## 8.3. Series and Power Series

### Example 8.3.9 (b): Power Series Center

The power series

*2*has center^{-n}(x-1)^{n}*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

x^{n}=^{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}= 22 (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}= 4/32 (x-2)^{n}= 2 (3/2-2)^{n}= 2 (-1/2)^{n}= 2^{1}/_{1+1/2}= 2 * 2/3 = 4/3