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

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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? 

 

Calculation:
Steps
Description
Σ
00
2n
1
1
(x – 3)2n-1
(х - 3) (х — 3)2»
(x – 3) \x -
n=1
00
Using the Fact : Er" =
we get...
n=0
Rewriting the power series as a
1
Σ
(x – 3)2
function using the fact that
(х-3)
(х-3)
n=0
(r-32
(r-32
(1-3)2
(r-3)?
(r-3)²–1
(x-3)
(r-3)²–1
Σ
x-3
(r-32-1
x2-6x+8
n=0
(r-3)2
1
The function for power series >
x-3
is
x2-6x+8
n=1 (x – 3)2n-1
x-3
→ f(x) =
x2-6x+8
||
Transcribed Image Text:Calculation: Steps Description Σ 00 2n 1 1 (x – 3)2n-1 (х - 3) (х — 3)2» (x – 3) \x - n=1 00 Using the Fact : Er" = we get... n=0 Rewriting the power series as a 1 Σ (x – 3)2 function using the fact that (х-3) (х-3) n=0 (r-32 (r-32 (1-3)2 (r-3)? (r-3)²–1 (x-3) (r-3)²–1 Σ x-3 (r-32-1 x2-6x+8 n=0 (r-3)2 1 The function for power series > x-3 is x2-6x+8 n=1 (x – 3)2n-1 x-3 → f(x) = x2-6x+8 ||
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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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