Calculus II: Integrations and Series

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1 Clculus II: Integrtions nd Series August 7, 200 Integrls Suppose we hve generl function y = f(x) For simplicity, let f(x) > 0 nd f(x) continuous Denote F (x) = re under the grph of f in the intervl [,x] Then we hve, for some vlue z in the intervl [x, x + δ] F (x + δ) F (x) = f(z)δ, or F (x + δ) F (x) = f(z) δ As δ goes to 0, z goes to x, nd we hve F F (x + δ) F (x) (x) = lim δ 0 δ So F is ntiderivtive of f, nd we denote F (x) = x = lim z x f(z) = f(x) f(t)dt This is definite integrl of the function f from to x f is clled the integrnd We lso hve indefinite integrl Tht is, for n rbitrry constnt C, Properties of integrls: f(x) = F (x) + C For ny constnt c nd ny integrble function f(x), cf(x)dx = c f(x)dx 2 For ny two integrble functions f(x) nd g(x), [f(x) ± g(x)]dx = f(x)dx ± g(x)dx

2 2 Some useful results Integrtion by prts Given the existence of ll integrtions, we hve b f(x)g (x)dx = f(x)g(x) b b f (x)g(x)dx L Hospitl Rule If limf(x) = limg(x) = 0 or ±, limf (x)/g (x) exists, x c x c x c then f(x) lim x c g(x) = lim f (x) x c g (x) The Leibnitz s rule It gives formul for differentition of definite integrl whose limits re functions of the differentil vribles, d dz b(z) (z) f(x, z)dx = b(z) (z) 3 Applictions 3 Newton-Rphson method d dz f(x, z)dx+f(b(z), z) d dz b(z) f((z), z) d dz (z) Suppose we wnt to solve for the roots of eqution f(x) = 0 We strt with initil guess x 0 2 Clculte x i+ = x i f(x i) f (x i, for i = 0,, ) 3 Stop the itertion until x i+ x i < ɛ, where ɛ is prespecified smll positive number, sy 0 4 Otherwise go bck to Step 2 Exercise: Solve for x: f(x) = x 3 2x x 220 = 0 Hint: Check tht f(2) = 26 nd f(3) = 23 Use initil vlue x 0 = 2 32 Locl extrem If function f(x) hs either locl mximum or locl minimum t some vlue x 0, nd f(x) is differentible t x 0, then the tngent line must be horizontl, or f (x 0 ) = 0 We cll x 0 locl extrem if it is either locl mximum or locl minimum On the other hnd, not ll solutions of f (x) = 0 is locl extrem Properties: 2

3 If f(x) hs locl extrem t x = x 0 nd f(x) is differentible t x 0, then f (x 0 ) = 0 2 If f (x) > 0 for every x on some intervl I, then f(x) is incresing on the intervl 3 If f (x) < 0 for every x on some intervl I, then f(x) is decresing on the intervl 4 If f (x) = 0 for every x on some intervl I, then f(x) is constnt on the intervl 5 If f (x) > 0 for ll x on some intervl I, then f(x) is concve up on I 6 If f (x) < 0 for ll x on some intervl I, then f(x) is concve down on I 7 If f (x 0 ) = 0, f (x 0 ) < 0 nd f (x) is continuous in region round x = x 0, then x = x 0 is locl mximum 8 If f (x 0 ) = 0, f (x 0 ) > 0 nd f (x) is continuous in region round x = x 0, then x = x 0 is locl minimum 9 If f (x 0 ) = 0, f (x 0 ) = 0 nd f (x) is continuous in region round x = x 0, then x = x 0 cn be locl mximum, locl minimum, or neither Exercise: Study f(x) = x 3 2x x 220 = 0 Clculte the first nd second order derivtive t x = 5, 6, 7, 8, 9 Describe wht you find 33 Tylor series Derivtive of the derivtive is clled the second-order derivtive f (2) (x) = f (x) = [f (x)] = d ( ) dy dx dx Similrly, we cn define the n-th order derivtive nd denote it by f (n) The Tylor series of function f(x), which is infinitely differentible in neighborhood of, is power series: f(x) f() + f ()(x ) + f () 2! 3 (x ) 2 + f (3) () (x ) 3 + 3!

4 Some importnt series: e x = log( + x) = x n n! = + x + x2 2! + x3 3! + n+ xn ( ) n for < x n=0 n= x m+ m m = x n for x, m =, 2, n=0 x = x n for x < n=0 x m x = n=m x n for x <, m =, 2, 4 Sequences nd limits A sequence is list of numbers written in specific order This list my hve finite, or infinite number of terms We denote n infinite sequence by {, 2,, n, n+, }, or { n } We sy tht lim n = L if for every ɛ > 0 there is n integer N such tht n L < ɛ whenever n > N We sy tht lim n = if for every M > 0 there is n integer N such tht n > M whenever n > N If lim n exists nd is finite, we sy the sequence is convergent If lim does not exist or is infinite, we sy the sequence diverges Properties: lim ( n ± b n ) = lim n ± lim b n 2 lim c n = c lim n 4 n

5 3 lim ( n b n ) = lim n lim b n 4 lim n bn = lim n lim bn if lim b n 0 Squeeze theorem If n c n b n for ll n > N nd lim n = lim b n = L, then lim c n = L Exercise Consider the sequence {r n } Prove tht it converges if < r nd diverges for ll other vlues of r Also { 0 if < r < lim rn = if r = We hve the following terminology: If n > n+ for every n, we sy the sequence is decresing 2 If n < n+ for every n, we sy the sequence is incresing 3 If { n } is either incresing or decresing, then we sy the sequence is monotonic 4 If there exists m such tht n > m for ll n, we sy the sequence is bounded below, nd m is lower bound of the sequence 5 If there exists M such tht n < M for ll n, we sy the sequence is bounded bove, nd M is upper bound of the sequence 6 If the sequence is both bounded bove nd bounded below, we sy the sequence is bounded Theorem If { n } is bounded nd monotonic, then { n } is convergent 5

6 5 Series 5 Introduction nd some tests We strt with sequence { n } Define the following: s = s 2 = + 2 s n = n = n i= i The s n re clled prtil sums nd they form sequence {s n } We wnt to look t the limit of the sequence of prtil sums {s n }, or lim s n = i= i If this limit exists nd is finite, we sy the series i= i convergent Properties If n nd b n re both convergent series, then c n is convergent, nd c n = c n 2 ( n ± b n ) is convergent, nd ( n ± b n ) = n ± b n Exercise Index shift Write n 2 n= s series tht strt t n = 3 3 n+ Exercise Determine whether the following series re divergent or convergent n= n 2 n=2 n 2 3 n= ( )n 4 n= 3 (n ) Theorem If n converges, then lim n = 0 Divergence test If lim n 0, then n diverges 6

7 52 Common series nd more tests Geometric series n= rn This series is convergent when < r <, nd it converges to r Telescoping series n= n 2 +3n+2 Hrmonic series n= This series is divergent We notice n n= n > On the other hnd, n= n= n = + 2 n=2 n 2 x dx = ln x = is convergent We notice x dx = 2 x n 2 < + = 2 Integrl test Suppose f(x) is continuous, positive nd decresing on the intervl [k, ), nd f(n) = n Then If f(x)dx is convergent, so is k n=k n 2 If f(x)dx is divergent, so is k n=k n The p series test For k > 0, n=k p n p converges if p > nd diverges if Comprison test Suppose we hve two series n nd b n, n > 0, b n > 0 for ll n Also n b n for ll n Then If b n is convergent, so is n 2 If n is divergent, so is b n Limit comprison test Suppose we hve two series n nd b n, n > 0, b n > 0 for ll n Define n c = lim b n If c is positive nd finite, then either both series converge or both series diverge Alternting series test Suppose we hve series n, nd either n = ( ) n b n or n = ( ) n+ b n where b n 0 for ll n If 7

8 lim b n = 0, nd 2 {b n } is decresing sequence, then series { n } is convergent Rtio test Suppose we hve series n Define L = lim n+ n Then if L >, n is divergent 2 if L <, n is convergent Root test Suppose we hve series n Define Then if L >, n is divergent 2 if L <, n is convergent Fct lim n /n = L = lim n /n 53 Strtegies for convergence tests With quick glnce does it look like the series terms do not converge to zero in the limit, ie does lim n 0? If so, the series is divergent 2 Is the series p-series ( n= n p ) or geometric series ( n= rn )? If so use the fct tht p-series will only converge if p > nd geometric series will only converge if r < Remember s well tht often some lgebric mnipultion is required to get geometric series into the correct form 3 Is the series similr to p-series or geometric series? If so, try the Comprison Test 8

9 4 Is the series rtionl expression involving only polynomils or polynomils under rdicls (ie frction involving only polynomils or polynomils under rdicls)? If so, try the Comprison Test nd/or the Limit Comprison Test Remember however, tht in order to use the Comprison Test nd the Limit Comprison Test the series terms ll need to be positive 5 Does the series contin fctorils or constnts rised to powers involving n? If so, then the Rtio Test my work Note tht if the series term contins fctoril then the only test tht weve got tht will work is the Rtio Test 6 Cn the series terms be written in the form n = ( ) n b n or n = ( ) n+ b n? If so, then the Alternting Series Test my work 7 Cn the series terms be written in the form n = (b n ) n? If so, then the Root Test my work 8 If n = f(n) for some positive, decresing function f(x) nd is esy to evlute f(x)dx then the Integrl Test my work 6 Evlution of common series We hve the following results: n i= i = n(n+) 2 2 n i= i2 = n(n+)(2n+) 6 3 n i=0 rn = rn+ 4 n i=0 ( n i for r r ) = 2 n 5 ( ) n+ n= = ln 2 n 6 x n n=0 n! = e x for ny < x < 7 lim ( + n) n = e nd lim ( n ) n = e 9

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