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Unformatted text preview: You might be thinking: Hey what about taking derivatives and integrals of 1 1 x , For example we know 1 1 x = 1 (1 x ) 2 and Z 1 1 x dx = ln(1 x ) + C Can we get power series for these new functions? The answer is yes! But first we need a theorem that tells us how to differentiate and integrate power series. Math 9C Summer 2011 (UCR) ProNotes October 25, 2011 1 / 9 Theorem If the power series f ( x ) = ∞ X k =0 a k ( x a ) k has radius of convergence R > 0, then f ( x ) is differentiable on the interval ( a R , a + R ) and I We can differentiate term by term ∞ X k =0 a k ( x a ) k = ∞ X k =0 a k ( x a ) k = ∞ X k =1 ka k ( x a ) k 1 Notice how k starts at 1 in the new series. I We can integrate term by term Z ∞ X k =0 a k ( x a ) k dx = ∞ X k =0 Z a k ( x a ) k dx = ∞ X k =0 a k k + 1 ( x a ) k +1 + C where C is some constant. The radii of convergence for these new power series is also R . Math 9C Summer 2011 (UCR) ProNotes October 25, 2011 2 / 9 Example 3 Find a power series expansion for the function f ( x ) = 1 (1 x ) 2 centered about x = 0....
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 Fall '11
 c
 Derivative, Integrals, Power Series, Taylor Series, Mathematical Series

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