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ENGG2012B Lecture 21 Power series approach to the complex exponential function

ENGG2012B Lecture 21 Power series approach to the complex exponential function. Kenneth Shum. Analogy. Approximation of a fraction by decimal numbers. Approximation of a rational function by power series. Approximation of 1/(1-x) near x=0. 1 1+x 1+x+x 2 1+x+x 2 +x 3 1+x+x 2 +x 3 +x 4

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ENGG2012B Lecture 21 Power series approach to the complex exponential function

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  1. ENGG2012BLecture 21Power series approach to the complex exponential function Kenneth Shum ENGG2012B

  2. Analogy Approximation of a fractionby decimal numbers Approximation of a rational function by power series Approximation of 1/(1-x) near x=0. 1 1+x 1+x+x2 1+x+x2+x3 1+x+x2+x3+x4 1+x+x2+x3+x4+x5 … • Approximations of 2/3 • 0.6 • 0.66 • 0.666 • 0.6666 • 0.66666 • 0.666666 • … ENGG2012B

  3. Analogy Approximation of a transcendental number by decimal numbers Approximation of a transcendental function by power series sin(x) x x – x3/3! x – x3/3! + x5/5! – x7/7! x – x3/3! + x5/5! – x7/7! + x9/9! … • Approximation of  • 3 • 3.1 • 3.14 • 3.141 • 3.1415 • 3.14159 • … ENGG2012B

  4. Analogy Accuracy = number of decimal places Accuracy = number of terms 1 + x + x2 + x3 + O(x4) 2 – 3x + O(x2) 1 + x + x2 + x3 + O(x4) + 2 – 3x + O(x2) = 3 – 2x + O(x2) • 2.351 (3 decimal places) • 1.8 (1 decimal places) • 2.351 + 1.4 = 4.2 (1 decimal places) higher-order terms Addition with carry Addition without carry ENGG2012B

  5. Multiplication of power series • Multiplication of two power series is essentially the same as convolution. (a0 + a1x + a2x2 + a3x3+…)(b0 + b1x+ b2x2 + b3x3 + …)= a0b0 + (a0b1 +a1 b0)x + (a0b2+a1b1+a2b0)x2 + … ENGG2012B

  6. Taylor series • Given a smooth function f(x), we derive the Taylor series expansion at x = x0, a0 + a1(x – x0) + a2(x – x0)2 + a3(x – x0)3 + … by setting ak= f(k)(x0) / k!, for k=0,1,2,…. • We need to answer two questions: • Does the derived Taylor series converges? • If it converges, does it converges to the function f(x)? ENGG2012B

  7. Example • Find the Maclaurin series of f(x) = (1+x)1/2. y y2 = 1+x x ENGG2012B

  8. Method 1 • Differentiate repeatedly and apply the formula ak= f(k)(x0) / k!, for k=0,1,2,…. • f(x) = (1+x)1/2  a0 = 1 ( the y-intercept) • f’(x) = (1/2)(1+x)–1/2 a1 = 1/2 (the slope at x=0) • f’’(x) = (1/2) (–1/2)(1+x)–3/2 a2 = –1/8 • f’’’(x) = (1/2) (–1/2)(–3/2)(1+x)–5/2 a3 = 1/16 • f’’’’(x) = (1/2) (–1/2)(–3/2) (–5/2)(1+x)–6/2 a4 =–5/128 • … by mathematical induction ak = (1/k!)(1/2) (–1/2)(–3/2)…(–(2k–1)/2). • If we extend the definition binomial coefficients to real number x,we can write the Maclaurin series expansion as ENGG2012B

  9. Method 2 • Let the required power series be • Solve for the ai’s by equating the coefficients in • a02 = 1  a0= 1, and we have to choose a0=1. • 2a0a1 = 1  a1 = 1/(2a0) = 1/2. • 2a0a2 + a12 = 0  a2 = (–a12)/(2a0) = –1/8. • 2a0a3 + 2a1a2 = 0  a3 = (–a1a2)/a0 = 1/16. • … ENGG2012B

  10. Other methods of obtaining power series expansion of a function (I) • From the example in last Friday’s lecture, Differentiate both sides to get the power series expansion of 1/(2-x)2 But does itconverge for all x? ENGG2012B

  11. Taylor series may not converge • Consider the power series • g(2) is infinite because 1/(2-x) is undefined at x=2. g(x) is undefined at x=2 divergent divergent convergent 0 -2 -3 -1 1 2 3 x ENGG2012B

  12. Geometric series In the geometric series, 1/(1– x) = 1+x+x2+x3+x4+x5+x6+… equality holds when |x| < 1.The R.H.S. diverges if |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 divergent. ENGG2012B

  13. Other methods of obtaining power series expansion of a function (II) • Integrating both side of 1/(1– x) = 1+x+x2+x3+x4+x5+x6+… we get for x between -1 and 1. ENGG2012B

  14. Question • What is the Maclaurin series of ENGG2012B

  15. The Taylor series may not convergeeven if the function is well-defined • Use geometric series • It converges if |x|<1, but diverges if |x|>1. ENGG2012B

  16. Why does it fail to converge? • The reason is revealed if we extend the domain to complex number where z may be a complex number. Im z = j divergent convergent Re The value of 1/(1+z2)is infinite if z=j z = - j ENGG2012B

  17. Complex power series • The variable z may be real or complex number. • The coefficients a0, a1, a2,.. may be real or complex numbers. ENGG2012B

  18. Complex geometric series • For any complex number z with |z|<1,the series 1+z+z2+z3+z4+… converges to . • Example: set z=(1+j)/2. |z| = 1/sqrt(2) • 1+z = 1.5+0.5j • 1+z+z2 = 1.5+j • 1+z+z2+z3 = 1.25+1.25j • 1+z+z2+z3+ z4= 1+1.25j • 1+z+z2+z3+z4+z5= 0.875 + 1.125j • 1+z+z2+z3+z4+z5+z6= 0.875 + j • 1+z+z2+z3+z4+z5+z6+z7= 0.9375 + 0.9375j • If we add infinitely many terms, it will converge to 1 + j. ENGG2012B

  19. Recall from calculus: limit of series • 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. • If no such L exists, the infinite series is said to be divergent. ENGG2012B

  20. Picture for the convergence ofa sequence of complex numbers Complex infinite series Real infinite series Im Complex plane S1 S2 S0 L L- L+ L  Re ENGG2012B

  21. CONVERGENCE TESTS ENGG2012B

  22. A necessary condition for convergence If a series  aiconverges, then | ai | converges to zero.This means that for any arbitrarily small >0, we can find a sufficientlylarge integer N such that | ai | <  for all i  N. N This conditionis not sufficientbecause we havethe divergent series  1/i. 2 ENGG2012B

  23. The converse is false • Conversely if |ai| approaches 0 as i approaches infinity, the series a1+ a2+ a3 + … may or may not converges. • The harmonic series 1+1/2+1/3+1/4+1/5+1/6+… is divergent. • However, 1 – 1/2 + 1/3 – 1/4 + 1/5 – 1/6 +… is convergent. ENGG2012B

  24. Absolute and conditional convergence • 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. ENGG2012B

  25. Examples divergent conditionally convergent absolutely convergent ENGG2012B

  26. Cauchy convergence criterion http://en.wikipedia.org/wiki/Cauchy%27s_convergence_test • This is a necessary and sufficient condition for convergence of series. • Given any arbitrarily small >0, we can find a sufficientlylarge integer N such that | am +am +1+ am+2 +… +an| <  for all m, n  N. (m < n) for any >0 The absolute value ofthe sum of the terms in the window from m to n is less than  If the “window size”is one, it reducesto the necessarycondition in p.15. m N n ENGG2012B

  27. Augustin-Louis Cauchy (1789-1857) • French mathematican. • Introduced the epsilon-delta argument in calculus. • The Cauchy-Riemann conditionin complex analysis. http://en.wikipedia.org/wiki/Augustin-Louis_Cauchy ENGG2012B

  28. Cauchy’s convergence criterion applied to geometric series • It is known that the series 1+0.5+0.52+0.53+0.53+0.54+ … is convergent, but we can still apply Cauchy’s convergence criterion as an illustration. • Let  be an arbitrarily small and positive real number. • Let N be the smallest integer such that 0.5N < . • Then for any integer m and n > N, with m<n, we have 0.5m + 0.5m+1 +… + 0.5n = (0.5m-0.5n+1) /(1-0.5)< 0.5m /0.5 < 0.5m-1 < 0.5N< . • We can conclude from Cauchy’s convergence criterion that this geometric series converges. ENGG2012B

  29. Theorem Absolute convergence implies convergence • Proof by Cauchy’s convergence criterion • Suppose z1 + z2 + z3 + … converges absolutely. By definition, |z1| + |z2|+ |z3| + … is convergent. • Want to show z1 + z2 + z3 + … is convergent. • Let  be an arbitrarily small positive number. • There exists a large integer N, such that |zm|+|zm+1|+…+|zn|   for all m, n  N, with m <n. • But |zm+zm+1+…+zn|  |zm|+|zm+1|+…+|zn| by the triangular inequality for complex numbers. • Therefore |zm+zm+1+…+zn|   for all m, n  N, with m <n. ENGG2012B

  30. Important property • Letbe a power series, whose coefficients are complex numbers. • If p(z) converges at z=z0, then p(z) converges absolutely for all z in the circle |z| < |z0|. Im z0 Re convergent Since p(z) is absolutely convergent,it is also convergent. ENGG2012B

  31. Proof by Cauchy’s convergence criterion • Suppose that p(z0) converges, and|z0|>0. Let  be any positive number less than 1. Want to prove that p(z) converges absolutely for |z|< |z0|. • The absolute value |ai z0i| tend to zero as i approaches infinity. • We can find a large integer M such that |ai z0i| < 1 for all i M.  |ai| < |z0|–i for all i M. • Let  be an arbitrarily small positive number. • Increase M if necessarily so that M/(1– ) < . (this is possible because  <1) • Consider a complex number z satisfying |z|<  |z0|. • For any m<n be two integers larger than M, we have |amzm| + |am+1zm+1| +…+ |anzn| = |am | | z |m + |am+1 | | z |m+1 +…+ |an | | z |n < |z0|–m | z |m + |z0|–(m+1) | z |m+1 +…+ |z0|–n | z |n <|z0|–m | z |m + |z0|–(m+1) | z |m+1 + |z0|–(m+2) | z |m+2 +… = |z/z0|m/ (1 – |z/z0| ) < |z/z0|M/ (1 – |z/z0| ) < M/ (1 –  ) <  • Hence p(z) = a0 + a1z1 + a2z2+… is absolutely convergent. Im z0 Re  |z0| ENGG2012B

  32. Modus Tollens of the important theorem • If the power series diverges at z=z1, then p(z) diverges whenever |z| > |z1| From Logic:A implies B  not B implies not A Im Re z1 ENGG2012B

  33. Region of convergence for power series in general For any power series there exists a radius R, such that • the power series converges if |z – c| < R, • the power series diverges if |z – c| > R. Im divergent convergent Re The region of convergence of a power serieshas the shape of a disc. The radius is calledthe radius of convergence. R ENGG2012B

  34. Why? • Consider three cases. • Case 1: The power series converges everywhere. In this case, R = infinity. The whole complex plane is the region of convergence. • Case 2: The power series converges only at the origin z=0. In this case R = 0. The region of convergence contains a single point. ENGG2012B

  35. Why? (cont’d) • Case 3: the power series converges for some complex number z0 and diverges for some complex number z1. We must have|z0| |z1|, otherwise it would violate the “important property” in p.23. • If |z0|= |z1|, then the results follows immediately by the “important property”. • If|z0|< |z1|, there is an annulus of uncertainty|z0|<z<|z1|. Pick a complex number z’ such that |z’| = (|z0|+|z1|)/2. If the power series converges at z’, then the area of convergence can be enlarged to |z| < |z’|.If the power series diverges at z’, then the area of divergence can be enlarged to |z| > |z’|.In any case, we shrink the annulus of uncertainty. If we repeat this argument infinitely many times, the difference between|z0| and|z1|can be made arbitrarily small. This will give the desired radius of convergence. divergent z1 z0 convergent uncertain ENGG2012B

  36. Region of convergence for Taylor series • Coefficients ai are complex numbers. c is the centre, and variable z is complex-valued. There exists a radius R, such that • the series converges if |z – c| < R, • the series diverges if |z – c| > R. • The region of convergence hasthe shape of a disc. • The radius is called the radiusof convergence divergent Im convergent R c Re ENGG2012B

  37. Behavior on the boundary of the region 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 ENGG2012B

  38. Convergence test: comparison test Let z1 + z2 + z3 + z4 + … be a given infinite series. (z1, z2, z3, … are real or complex numbers) If we can find non-negative real numbers b1, b2, b3, b4, … such that • |zi|  bi for all i, and // zi is dominated by bi • b1 + b2 + b3 + b4 + … converges, then z1 + z2 + z3 + z4 + … converges absolutely. http://en.wikipedia.org/wiki/Comparison_test ENGG2012B

  39. Last time: the region of convergence is proved by the comparison test Want to see whetherconverges. Try to find a which is convergent and dominates bi zi i If b1 + b2 + b3 + b4 + … convergesthen |z1| + |z2| + |z3| + |z4| + … converges -bi Note: If we cannot find such a series  bi,it does not mean that the series  zi is divergent. Even if we can find such a series  bi, the test does not tell where it converges to. ENGG2012B

  40. Proof by Cauchy’s convergence criterion (CCC) • Because b1 + b2 + b3 + b4 + … converges, for any  >0, we can find a large integer N such that |bm| + |bm+1|+ …|bn|  for all m, n N (m<n). (this is true by CCC) • By the assumption that |zi| |bi| for all i, we have |zm|+|zm+1|+ … +|zn| |bm|+|bm+1|+ …+|bn|  • Hence |z1| + |z2| + |z3| + |z4| + … converges, by CCC again. ENGG2012B

  41. Convergence test: 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. For simplicity, we assume that the limit exists. ENGG2012B

  42. Proof by comparing with geometric series • Suppose that < 1. • Pick a number  such that <  <1. • Because the |zn+1/zn| tends to , we have |zn/zn+1| <  for all sufficiently large n. •  a large integer N such that |zn+1/zn| <  for all n  N. • Therefore |zn+1| < |zn|  for all n  N. • By the comparison test, |zN| + |zN+1| + |zN+2| + … converges • |zN+1| <  |zN|, |zN+2| < 2|zN|, |zN+3| < 3|zN|. • Take bn = |zN| n in the comparison test. • |zN|( +2 +3 +…) is a convergent geometric series. • Hence zN + zN+1 + zN+2+ … converges by comparison test • Hence z1 + z2 + z3+ z4 + … + zN + zN+1 + zN+2+ … converges. ENGG2012B

  43. Proof (cont’d) • Suppose that  > 1. • Because the |zn+1/zn| tends to  and  > 1, |zn+1/zn| > 1 for all sufficiently large n. • |zn+1| > |zn| for all sufficiently large n. • |zn| do not converge to zero as n goes to infinity. • Hence, z1 + z2 + z3 + … is divergent. • If  = 1, there is no conclusion, because •  (1/n) is divergent •  (1/n2) is convergent, and converges to 2/6. ENGG2012B

  44. Three examples R= • The power series converges for all complex numbers z. The radius of convergence is infinity. • It is because for each z, the ratio of consecutive terms as n approaches infinity. By the ratio test, this power series converges for every complex number z. ENGG2012B

  45. Three examples (cont’d) R=1 • : radius of convergence = 1. It converges at the point z= –1, but diverges for all |z|>1. • : radius of convergence is 0, because it diverges everywhere except z=0. R=0 ENGG2012B

  46. Three properties for power series • A power series can be differentiated term by term within the region of convergence. • A power series can be integrated term by term within the region of convergence • The order of summing the terms in a power series does not matter. We can re-arrange the order of summation freely for power series. All of these properties are not true for series in general. ENGG2012B

  47. COMPLEX EXPONENTIAL FUNCTION ENGG2012B

  48. Eight Wonders of the Mathematical World • http://www.youtube.com/watch?v=lxAZ_FLudKc ENGG2012B

  49. Logical flow Real-valued exponential function Power series expansionconverges for all xby comparison test Extend the domainof the function tocomplex numbers.It converges for all complexnumber z. Define the complex exponentialfunction by ez is a short-hand notation for exp(z). ENGG2012B

  50. The expected properties of exponential function For any complex numbers z1 and z2, Proof by multiplying the L.H.S.: ENGG2012B

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