Math 1B Section 112 Quiz #9Thursday, 25 October 2007GSI: Theo Johnson-Freydhttp://math.berkeley.edu/∼theojfName:1. (3 pts) Write∞Xn=1xnn2+ n=x2+x26+x312+x420+x530+x642+ . . .in terms of elementary functions. (Hint: Partial fractions) For what values of x isyour solution justified?This problem was a worksheet problem that I ran out of space for. I was interestedin seeing how folks would solve it. Ultimately I decided to make the problem out of 2points.We can start by finding the interval of convergence (if the series doesn’t converge, thenany manipulations we do won’t be justified). By the ratio test, this series convergesiflimn→∞xn+1(n + 1)2+ (n + 1)n2+ nxn= |x| < 1Moreover, by comparison withP1/n2, this series converges absolutely at the end-points ±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?1n2+ n=1n−1n + 1by partial fractions∞Xn=1xnn2+ n=∞Xn=1xnn−∞Xn=1xnn + 1=Zx0∞Xn=1tndt −1x∞Xn=1xn+1n + 1=Zx0dt1 − t−1x∞X2xnn= − ln(1 − x) −1x −x +∞X1xnn!1= − ln(1 − x) + 1 +1xln(1 − x)This argument is justified whenever everything is defined and all the sub-seriesconverge. I.e. not at x = 1 or x = 0. But we can interpret − ln(1−x)+1+1xln(1−x)as being 0 when x = 0 (this is the limit, in any case), and the x = 1 end-point mustbe correct by continuity.22. (3 pts) Find the Taylor series expansion of sin(x) centered at c = π/2. What is theinterval of convergence for this series?There is a sneaky way to do this problem:sin(x) = cos (x − π/2) =∞Xn=0(−1)n(2n)!(x − π/2)2n, which converges everywhere.But here’s the non-sneaky way:n sin(n)(x) sin(n)(π2)0 sin(x) 11 cos(x) 02 − sin(x) −13 − cos(x) 04 sin(x) 15 cos(x) 06 − sin(x) −17 − cos(x) 0.........2n (−1)n2n + 1 0.........Thus, the Taylor series for sin(x) centered at π/2 is∞Xn=0(−1)n(2n)!(x − π/2)2nand the ratio test gives that this converges everywhere.33. (4 pts) For the following power series(a) find the general nth term (i.e. write it asP∞n=0(something)(b) find the radius of convergence(c) check whether the series converges at the endpointsso that you can determine for which x the series• converges absolutely• converges conditionally• diverges.14+2x9+3x216+4x325+5x436+6x549+ . . .We recognize the pattern as∞Xn=0n + 1(n + 2)2xn. Then the ratio test giveslimn→∞(n + 2)xn+1(n + 3)2×(n + 2)2(n + 1)xn= limn→∞(n + 2)3(n + 1)(n + 3)2|x| = 1 · |x|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 alternatingseries test. At x = 1 the series diverges (for instance, by limit comparison test withP1n). Thus, the convergence at x = −1 must be conditional
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