121616949-math.275

# 121616949-math.275 - 10.9 Calculus with Power Series 261...

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10.9 Calculus with Power Series 261 because this is just a geometric series with x replaced by ( x +2) / 3. Multiplying both sides by 1 / 3 gives X n =0 ( x + 2) n 3 n +1 = 1 1 - x , the same function as before. For what values of x does this series converge? Since it is a geometric series, we know that it converges when | x + 2 | / 3 < 1 | x + 2 | < 3 - 3 < x + 2 < 3 - 5 < x < 1 . So we have a series representation for 1 / (1 - x ) that works on a larger interval than before, at the expense of a somewhat more complicated series. The endpoints of the interval of convergence now are - 5 and 1, but note that they can be more compactly described as - 2 ± 3. Again we say that 3 is the radius of convergence, and we now say that the series is centered at
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Unformatted text preview: A power series centered at a has the form ∞ X n =0 a n ( x-a ) n , with the understanding that a n may depend on n but not on x . Exercises Find the radius and interval of convergence for each series. 1. ∞ X n =0 nx n ⇒ 2. ∞ X n =0 x n n ! ⇒ 3. ∞ X n =1 n ! n n x n ⇒ 4. ∞ X n =1 n ! n n ( x-2) n ⇒ 5. ∞ X n =1 ( n !) 2 n n ( x-2) n ⇒ 6. ∞ X n =1 ( x + 5) n n ( n + 1) ⇒ 10.9 Calculus with Power Series Now we know that some functions can be expressed as power series, which look like inﬁnite polynomials. Since calculus, that is, computation of derivatives and antiderivatives, is easy...
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• Fall '07
• JonathanRogawski

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