06OperationsPowerSeriesAMS

06OperationsPowerSeriesAMS - Operations on Power Series...

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Unformatted text preview: Operations on Power Series Let’s start with 2 power series about x0 : ∞ f ( x) = n=0 ∞ a n ( x − x0 ) n = a 0 + a 1 ( x − x0 ) 1 + a 2 ( x − x0 ) 2 + a 3 ( x − x0 ) 3 + . . . b n ( x − x0 ) n = b 0 + b 1 ( x − x0 ) 1 + b 2 ( x − x0 ) 2 + b 3 ( x − x0 ) 3 + . . . n=0 g ( x) = each of which converge absolutely for x ∈ (x0 − R , x0 + R). c ∈ R = (−∞, +∞) , x ∈ ( x0 − R , x 0 + R ) , def Let: m ∈ N = {1, 2, 3, . . . } β ∈ ( x0 − R , x0 + R ) def α ∈ ( x 0 − R , x0 + R ) , Then (note we excluded the endpoints of (x0 − R , x0 + R), ie. we excluded x = x0 ± R since things sometimes don’t hold at the endpoints): ∞ f ( x) + g ( x) = (∗) n=0 ∞ ( a n + b n ) ( x − x0 ) n ( a n − b n ) ( x − x0 ) n n=0 ∞ f ( x) − g ( x) = (∗) c f ( x) = ( x − x0 ) ∞ m (∗) n=0 ∞ c a n ( x − x0 ) n ∞ f ( x) = (∗) n=0 ∞ a n ( x − x0 ) ( x − x0 ) Dx (an (x − x0 )n ) = n=0 ∞ m n = n=0 ∞ an (x − x0 )m+n Dx [f (x)] = Dx n=0 a n ( x − x0 ) n So (∗) = nan (x − x0 )n−1 n=0 Dx [f (x)] = n=1 ∞ nan (x − x0 )n−1 β β β ∞ f (t) dt = α α n=0 a n ( t − x0 ) n β dt = (∗) n=0 ∞ (an (t − x0 )n ) dt = α an (t − x0 )n+1 n+1 n=0 ∞ t=β t=α So α f (t) dt = an (β − x0 )n+1 − (α − x0 )n+1 n+1 n=0 Furthermore, f (x) · g (x) is what you think it should be for x ∈ (−R , R). If b0 = 0, (x) then f (x) is what you think it should be but for only x sufficiently small enough. g ...
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