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MIT 18 01 - Lecture Notes

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� � � 18.01 Calculus Jason Starr Fall 2005 Lecture 31. December 8, 2005 Practice Problems. Course Reader: 7B-4, 7B-6, 7C-1, 7C-5, 7D-1, 7D-2. 1. Power series. Given a real number a and a sequence of real numbers (cn)n≥0, there is an associated expression, called a power series about x = a, ∞cn(x − a)n = c0 + c1(x − a) + c2(x − a)2 + . . . n=0 For every choice of a real number x, the power series gives a usual series. In particular, for the choice x = a, the series has only 1 nonzero term, thus converges to c0. Question. Given a power series, for which real numbers x does the corresponding series absolutely converge? Examples. 1. Consider the power series, ∞30 + 11 x 1 + 22 x 2 + 33 x = n x .+ · · · n n n=1 Of course this converges to 0 for x = 0. But for any x other than 0, the sequence nnxn = (nx)n diverges. Therefore the series does not converge. In other words, the series converges only for x = 0. 2. Consider the power series, ∞1 + x + x 2 = x .+ · · · n n=1 This is a geometric series. From the last lecture, the series converges absolutely for |x| < 1 and diverges if x≥ 1.| | 3. Consider the power series, � 1∞31 + x + x 2/2 + x /3! + · · · = n! x n . n=0�� � � � 18.01 Calculus Jason Starr Fall 2005 The ratio of the nth and (n + 1)st terms in the series is, x n(x n+1/(n + 1)!)/(x /n!) = . n + 1 For fixed x, as n grows, this sequence of ratios converges to 0, which is less than 1. Therefore, by the ratio test, for every choice of x the series converges. These 3 examples illustrate the whole range of p ossibilities. Theorem. Let ∞ cn(x − a)n be a power series about x = a. Exactly one of the following hold. n=0 (i) For every x different from a, the series does not converge absolutely. (ii) There exists a real number R such that the series converges absolutely if x − a < R and| |does not converge absolutely if x − a > R.| | (iii) For every real number x, the series converges absolutely. The real number R occuring in Case (ii) is called the radius of convergence. By convention, in Case (i) the radius of convergence is defined to be R = 0. By convention, in Case (iii) the radius of convergence is defined to be R = ∞. This allows us to replace the original question by a more precise question. Question. Given a power series, what is the radius of convergence? Although there is no single answer to this question, in many interesting cases the ratio or root test gives an answer. 2. Analytic functions. If the radius of convergence R of a power series cn(x − a)n is positive, then the power series defines a function on the interval (a− R, a + R), ∞f(x) = cn(x − a)n . n=0 A function defined in this manner is called an analytic function. This is the real significance of power series: they give important examples of functions that cannot be describ ed in a more direct manner. Analytic functions have nice analytic properties (whence the name). For instance, it is a theorem (proved in 18.100) that an analytic function f(x) is differentiable and the derivative has a power series converging absolutely with the same radius R, ∞ ∞f�(x) = cnn(x − a)n−1 = (m + 1)cm+1(x − a)m . n=0 m=0 We can iterate the theorem, i.e., f�(x) is differentiable and f��(x) has a power series converging absolutely with radius R. Iterating k times, the function f(x) is k-times differentiable and its kth derivative has a power series, ∞f(k)(x) = � (n + k)! cn+k (x − a)n . n! n=0� 18.01 Calculus Jason Starr Fall 2005 In particular, every derivative of f (x) is defined. A function with this property is called infinitely differentiable or smooth. Thus, every analytic function is infinitely differentiable. This is only 1 of many useful properties of analytic functions. Which functions f (x) are analytic functions? By the last paragraph, if f (x) is analytic, then it is infinitely differentiable. Are there other restrictions? Can more than 1 power series about x = a give rise to the same analytic function? To answer both of these questions, consider the analytic function defined by a power series, ∞f (x) = cn(x − a)n . n=0 Plugging in x = a gives the equation, f (a) = c0 + c1(a − a) + c2(a − a)2 = c0 + 0 + 0 + ··· = c0.+ ··· Thus the first coefficient of the power series is simply, c0 = f (a). Moreover, from the power series for the kth derivative, f (k)(a) = k!ck + (k + 1)!/1!ck+1(a − a) + (k + 2)!/2!ck+2(a − a)2 = k!ck + 0 + 0 + ··· = k!ck .+ ··· Solving for ck , the kth coefficient of the power series is, ck = f (k)(a)/k!. Therefore, the power series defining f (x) is, � f (n)(a)∞f (x) = n! (x − a)n . n=0 In particular, this series is unique. This answers the second question. Two absolutely convergent power series about x = a give the same analytic function if and only if the power series are themselves equal (i.e., the corresponding coefficients of the 2 series are equal). Moreover, this gives us alot of information about the first question. For an infinitely differentiable function f (x) defined at a point x = a, there is a very important power series, the Taylor series expansion of f (x) about x = a, �∞ n=0 f (n) (a) n! (x − a)n . If f (x) is analytic, then the Taylor series converges absolutely to f (x). This reduces the original question to 2 new questions. Does the Taylor series have a positive radius of convergence? If so, does the analytic function defined in this way equal the original function f (x)?� 18.01 Calculus Jason Starr Fall 2005 The radius of convergence question is precisely the radius of convergence question posed earlier. As there, the answer can often be found by using the ratio or root tests. The second question is yes in every practical case. There are examples of infinitely differentiable functions where the Taylor series has a positive radius of convergence, but does not converge to the original function. However, every example is somewhat contrived; they rarely come up “in nature”. Just for completeness, here is an example of one of these pathological functions, e−1/x2 , x = 0,f (x) = �0 x = 0 3. Algorithm for computing Taylor series. The method for finding the Taylor series of a function is always the same. For definiteness, consider the Taylor series expansion of f (x) =


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MIT 18 01 - Lecture Notes

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