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Unformatted text preview: = -11 x(-1,1)! = ex(-∞,∞)nx2n+12n+1! = sin(x)(-∞,∞)nx2n2n! = cos(x)(-∞,∞)nx2n+1(2n+1) = tan-1(x)[-1,1]nn = -ln(1-x)[0,2)nx2n22n(n!)2 = J(x)Bessel function (-∞,∞)ddxn=0 cnxn∞= = ∞-n 0 ncnxn 1n=0 cnxn∞= = ∞++n 0 cnn 1xn 1Error bound for a series:0 S-S≤n(n to oo) f(x)dx ; S≤∫n S S≤ ≤n+ (n to oo) f(x)dx∫Comparison test for series Alternating Series tests for convergence Suppose ansatisfies: 1. An>0 2. An+1<an 3. Lim (n->oo) an= 0Then (-1)∑nanconvergesAlternating Series Estimation Theorem If S = = ∞(- )n 11n-1bnis the sum of an alternating series that satisfies 1. 0 b≤n+1b≤n2. →∞limnbn= 0 then Rn= S – Snb≤n+1Difference Equation pn+1= kpn(1-pn) Constant solutions: pn=0 and pn= (k-1)/kLimit Comparison Test Suppose a∑nand b∑nare series with positive terms. If →∞(limnan/bn) = c, where c is a finite number, then either both series converge or both series diverge....
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This note was uploaded on 04/01/2008 for the course MATH 156 taught by Professor Diniz-behn during the Fall '07 term at University of Michigan.
- Fall '07