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## Example 8.3.9 (b): Power Series Center

The power series 2-n(x-1)n has center c = 1 and radius of convergence r = 2 (confirm). Re-center the series at c = 2. What is the new radius of convergence?

To solve this, we'll make a slight detour: we know that

xn = 1/1-x

is our familiar geometric series. Substituting x = (t-1)/2 on both sides we get:

2-n(t-1)n = 1/1-(t-1)/2 = 2/3-t

Therefore we have a simple expression (on the right) for our original series (on the left). To center something at c=2 we need to get t-2 involved somehow. Thus:

2-n(t-1)n = 2/3-t = 2/3-(t-2+2) = 2/1-(t-2) = 2 (t-2)n

The center of this series is, as requested, c = 2 and the radius of convergence is r = 1 (as is easily confirmed). Geometrically, the regions of convergence overlap as shown below:

As you can see, inside the blue circle we have two power series representations. Of course they should give the same value, as you can confirm in special cases:

• Let x = 2:
• 2-n(x-1)n = 2-n(2-1)n = 2-n(1)n = (1/2)n = 1/1-1/2 = 2
• 2 (x-2)n = 2 (2-2)n = 2 * 1 = 2
• Let x=3/2:
• 2-n(x-1)n = 2-n(3/2-1)n = 2-n(1/2)n = (1/4)n = 1/1-1/4 = 4/3
• 2 (x-2)n = 2 (3/2-2)n = 2 (-1/2)n = 2 1/1+1/2 = 2 * 2/3 = 4/3
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