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Unformatted text preview: 1 , 7 .
22 1
Now at x = − 2 the series becomes Solution: this is the same power series ∞
∞
n=1 n2
(2x − 3)n =
n
4 ∞
n=0 n2
3
x−
2n
2 n
n=1 we looked at in Example 3. There we saw that while at x = 7
2 R = 2, so the series converges for x in the (−1)n n2 , the series becomes
∞ n2 . interval n=1 3
3
− 2, + 2
2
2 i.e., in 17
,
−,
22 Both of these diverge by the Divergence test.
Thus the interval of convergence of the series
is − 1 ,
2 but diverges for x outside 7
2 . Power series representations of many functions follow once we know representations of a few
functions like the geometric series. We often use diﬀerentiation or integration, as in the next lecture,
but sometimes simple algebra is all that is required. Example 5: ﬁnd a power series So we now simply replace y by −y/3 in the previous representation of the function
f (y ) = representation. Thus y2
.
y+3 y2 1
3+y = y2
y y2
(−y )n
1 − + 2 + ...+
+ ... .
3
33
3n Consequently,
Solution: we know that
1 + y + y2 + y3 + y4 + . . . + yn + . . . = 1
1−y But
y2
1
= y2
3+y
3+y f (y ) = y2
y2
=
y+3
3 1
1 − (−y/3) n=0 . n=0 (−1)n ∞ =
y2
=
3 ∞ (−1)n y n+2
.
3n+1 y
3 n...
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This note was uploaded on 06/05/2011 for the course MATH 408 D taught by Professor Gilbert during the Spring '11 term at University of Texas at Austin.
 Spring '11
 Gilbert
 Power Series

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