Basically we can differentiate the power series term

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Basically we can differentiate the power series term-by-term as follows. d d x [ f ( x )] = d d x " X n =0 c n ( x - a ) n # = d d x h c 0 + c 1 ( x - a ) + c 2 ( x - a ) 2 + · · · i f 0 ( x ) = X n =0 d d x [ c n ( x - a ) n ] = c 1 + 2 c 2 ( x - a ) + 3 c 3 ( x - a ) 2 + · · · = X n =1 nc n ( x - a ) n - 1 Working backwards we can also integrate term-by-term too. Z f ( x ) d x = Z " X n =0 c n ( x - a ) n # d x = Z h c 0 + c 1 ( x - a ) + c 2 ( x - a ) 2 + · · · i d x = X n =0 Z [ c n ( x - a ) n ] d x = C + c 0 ( x - a ) + c 1 ( x - a ) 2 2 + c 2 ( x - a ) 3 3 + · · · = C + X n =0 c n ( x - a ) n +1 n + 1 Let’s take a simple example for integration. For this example I’d like to introduce two simple power series. The power series for sine is: sin x = x - x 3 3! + x 5 5! - · · · = X n =0 ( - 1) n x 2 n +1 (2 n + 1)! , and the power series for cosine is: cos x = 1 - x 2 2! + x 4 4! - · · · = X n =0 ( - 1) n x 2 n (2 n )! . Both of these series converge for all x . We know that Z cos x d x = sin x + C, 6
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so let’s give it a try using the series instead. Z cos x d x = sin x + C Z 1 - x 2 2! + x 4 4! - · · · d x = x - x 3 3! + x 5 5! - · · · + C Yes, exactly as expected. Now if were differentiate sine we should get cosine. d d x [sin x ] = cos x d d x x - x 3 3! + x 5 5! - · · · = 1 - x 2 2! + x 4 4! - · · · Yes, as expected once again. Here’s another example, but this one is related to 1 1 - x = 1 + x + x 2 + x 3 + x 4 + · · · = X n =0 x n . If we differentiate this series we get d d x 1 1 - x = d d x 1 + x + x 2 + x 3 + x 4 + · · · = d d x " X n =0 x n # 1 (1 - x ) 2 = 1 + 2 x + 3 x 2 + 4 x 3 + · · · = X n =1 nx n - 1 = X n =0 ( n + 1) x n Now we have a series expansion for 1 (1 - x ) 2 = X n =1 nx n - 1 = X n =0 ( n + 1) x n . The interval of convergence is ( - 1 , 1). It might be useful to graph f ( x ) = 1 (1 - x ) 2 , and finite number of terms from the power series, let’s say p ( x ) = 1 + 2 x + 3 x 2 + 4 x 3 + 5 x 4 + 6 x 5 + 7 x 6 + 8 x 9 + 9 x 10 + 10 x 11 , on the same axis to see how they fit each other. We’re expecting a good fit only on the interval ( - 1 , 1). Certainly, taking more terms would create a better fit, but the point here is that we’re getting a fit only on the interval ( - 1 , 1). You might want to play around with graphing f ( x ) and various n ’s for n X i =0 ( i + 1) x i to see what happens as n increases. 7
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-1 0 1 1 2 3 4 5 6 7 Figure 3: Partial graphs of f ( x ) [black], p ( x ) [red]. Not drawn to proper aspect ratio. Tricks abound, and finding power series can be quite elegant, or thuggish. Well, for example, knowing the power series for 1 1 - x = 1 + x + x 2 + x 3 + x 4 + · · · can lead us to the power series for 1 1 + x 2 = 1 1 - ( - x 2 ) = 1 - x 2 + x 4 - x 6 + · · · And since we know (possibly remember) that Z 1 1 + x 2 d x = arctan x + C, we can find a power series for arctangent by integrating the power series. Z 1 1 + x 2 d x = arctan x + C Z ( 1 - x 2 + x 4 - x 6 + · · · ) d x = arctan x + C 1 C 2 + x - x 3 3 + x 5 5 - x 7 7 + · · · = arctan x + C 1 C 3 + x - x 3 3 + x 5 5 - x 7 7 + · · · = arctan x 8
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To find the constant let’s take an easy value for x , let’s say x = 0.
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