# E in 17 22 both of these diverge by the divergence

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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.

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