1 1 2n Σ (x – 3)2n-1 (х — 3) (х — 3)2я (x – 3) х — 3 n=| Σ Using the Fact : x" = , we get.. n=0 Σ Rewriting the power series as a function using the fact that (x – 3)2 (x-3y? n=0 (r-32 (r-32 (r-32 (x-3)? (x-3)²–1 Ex" = Σ (х-3) r-3 %3D %3D (r-32-1 (х-3) (x-3)?-1 x2-6x+8 n=0 (r-32 1 The function for power series is 3 x2-6x+8 n=I (x – 3)2n-1 → f(x) = x-3 x2-6x+8
I have done a bunch of these of problems, but this one step looks wrong in the solution for this problem. https://www.bartleby.com/solution-answer/chapter-62-problem-71e-calculus-volume-2-17th-edition/9781938168062/in-the-following-exercises-express-each-series-as-a-rational-function-71-n11-x3-2n1/b209f360-2097-11e9-8385-02ee952b546e
On the very first step when I distribute the (x-3)^(2n-1) exponent in the denominator to be (x-3)^2n and (x-3)^-1, the (x-3)^-1, should flip up to the numerator (not stay in the denominator as (x-3)), but it doesn't according to the answer key. And the answer is right but that first step LOOKS wrong. Can you help explain that exponent distribution?
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This solution was very helpful. I see the part I was missing was shifting the index from n=1 to n=0, so that it is possible to create the rational function in the form a/(1-r). Is there an easy procedure to go from n=1 to n=0 without working out the terms to see what the new sum is? I have seen the generalized procedure for shifting from any n to any other n, but is there a quick way to go from n=1 to n=0 for a power series?
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