order Bessel function of the first kind K n = nth-order modified Bessel function of the second kind. H (i) = th-order Hankel function of the ith type. kshum3

J  (x) There is a parameter  called the “order”. The  th-order Bessel function of the first kind – Two different definitions: – Defined as the solution to the differential equation – Defined by power series: kshum4

Gamma function  (x) Gamma function is the extension of the factorial function to real integer input. – Definition by integral Property :  (1) = 1, and for integer n,  (n)=(n – 1)! kshum5

Examples The 0-th order Bessel function of the first kind The first order Bessel function of the first kind kshum6

INFINITE SERIES kshum7

Infinite series Geometric series – If a = 1 and r= 1/2, – If a = 1 and r = … – If a = 1 and r = – 1 1 – – – 1 + … – If a = 1 and r = … kshum8 = 1 diverges

Formal definition for convergence Consider an infinite series – The numbers a i may be real or complex. Let S n be the nth partial sum The infinite series is said to be convergent if there is a number L such that, for every arbitrarily small  > 0, there exists an integer N such that The number L is called the limit of the infinite series. kshum9

Geometric pictures kshum10 Complex infinite series Complex plane Re Im  L Real infinite series L L+  L-  S0S0 S1S1 S2S2

Convergence of geometric series If |r|<1, then converges, and the limit is equal to. kshum11

Easy fact If the magnitudes of the terms in an infinite series does not approach zero, then the infinite series diverges. But the converse is not true. kshum12

Harmonic series kshum13 is divergent

But kshum14 is convergent

Terminologies An infinite series z 1 +z 2 +z 3 +… is called absolutely convergent if |z 1 |+|z 2 |+|z 3 |+… is convergent. An infinite series z 1 +z 2 +z 3 +… is called conditionally convergent if z 1 +z 2 +z 3 +… is convergent, but |z 1 |+|z 2 |+|z 3 |+… is divergent. kshum15

Examples is conditionally convergent. is absolutely convergent. kshum16

Convergence tests Some sufficient conditions for convergence. Let z 1 + z 2 + z 3 + z 4 + … be a given infinite series. (z 1, z 2, z 3, … are real or complex numbers) 1.If it is absolutely convergent, then it converges. 2.(Comparison test) If we can find a convergent series b 1 + b 2 + b 3 + … with non-negative real terms such that |z i |  b i for all i, then z 1 + z 2 + z 3 + z 4 + … converges. kshum17

Convergence tests 3.(Ratio test) If there is a real number q < 1, such that for all i > N (N is some integer), then z 1 + z 2 + z 3 + z 4 + … converges. If for all i > N,, then it diverges kshum18

Convergence tests 4.(Root test) If there is a real number q < 1, such that for all i > N (N is some integer), then z 1 + z 2 + z 3 + z 4 + … converges. If for all i > N,, then it diverges. kshum19

Derivation of the root test from comparison test Suppose that for all i  N. Then for all i  N. But is a convergent series (because q<1). Therefore z 1 + z 2 + z 3 + z 4 + … converges by the comparison test. kshum20

Application Given a complex number x, apply the ratio test to The ratio of the (i+1)-st term and the i-th term is Let q be a real number strictly less than 1, say q=0.99. Then, Therefore exp(x) is convergent for all complex number x. kshum21

Application Given a complex number x, apply the root test to The ratio of the (i+1)-st term and the i-th term is Let q be a real number strictly less than 1, say q=0.99. Then, Therefore exp(x) is convergent for all complex number x. kshum22

Variations: The limit ratio test If an infinite series z 1 + z 2 + z 3 + …, with all terms nonzero, is such that Then 1.The series converges if  < 1. 2.The series diverges if  > 1. 3.No conclusion if  = 1. kshum23

Variations: The limit root test If an infinite series z 1 + z 2 + z 3 + …, with all terms nonzero, is such that Then 1.The series converges if  < 1. 2.The series diverges if  > 1. 3.No conclusion if  = 1. kshum24

Application Let x be a given complex number. Apply the limit root test to The nth term is The nth root of the magnitude of the nth term is kshum25

Useful facts Stirling approximation: for all positive integer n, we have kshum26 J 0 (x) converges for every x

POWER SERIES kshum27

General form The input, x, may be real or complex number. The coefficient of the nth term, a n, may be real or complex number. kshum28

Approximation by tangent line kshum29 x = linspace(0.1,2,50); plot(x,log(x),'r',x, log(0.6)+(x-0.6)/0.6,'b') grid on; xlabel('x'); ylabel('y'); legend(‘y = log(x)’, ‘Tangent line at x=0.6‘)

Approximation by quadratic kshum30 x = linspace(0.1,2,50); plot(x,log(x),'r',x, log(0.6)+(x-0.6)/0.6-(x-0.6).^2/0.6^2/2,'b') grid on; xlabel('x'); ylabel('y') legend(‘y = log(x)’, ‘Second-order approx at x=0.6‘)

Third-order kshum31 x = linspace(0.05,2,50); plot(x,log(x),'r',x, log(0.6)+(x-0.6)/0.6-(x-0.6).^2/0.6^2/2+(x-0.6).^3/0.6^3/3,'b') grid on; xlabel('x'); ylabel('y') legend('y = log(x)', ‘Third-order approx at x=0.6')

Fourth-order kshum32 x = linspace(0.05,2,50); plot(x,log(x),'r',x, log(0.6)+(x-0.6)/0.6-(x-0.6).^2/0.6^2/2+(x-0.6).^3/0.6^3/3-(x-0.6).^4/0.6^4/4,'b') grid on; xlabel('x'); ylabel('y') legend(‘y = log(x)’, ‘Fourth-order approx at x=0.6‘)

Fifth-order kshum33 x = linspace(0.05,2,50); plot(x,log(x),'r',x, log(0.6)+(x-0.6)/0.6-(x-0.6).^2/0.6^2/2+(x-0.6).^3/0.6^3/3-(x-0.6).^4/0.6^4/4+(x-0.6).^5/0.6^5/5,'b') grid on; xlabel('x'); ylabel('y') legend(‘y = log(x)’, ‘Fifth-order approx at x=0.6‘)

Taylor series Local approximation by power series. Try to approximate a function f(x) near x 0, by a 0 + a 1 (x – x 0 ) + a 2 (x – x 0 ) 2 + a 3 (x – x 0 ) 3 + a 4 (x – x 0 ) 4 + … x 0 is called the centre. When x 0 = 0, it is called Maclaurin series. a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 + a 6 x 6 + … kshum34

Taylor series and Maclaurin series kshum35 Brook Taylor English mathematician 1685—1731 Colin Maclaurin Scottish mathematician 1698—1746

Examples kshum36 Geometric series Exponential function Sine function Cosine function More examples at

How to obtain the coefficients Match the derivatives at x =x 0 Set x = x 0 in f(x) = a 0 +a 1 (x – x 0 )+a 2 (x – x 0 ) 2 +a 3 (x – x 0 )  a 0 = f(x 0 ) Set x = x 0 in f’(x) = a 1 +2a 2 (x – x 0 ) +3a 3 (x – x 0 ) 2 +…  a 1 = f’(x 0 ) Set x = x 0 in f’’(x) = 2a 2 +6a 3 (x – x 0 ) +12a 4 (x – x 0 ) 2 +…  a 2 = f’’(x 0 )/2 – In general, we have a k = f (k) (x 0 ) / k! kshum37

Example f(x) = log(x), x 0 =0.6 First-order approx. log(0.6)+(x – 0.6)/0.6 Second-order approx. log(0.6)+(x – 0.6)/0.6 – (x – 0.6) 2 /(2· ) Third-order approx. log(0.6)+(x–0.6)/0.6 – (x–0.6) 2 /(2· ) +(x–0.6) 3 /(3· ) kshum38

Example: Geometric series Maclaurin series 1/(1– x) = 1+x+x 2 +x 3 +x 4 +x 5 +x 6 +… Equality holds when |x| < 1 If we carelessly substitute x=1.1, then L.H.S. of 1/(1– x) = 1+x+x 2 +x 3 +x 4 +x 5 +x 6 +… is equal to -10, but R.H.S. is not well-defined. kshum39

Radius of convergence for GS For the geometric series 1+z+z 2 +z 3 +…, it converges if |z| 1. We say that the radius of convergence is 1. 1+z+z 2 +z 3 +… converges inside the unit disc, and diverges outside. kshum40 complex plane

Convergence of Maclaurin series in general If the power series f(x) converges at a point x 0, then it converges for all x such that |x| < |x 0 | in the complex plane. kshum41 x0x0 Re Im converge Proof by comparison test

Convergence of Taylor series in general If the power series f(x) converges at a point x 0, then it converges for all x such that |x – c| < |x 0 – c| in the complex plane. kshum42 x0x0 Re Im converge Proof by comparison test also c R

Region of convergence The region of convergence of a Taylor series with center c is the smallest circle with center c, which contains all the points at which f(x) converges. The radius of the region of convergence is called the radius of convergence of this Taylor series. kshum43 Re Im converge c R diverge

Examples : radius of convergence = 1. It converges at the point z= –1, but diverges for all |z|>1. exp(z): radius of convergence is , because it converges everywhere. : radius of convergence is 0, because it diverges everywhere except z=0. kshum44

Behavior on the circle of convergence On the circle of convergence |z-c| = R, a Taylor series may or may not converges. All three series  z n,  z n /n, and  z n /n 2 Have the same radius of convergence R=1. But  z n diverges everywhere on |z|=1,  z n /n diverges at z= 1 and converges at z=– 1,  z n /n 2 converges everywhere on |z|=1. kshum45 R

Summary Power series is useful in calculating special functions, such as exp(x), sin(x), cos(x), Bessel functions, etc. The evaluation of Taylor series is limited to the points inside a circle called the region of convergence. We can determine the radius of convergence by root test, ratio test, etc. kshum46