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ENGG2013 Unit 21 Power Series. Apr, 2011. Charles Kao. Vice-chancellor of CUHK from 1987 to 1996. Nobel prize laureate in 2009. K. C. Kao and G. A. Hockham, " Dielectric-fibre surface waveguides for optical frequencies ," Proc. IEE, vol. 133, no. 7, pp.1151–1158, 1966.
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ENGG2013 Unit 21Power Series Apr, 2011.
Charles Kao • Vice-chancellor of CUHK from 1987 to 1996. • Nobel prize laureate in 2009. K. C. Kao and G. A. Hockham, "Dielectric-fibre surface waveguides for optical frequencies," Proc. IEE, vol. 133, no. 7, pp.1151–1158, 1966. “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.” kshum
Special functions From the first paragraph of Prof. Kao’s paper (after abstract), we see • Jn = nth-order Bessel function of the first kind • Kn = nth-order modified Bessel function of the second kind. • H(i)= th-order Hankel function of the ith type. kshum
J(x) • There is a parameter called the “order”. • The th-order Bessel function of the first kind • http://en.wikipedia.org/wiki/Bessel_function • Two different definitions: • Defined as the solution to the differential equation • Defined by power series: kshum
Gamma function (x) • Gamma function is the extension of the factorial function to real integer input. • http://en.wikipedia.org/wiki/Gamma_function • Definition by integral • Property : (1) = 1, and for integer n, (n)=(n – 1)! kshum
Examples • The 0-th order Bessel function of the first kind • The first order Bessel function of the first kind kshum
INFINITE SERIES kshum
Infinite series • Geometric series • If a = 1 and r= 1/2, • If a = 1 and r = 1 1+1+1+1+1+… • If a = 1 and r = – 1 1 – 1 + 1 – 1 + 1 – 1 + … • If a = 1 and r = 2 1+2+4+8+16+… = 1 diverges diverges diverges kshum
Formal definition for convergence • Consider an infinite series • The numbers aimay be real or complex. • Let Sn 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. kshum
Geometric pictures Complex infinite series Real infinite series Im Complex plane S1 S2 S0 L L- L+ L Re kshum
Convergence of geometric series • If |r|<1, then converges, and the limit is equal to . kshum
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. kshum
Harmonic series is divergent kshum
But is convergent kshum
Terminologies • An infinite series z1+z2+z3+… is called absolutely convergent if |z1|+|z2|+|z3|+… is convergent. • An infinite series z1+z2+z3+… is called conditionally convergent if z1+z2+z3+… is convergent, but |z1|+|z2|+|z3|+… is divergent. kshum
Examples is conditionally convergent. is absolutely convergent. kshum
Convergence tests Some sufficient conditions for convergence. Let z1 + z2 + z3 + z4 + … be a given infinite series. (z1, z2, z3, … are real or complex numbers) • If it is absolutely convergent, then it converges. • (Comparison test) If we can find a convergent series b1 + b2 + b3 + … with non-negative real terms such that |zi| bi for all i, then z1 + z2 + z3 + z4 + … converges. http://en.wikipedia.org/wiki/Comparison_test kshum
Convergence tests • (Ratio test) If there is a real number q < 1, such that for all i > N (N is some integer), then z1 + z2 + z3 + z4 + … converges. If for all i > N , , then it diverges http://en.wikipedia.org/wiki/Ratio_test kshum
Convergence tests • (Root test) If there is a real number q < 1, such that for all i > N (N is some integer), then z1 + z2 + z3 + z4 + … converges. If for all i > N , , then it diverges. http://en.wikipedia.org/wiki/Root_test kshum
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 z1 + z2 + z3 + z4 + … converges by the comparison test. kshum
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. kshum
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. kshum
Variations: The limit ratio test • If an infinite series z1 + z2 + z3 + … , with all terms nonzero, is such that Then • The series converges if < 1. • The series diverges if > 1. • No conclusion if = 1. kshum
Variations: The limit root test • If an infinite series z1 + z2 + z3 + … , with all terms nonzero, is such that Then • The series converges if < 1. • The series diverges if > 1. • No conclusion if = 1. kshum
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 kshum
Useful facts • Stirling approximation: for all positive integer n, we have J0(x) converges for every x kshum
POWER SERIES kshum
General form • The input, x, may be real or complex number. • The coefficient of the nth term, an, may be real or complex number. http://en.wikipedia.org/wiki/Power_series kshum
Approximation by tangent line 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‘) kshum
Approximation by quadratic 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‘) kshum
Third-order 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') kshum
Fourth-order 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‘) kshum
Fifth-order 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‘) kshum
Taylor series Local approximation by power series. Try to approximate a function f(x) near x0, by a0 + a1(x – x0) + a2(x – x0)2 + a3(x – x0)3 + a4(x – x0)4 + … x0 is called the centre. When x0 = 0, it is called Maclaurin series. a0 + a1x + a2 x2 + a3 x3 + a4x4 + a5x5 + a6x6 + … kshum
Taylor series and Maclaurin series Brook Taylor English mathematician 1685—1731 Colin Maclaurin Scottish mathematician 1698—1746 kshum
Examples Geometric series Exponential function Sine function Cosine function More examples at http://en.wikipedia.org/wiki/Maclaurin_series kshum
How to obtain the coefficients • Match the derivatives at x =x0 • Set x = x0 in f(x)= a0+a1(x – x0)+a2(x – x0)2 +a3(x – x0)3+... • a0= f(x0) • Set x = x0 in f’(x)= a1+2a2(x – x0)+3a3(x – x0)2+… a1= f’(x0) • Set x = x0 in f’’(x)= 2a2+6a3(x – x0) +12a4(x – x0)2+… • a2= f’’(x0)/2 • In general, we have ak= f(k)(x0) / k! kshum
Example f(x) = log(x), x0=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· 0.62) Third-order approx. log(0.6)+(x–0.6)/0.6 – (x–0.6)2/(2· 0.62) +(x–0.6)3/(3· 0.63) kshum
Example: Geometric series Maclaurin series 1/(1– x) = 1+x+x2+x3+x4+x5+x6+… Equality holds when |x| < 1 If we carelessly substitute x=1.1, then L.H.S. of 1/(1– x) = 1+x+x2+x3+x4+x5+x6+… is equal to -10, but R.H.S. is not well-defined. kshum
Radius of convergence for GS complex plane For the geometric series 1+z+z2+z3+… , it converges if |z|< 1, but diverges when |z| > 1. We say that the radius of convergence is 1. 1+z+z2+z3+… converges inside the unit disc, and diverges outside. kshum
Convergence of Maclaurin series in general • If the power series f(x) converges at a point x0, then it converges for all x such that |x| < |x0| in the complex plane. Im converge Re x0 Proof by comparison test kshum
Convergence of Taylor series in general • If the power series f(x) converges at a point x0, then it converges for all x such that |x – c| < |x0 – c| in the complex plane. Im converge R c Re x0 Proof by comparison test also kshum
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. Im diverge converge R c Re kshum
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. kshum
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 zn, zn/n, and zn/n2 Have the same radius of convergence R=1. But zn diverges everywhere on |z|=1, zn /n diverges at z= 1 and converges at z=– 1 , zn/n2 converges everywhere on |z|=1. R kshum
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. kshum