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Math 104 - Calculus I Part 8 Power series, Taylor series Series First .. a review of what we have done so far: 1. We examined series of constants and learned that we can say everything there is to say about geometric and telescoping series. 2. We developed tests for convergence of series of constants. 3. We considered power series, derived formulas and other tricks for finding them, and know them for a few functions. 4. We used the ratio tests to determine intervals on which power series converge, and use the other tests to check convergence at the endpoints of the intervals. Geometric series a/(1-r) = a + ar + ar2 + ar3 + ... provided |r|< 1. We often use partial fractions to detect telescoping series, for which we can calculate explicitly the partial sums sn Tests for convergence for series of constants Fundamental divergence test (nth term must go to zero for convergence to be possible) Integral test Comparison and limit comparison tests Ratio test Root test Alternating series test Power series f(x) = a0 + a1 x + a2 x2 + a3 x3 + ... where an = f(n)(0)/n! We know the series for ex, sin(x), cos(x), 1/(1-x), and a few related functions. Convergence of power series Before we get too excited about finding series, let's make sure that, at the very least, the series converge. Next week, we'll deal with the question of whether they converge to the function we expect. But for now, we'll assume that if they converge, they converge to the function they "came from". (Strictly speaking, this is not always true -- but it is true for a large class of functions, which includes nearly all the ones encountered in basic science and mathematics. This fact was not fully appreciated until the early part of the twentieth century.) Fortunately, most of the question of whether power series converge is answered fairly directly by the ratio test. Recall that... for a series of constants bn , we have that the n 0 series converges (absolutely) if the the limit of the absolute value of bbn1 is less than one, diverges if n the limit is greater than one, and the test is indeterminate if the limit equals one. To use the ratio test on power series, just leave the x there and calculate the limit for each value of x. This will give an inequality that x must satisfy in order for the series to converge. For the series for the exponential function... x For e , we carry out the ratio test as follows : ( n 1 ) bn 1 x n! x . No matter wha t x n n 1 bn ( n 1 )! x x is, we have lim 0. Since the ratio limit is n n 1 less than one for all x, the series converges for all x. Your turn... Calculate the series for the function sin(x) and determine for which x the series converges. Here’s a more interesting example xn x ( n 1) n nx n - - The ratio test this time gives (n 1) x n n 1 . n 1 The limit of this as n tends to infinity is simply x. The conclusionis that theseries convergesif x 1 and diverges if x 1 (reminiscent of the geometric series). What remains... is to check thepoints x 1 and x -1. For x 1, the series becomes 1 , the harmonic series, which n n 1 we know to be divergent. For x 1, the series becomes ( 1) n n , the alternating n 1 harmonic series, which weknow to be (conditionally) convergent. Final Conclusion The series xn n convergesif - 1 x 1 n 1 and diverges otherwise. OK, your turn... For which values of x does the series ( 2 x )n n2 converge? n 1 A. -1 < x < 1 B. -2 < x < 2 C. 1/2 < x < 1/2 D. -2 < x < 2 E. -1/2 < x < 1/2 One more... For which values of x does the series nx n converge? n 1 A. -1 < x < 1 B. -1 < x < 1 C. 1 < x < 1 D. -1 < x < 1 E. 0 < x < 1 From these examples, ...it should be apparent that power series converge for values of x in an interval that is centered at zero, i.e., an interval of the form [-a, a] , (-a, a], [-a, a) or (-a, a) (where a might be either zero or infinity). The interval is called the interval of convergence and the number a is called the radius of convergence . Let’s go back To finding series of functions: There are two ways: 1. Use the formula an f n (0)/n! We've found series for e x and sin( x) : x 2 x3 x 4 xn e 1 x ... x 2! 3! 4! n 0 n! x3 x5 x7 (1) n x ( 2 n 1) sin( x) x ... 3! 5! 7! (2n 1)! We could do other series this way,but method 2 is more fun : The other way 2. Start from known series and use algebraic and/or analytic manipulation to get others : Examples : Substitute x 2 for x everywhere in the e x series to get : x 4 x 6 x8 x (2n) 1 x 2 ... 2 e( x ) 2! 3! 4! n 0 n! Try this... Take the derivative of the series for sin(x) to get x2 x4 x6 ( 1) n x ( 2 n ) cos(x) 1 ... 2! 4! 6! n 0 (2n)! Integrate both sides of the geometric series from 0 to x to get: x x x x x dt 1dt tdt t dt t dt ... 1 2 3 0 1-t 0 0 0 0 x 2 x3 x 4 - ln(1 - x) x ... 2 3 4 Negate both sides and replace x by (-x) everywhere to get: 2 3 4 x x x ln(1 x) x ... 2 3 4 (This shows that the fourth series sums to ln(2)). Start from the geometric series again... And substitute x 2 for x everywhere it appears to get 1 1 x 2 x 4 x 6 x 8 x10 ... 1 x2 Now integrate both sides from 0 to x to get : x 3 x 5 x 7 x 9 x11 arctan(x) x ... 3 5 7 9 11 (This shows that thefirst series we saw earlier converges to .) 4 A challenge to think about... How to get the other one from previously ( n12 ) ? n 1 Application of Series 1. Limits: Series give a good idea of the behavior of functions in the neighborhood of 0: sin( x) We know for other reasons that lim 1 x 0 x We could do this by series: sin( x) (1) n x 2 n x2 x4 x6 lim lim lim 1 1 0 0... 1 n 0 ( 2n 1)! x 0 x x 0 x 0 3! 5! 7! This can be used on complicated limits... x sin( x) Calculate the limit: lim ( x3 ) x 0 1 e A. 0 B. 1/6 C. 1 D. 1/12 E. does not exist Application of series (continued) 2. Approximate evaluation of integrals: Many integrals that cannot be evaluated in closed form (i.e., for which no elementary anti-derivative exists) can be approximated using series (and we can even estimate how far off the approximations are). 1 e ( x2 ) Example: Calculate dx to the nearest 0.001. 0 We begin by... substituti ng - x 2 for x in the series we already know for e x , and integratin g it. This will give us a numerical series that 1 ( x2 ) converges to the answer : e 0 dx is approximat ed by 1 x4 x6 1 1 1 2 1 x dx ... 1 ... 0 2! 3! 3 5 2! 7 3! According to Maple... The last series is an alternating series with decreasing terms. We need to find the first one that is less than 0.0005 to ensure that the error will be less than 0.001. According to Maple: evalf(1/(7*factorial(3))), evalf(1/(9*factorial(4))),evalf( 1/(11*factorial(5))); .02380952381, .004629629630, .0007575757576 evalf(1/(13*factorial(6))); .0001068376068 Keep going... So it's enough to go out to the 5! term. We do this as follows: Sum((-1)^n/((2*n+1)*factorial(n)),n=0..5) = sum((-1)^n/((2*n+1) *factorial(n)),n=0..5); 5 n 0 ( 1) n ( 2 n 1) n! 31049 41580 evalf(%); .7467291967=.7467291967 and finally... 1 So we get that e ( x2 ) dx .747 to the 0 nearest thousandth. Again, according to Maple, the actual answer (to 10 places) is evalf(int(exp(-x^2),x=0..1)); .74669241330 Try this... Sum the first four nonzero terms to approximate 1 cos( 0 x )dx A. 0.7635 B. 0.5637 C. 0.3567 D. 0.6357 E. 0.6735

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