NAME____________________________________ SCHOOL__________________________________ TEST 3 Monday, April 14, 2008 Fax your answers to: Irvin Roy Hentzel at Fax 515-294-5454 Snail Mail your answers to: Irvin Roy Hentzel Department of Mathematics 432 Carver Hall Iowa State University Ames, Iowa 50011-2064 1. (i) Write the power series for each of these. (ii) Give the radius of convergence. (iii) Give an error bound for truncating the series at degree 7 on the interval ( -9/10, 9/10 ). 1 (a) -------- 1-x 2 3 n (i) 1+x+x +x + ... + x + ... (ii) R = 1 8 8 x (0.9) - (iii) Error = ------ <= ------- 1-r 0.1 (b) ln(1+x) 2 3 4 5 6 7 n+1 n (i) x - x /2 + x /3 - x /4 + x /5 + x /6 - x /7 ... (-1) x /n (ii) R = 1 (iii) You cannot use the alternating series estimate because when x is negative, the series is not alternating. We cannot use the alternating series error formula because the series is not alternating for negative x. f(x) = ln(1+x) f'(x) = (1+x)^(-1) f"(x) = -(1+x)^(-2) f"'(x) = (-1)(-2) (1+x)^(-3) f""(x) = (-1)(-2)(-3) (1+x)^(-4) f""'(x) = (-1)(-2)(-3)(-4) (1+x)^(-5) f"""(x) = (-1)(-2)(-3)(-4)(-5) (1+x)^(-6) f"""'(x) = (-1)(-2)(-3)(-4)(-5)(-6) (1+x)^(-7) f""""(x) = (-1)(-2)(-3)(-4)(-5)(-6)(-7) (1+x)^(-8) 8 7! 1 (0.9) (iii) Error <= ------------------- 8 8! (0.1) c) ArcTan[x] dy/dx = 1/(1+x^2) 2 4 6 8 n 2n dy/dx = 1 - x + x - x + x + ... + (-1) x 3 5 7 9 n 2n+1 (i) ArcTan[x] = x - x /3 + x /5 - x /7 + x /9 ... + (-1) x /(2n+1) ... (ii) R = 1 9 9 (0.9) (iii) Error < x /9 = --------- 9 x (d) e 2 3 4 5 6 7 n (i) 1 + x + x /2 + x /3! + x /4! + x /5! + x /6! + x /7! ... + x / n! + ... (ii) R = Infinity 8 0.9 c x e 8 (iii) Error <= e ------- < -------- (0.9) 8! 8! (e) Sin[x] 3 5 7 n 2n+1 (i) x - x /3! + x /5! - x /7! + ... (-1) x /(2n+1)! + ... (ii) R = Infinity 9 9 x (0.9) (iii) Error < ------- = ----- 9! 9! (g) Sinh[x] 3 5 7 2n+1 (i) x + x /3! + x /5! + x /7! + ... + x /(2n+1)! + ... (ii) R = Infinity 8 8 x (0.9) (iii) Error < Cosh[c] ------ <= Cosh[0.9] ------- 8! 8! OR 9 9 x (0.9) Error < Cosh[c] ------ <= Cosh[0.9] ------- 9! 9! 2 2. (a) Write the power series for y = Cos[ x ] . (b) What is its radius of convergence. (c) What is the error by truncating the series at degree 8 on the interval [-1,1]; 2 4 6 n 2n Cos[x] = 1 - x /2 + x /4! - x /6! .... + (-1) x /(2n)! + ... 2 4 8 n 4n (a) Cos[ x ] = 1 - x /2 + x /4! + (-1) x / (2n)! + ... (b) R = Infinity 8 (c) error <= x /6! < 1/6! 3. (a) Write the power series for y = Cos( x ). (b) What is its radius of convergence. (c) What degree polynomial is necessary to have an accuracy of 0.00001 on the interval (-3,3) ? 2 4 6 8 n 2n (i) 1 - x /2 + x /4! - x /6! + x /8! ... +(-1) x /(2n)! + ... <--------------Terms are Decreasing-----------------------> (ii) R = Infinity n = 1 2 3 4 5 6 7 n+1 = 2 3 4 5 6 7 8 2(n+1) = 4 6 8 10 12 14 16 4 6 8 10 12 14 16 3^(2n+1) 3 3 3 3 3 3 3 -------- ------- ---- ---- --- ---- ----- ----- (2n+1)! 4! 6! 8! 10! 12! 14! 16! -6 3.3 1.0 0.1 0.01 0.001 0.00005 2.05741 10 n = 7 So the degree of the polynomial is 14 <====================== 4. Find the radius of convergence of n Infinity 2 n SUM ------ x n=1 n 1+5 a n+1 n n n+1 2 n+1 1+5 1+5 ----- = ------ x * ----- = 2 x ---- = 2/5 x a n+1 n n n+1 n 1+5 2 x 1+5 R = 5/2 5. Prove that the series converges Infinity Sin[n] SUM -------- n=1 2 n | | Infinity | Sin[n] | Infinity 1 SUM |--------| <= SUM ----- n=1 | 2 | n=1 2 | n | n This is a p-series, p=2 and it converges. The original series is absolutely convergent by the comparison test. The original series converges by the absolute convergence theorem. 6. Prove that the series diverges Infinity 1 SUM ---------- n=2 n Log[n] 1 ---------- decreases to zero n Log[n] _ x->Infinity Infinity 1 | INT ---------- = Log[ Log[x] ] | = + Infinity n=2 x Log[x] _| x=2 By the Integral test, the original series diverges. th 7. State and prove the n term test for divergence. If Sum a converges, then Limit a = 0. n n Proof: Suppose Limit S = L n->Infinity n Limit a = Limit S - Limit S = L - L = 0 n n+1 n th 8. State and prove the n term test for convergence of an alternating series. If a series a decreases to zero, then n Infinity n SUM (-1) a converges n=0 n S = (a -a )+(a - a )+(a - a ) ... + a is increasing 2n 0 1 2 3 4 5 2n S = a -(a -a )-(a -a )-(a -a )... - (a - a )is bounded above. 2n 0 1 2 3 4 5 6 2n-1 2n Therefore S is increasing and bounded above so it converges. 2n If Limit S = L then n->Infinity 2n Limit S = Limit S + a = L + 0 n->Infinity 2n+1 n->Infinity 2n 2n+1 Thus the even sums and the odd sums converge to L so the series converges to L. 9. State and prove the integral test. If f is a positive decreasing function, then Infinity Infinity INT f(x) dx converges <===> SUM f(n) converges x=0 n=0 Proof Infinity Infinity Infinity Sum f(n) <= INT f(x) dx <= SUM f(n) n=1 x=0 n=0 Infinity Infinity If INT f(x) dx converges, then SUM f(n) x=0 n=1 Is positive and bounded and so converges. Infinity Infinity If SUM converges, then INT f(x) dx is bounded and converges. n=0 x=0 10. State the ratio test. For a positive term series a n+1 If Limit -------- = r, n->Infinity a n then if r < 1 the series converges. if r > 1 the series diverges. if r = 1 the theorem is inclusive.