Morphing Geometric Series Into Power Series Suppose we take the - PDF document

Morphing Geometric Series Into Power Series Suppose we take the geometric series 1 + r + r 2 + r 3 + . . . , which we 1 know converges to 1 − r for | r | < 1, and replace r by x : 1 1 + x + x 2 + x 3 + . . . converges to 1 − x for | x | < 1. We haven’t really changed anything, but 1 + x + x 2 + x 3 + . . . looks a little like a polynomial. It’s an example of a power series . Power Series ∞ a n ( x − c ) n is called a � Definition 1 (Power Series) . An expression n =0 power series centered at c . ∞ ∞ � � a n ( x − c ) n , we really mean a 0 + a n ( x − c ) n , Note: When we write n =0 n =1 since a 0 ( x − c ) 0 isn’t defined when x = c , but it’s more convenient to just write the sum starting from n = 0 . Power Series We will mostly consider power series centered at 0, written in the form

2 So assume the series converges for x = x 0 . It follows that a n x n 0 → 0 as n → ∞ , since otherwise the series could not converge. It follows that there is some bound B such that | a n x n 0 | < B for all n . Proof (Continued) Now let | x | < | x 0 | and examine the magnitude of a n x n . | a n x n | = n n � � � � x x � � � � | a n x n 0 | · ≤ B . � � � � x 0 x 0 � � � � n ∞ � � � � x x � � � � � Since B is a geometric series with common ratio � < 1, � � � � x 0 x 0 � � � n =0 ∞ � it must converge. By the Comparison Test, | a n x n | < ∞ , and thus n =0 ∞ a n x n is absolutely convergent. � n =0 Finding the Radius of Convergence For most power series, the easiest way to determine the radius of convergence is to use the ratio test. ∞ a n +1 x n +1 � � � � a n +1 � � � � � a n x n , we calculate Given a series � = x � . � � � � a n x n a n � � n =0 We find the limit as n → ∞ and find the values of x for which the limit is less than or equal to 1. One complication that can occur is that some coefficients a n equal 0. In this case, we look at the ratios of the adjacent terms that actually appear. Algebra and Calculus of Power Series Within their intervals of convergence, although possibly not at the endpoints, power series may be added, subtracted, multiplied, differ- entiated and integrated in the natural way. Specifically: ∞ ∞ a n x n and B ( x ) = b n x n in some open interval, � � Let A ( x ) = n =0 n =0 in the sense that the power series converge to A ( x ) and B ( x ) in the interval. Then: Algebra and Calculus of Power Series ∞ � ( a n + b n ) x n • A ( x ) + B ( x ) = n =0 ∞ � • A ( x ) − B ( x ) = ( a n − b n ) x n n =0

3 ∞ � ( � n • A ( x ) B ( x ) = i =0 a i b n − i ) x n n =0 ∞ • A ′ ( x ) = � na n x n − 1 n =1 ∞ a n x n +1 � � • A ( x ) dx = n + 1 + k n =0 Example: ln(1 + x ) Each of the following calculations can be done, based on the prop- erties of power series, whenever | x | < 1: 1 Start with 1 + x + x 2 + x 3 + · · · = 1 − x . 1 Replace x by − x to get 1 − x + x 2 − x 3 + · · · = 1 + x . Integrate to get ( x − x 2 2 + x 3 3 − x 4 4 + . . . ) + k = ln(1 + x ). Plugging in x = 0, we find k = 0 to obtain ln(1+ x ) = x − x 2 2 + x 3 3 − x 4 4 + . . . . The power series also converges (by the Alternating Series Test) for x = 1, giving ln 2 as the sum of the Alternating Harmonic Series : ln 2 = 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + . . . . arctan We can also obtain a power series converging to the arctangent func- 1 tion by starting with the power series 1 − x + x 2 − x 3 + · · · = 1 + x . All the following calculations again hold for | x | < 1. 1 Replacing x by x 2 gives 1 − x 2 + x 4 − x 6 + x 8 − x 10 + · · · = 1 + x 2 . Integrating: ( x − x 3 3 + x 5 5 − x 7 7 + . . . ) + k = arctan x . Plugging in x = 0 yields k = 0, so arctan x = x − x 3 3 + x 5 5 − x 7 7 + . . . for | x | < 1. Calculation of π arctan x = x − x 3 3 + x 5 5 − x 7 7 + . . . for | x | < 1. Here, too, the series actually converges for x = 1 and converges to arctan 1. Since arctan 1 = π 4 , we get the interesting series expansion:

4 π 4 = 1 − 1 3 + 1 5 − 1 7 + 1 9 − 1 11 + . . . . Obtaining a Series Which Converges to a Given Function Suppose we have a function f ( x ) and want to find a power series n =0 a n ( x − c ) n which converges to f ( x � ∞

5 Most of the time, we will center Taylor Series at 0, in which case the ∞ f ( n ) (0) � formula simplifies to T ( x ) = x n . These series are also known n ! n =0 as Maclaurin Series. Taylor Series We have shown that if a power series converges to a function, it must be the Taylor Series. On the other hand, the Taylor Series for a function does not always have to converge to that function although it often will . . . otherwise we wouldn’t bother with them. 1 The series previously shown to converge to 1 − x , ln(1 + x ) and arctan x for | x | < 1 therefore must be the Taylor Series, centered at 0, for those functions. Taylor Series for the Exponential Function Let f ( x ) = exp( x ) = e x . Since all the derivatives of the exponential function are the same, we have f ( n ) ( x ) = e x for all n and thus f ( n ) (0) = e 0 = 1 for all n . ∞ ∞ ∞ f ( n ) (0) x n 1 x n = n ! x n = � � � The Taylor Series is thus T ( x ) = n ! . n ! n =0 n =0 n =0 We will show this converges for all x and it converges to e x for all x , so we may write e x = 1 + x + x 2 2! + x 3 3! + x 4 4! + x 5 5! + . . . . Radius of Convergence To show where the Taylor Series for e x converges, we use the Ratio Test. ( n + 1) st term x n +1 / ( n + 1)! � � � � � � � � � = n th term � � � � x n /n ! � � � � x n +1 � � � ( n + 1)! · n ! x � � � � = � = � → 0 � � � � x n n + 1 � � as n → ∞ . By the Ratio Test, it follows that the series converges for all x . Another Derivation of the Taylor Series There’s a second way of coming up with the formula for a Taylor Series which naturally leads to an estimate for the error if one estimates the value of the function by a partial sum of the Taylor Series. Definition 3 (Taylor Polynomial) . The n th partial sum of a Taylor Series T ( x ) is denoted by T n ( x ) and is called a Taylor Polynomial.

6 � x 0 f ′ ( t ) dt . Using the Fundamental Theorem of Calculus, Consider x � � x 0 f ′ ( t ) dt = f ( t ) � = f ( x ) − f (0). � � 0 � x 0 f ′ ( t ) dt . It follows we can write f ( x ) = f (0) + If we repeatedly integrate by parts, we can obtain the terms of the Taylor Series. Integrating By Parts � x 0 f ′ ( t ) dt f ( x ) = f (0) + uv ′ dt = uv − u ′ v dt , taking � � We’ll use the version u = f ′ ( t ), v ′ = 1, so u ′ = f ′ ′ ( t ), v = t − x . Note the trick here: We could take v = t , but it turns out that doesn’t work well, while taking v = t − x works very nicely, and we obtain: � x � x x � f ′ ( t ) dt = f ′ ( t )( t − x ) f ′ ′ ( t )( t − x ) dt � − � � 0 0 0 � x = f ′ (0) x − f ′ ′ ( t )( t − x ) dt, 0 so � x f ( x ) = f (0) + f ′ (0) x − 0 f ′ ′ ( t )( t − x ) dt . Continuing In order to have a sum rather than a difference, we’ll rewrite that as � x f ( x ) = f (0) + f ′ (0) x + 0 f ′ ′ ( t )( x − t ) dt . Now integrate by parts again, taking u = f ′ ′ ( t ), v ′ = x − t , so u ′ = f ′ ′ ′ ( t ), v = − ( x − t ) 2 , and 2 � x f ′ ′ ( t )( x − t ) dt 0 � x x − ( x − t ) 2 − ( x − t ) 2 � �� � � = f ′ ′ ( t ) f ′ ′ ′ ( t ) � − dt � 2 2 � 0 0 = f ′ ′ (0) x 2 � x + 1 f ′ ′ ′ ( t )( x − t ) 2 dt 2 2 0 . Continuing We thus have f ( x ) = f (0) + f ′ (0) x + f ′ ′ (0) x 2 + 1 � x 0 f ′ ′ ′ ( t )( x − t ) 2 dt . 2 2 If we carry out another step, we obtain: