barcode in ssrs report POWER SERIES 11.39. Prove that both the power series in .NET framework

Creator QR in .NET framework POWER SERIES 11.39. Prove that both the power series

POWER SERIES 11.39. Prove that both the power series
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an xn and the corresponding series of derivatives
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nan xn 1
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have the same radius of convergence.
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Let R > 0 be the radius of convergence of an xn . Let 0 < jx0 j < R. Then, as in Problem 11.33, we can 1 choose N as that jan j < for n > N. jx0 jn Thus, the terms of the series jnan xn 1 j njan jjxjn 1 can for n > N be made less than corresponding jxjn 1 terms of the series n , which converges, by the ratio test, for jxj < jx0 j < R. jx0 jn Hence, nan xn 1 converges absolutely for all points x0 (no matter how close jx0 j is to R). If, however, jxj > R, lim an xn 6 0 and thus lim nan xn 1 6 0, so that nan xn 1 does not converge.
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Thus, R is the radius of convergence of nan xn 1 . Note that the series of derivatives may or may not converge for values of x such that jxj R.
11.40. Illustrate Problem 11.39 by using the series
1 X xn . n2 3n n 1
     u  xn 1 n2 3n  n2 jxj   lim  n 1  lim  jxj lim  u  n!1 n 1 2 3n 1 n  n!1 n!1 3 x  3 n 1 2 n so that the series converges for jxj < 3. convergence is 3 @ x @ 3. The series of derivatives is At x 3 the series also converges, so that the interval of
1 1 X nxn 1 X xn 1 2 n n 3n n 3 n 1 n 1
By Problem 11.25(a) this has the interval of convergence 3 @ x < 3. The two series have the same radius of convergence, i.e., R 3, although they do not have the same interval of convergence. Note that the result of Problem 11.39 can also be proved by the ratio test if this test is applicable. The proof given there, however, applies even when the test is not applicable, as in the series of Problem 11.22.
11.41. Prove that in any interval within its interval of convergence a power series a represents a continuous function, say, f x , b can be integrated term by term to yield the integral of f x , c can be di erentiated term by term to yield the derivative of f x .
We consider the power series an xn , although analogous results hold for an x a n . (a) This follows from Problem 11.33 and 11.34, and the fact that each term an xn of the series is continuous. (b) This follows from Problems 11.33 and 11.35, and the fact that each term an xn of the series is continuous and thus integrable. (c) From Problem 11.39, the series of derivatives of a power series always converges within the interval of convergence of the original power series and therefore is uniformly convergent within this interval. Thus, the required result follows from Problems 11.33 and 11.36.
If a power series converges at one (or both) end points of the interval of convergence, it is possible to establish (a) and (b) to include the end point (or end points). See Problem 11.42.
11.42. Prove Abel s theroem that if a power series converges at an end point of its interval of convergence, then the interval of uniform convergence includes this end point.
For simplicity in the proof, we assume the power series to be
1 X k 0
ak xk with the end point of its interval Then we must show that the
of convergence at x 1, so that the series surely converges for 0 @ x @ 1. series converges uniformly in this interval. Let Rn x an xn an 1 xn 1 an 2 xn 2 ;
Rn an an 1 an 2
To prove the required result we must show that given any  > 0, we can nd N such that jRn x j <  for all n > N, where N is independent of the particular x in 0 @ x @ 1.
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