Math 1B Section 112 Quiz 9 Thursday 25 October 2007 GSI Theo Johnson Freyd http math berkeley edu theojf Name 1 3 pts Write X n 1 xn x x2 x3 x4 x5 x6 n 2 6 12 20 30 42 n2 in terms of elementary functions Hint Partial fractions For what values of x is your solution justified This problem was a worksheet problem that I ran out of space for I was interested in seeing how folks would solve it Ultimately I decided to make the problem out of 2 points We can start by finding the interval of convergence if the series doesn t converge then any manipulations we do won t be justified By the ratio test this series converges if n2 n xn 1 lim x 1 n n 1 2 n 1 xn P Moreover by comparison with 1 n2 this series converges absolutely at the endpoints 1 If you got this far I gave 2 points 1 for radius of convergence How do you write the series in terms of nice functions 1 n X xn n2 n n2 n 1 1 1 by partial fractions n n 1 X xn X xn n n 1 n 1 n 1 Z xX 1 X xn 1 tn dt x n 1 0 n 1 n 1 Z x dt 1 X xn x n 0 1 t 2 1 ln 1 x x 1 x X xn 1 n ln 1 x 1 1 ln 1 x x This argument is justified whenever everything is defined and all the sub series converge I e not at x 1 or x 0 But we can interpret ln 1 x 1 x1 ln 1 x as being 0 when x 0 this is the limit in any case and the x 1 end point must be correct by continuity 2 2 3 pts Find the Taylor series expansion of sin x centered at c 2 What is the interval of convergence for this series There is a sneaky way to do this problem sin x cos x 2 X 1 n n 0 2n x 2 2n which converges everywhere But here s the non sneaky way sin n x sin n 2 sin x 1 cos x 0 sin x 1 cos x 0 sin x 1 cos x 0 sin x 1 cos x 0 2n 1 n 2n 1 0 n 0 1 2 3 4 5 6 7 Thus the Taylor series for sin x centered at 2 is X 1 n n 0 2n x 2 2n and the ratio test gives that this converges everywhere 3 3 4 pts For the following power series a find the general nth term i e write it as P n 0 something b find the radius of convergence c check whether the series converges at the endpoints so that you can determine for which x the series converges absolutely converges conditionally diverges 1 2x 3x2 4x3 5x4 6x5 4 9 16 25 36 49 X n 1 n We recognize the pattern as x Then the ratio test gives n 2 2 n 0 lim n n 2 2 n 2 xn 1 n 2 3 x 1 x lim n n 1 n 3 2 n 3 2 n 1 xn So this series converges absolutely when x 1 and diverges when x 1 Checking endpoints we see that when x 1 the series converges by the alternating series test At x 1 the series diverges for instance by limit comparison test with P1 n Thus the convergence at x 1 must be conditional 4
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