Review of some needed Trig Identities for Integrtion Your nswers should be n ngle in RADIANS rccos( 1 2 ) = rccos( - 1 2 ) = rcsin( 1 2 ) = rcsin( - 1 2 ) = Cn you do similr problems? Review of Bsic Concepts Double-ngle formuls Your nswer should involve trig functions of θ, nd not of 2θ cos(2θ) = sin(2θ) = Hlf-ngle formuls Your nswer should involve cos(2θ) cos 2 (θ) = sin 2 (θ) = Since cos 2 θ + sin 2 θ = 1, we know tht the corresponding reltionship beween: tngent (ie, tn) nd secnt (ie, sec) is cotngent (ie, cot) nd cosecnt (ie, csc) is du u Remember Your Clculus I Integrtion Bsics? u 0 = u n du n 1 = +C e u du = +C u du 1 = = cos u du = sec 2 u du = sec u tn u du = sin u du = csc 2 u du = csc u cot u du = tn u du = cot u du = sec u du = csc u du = 1 2 u 2 du >0 = 1 2 +u 2 du >0 = 1 u u 2 2 du >0 = Prof Girrdi Pge 1 of 11 Mth 142
Integrtion from Clculus II Integrtion by prts formul: u dv = To integrte f(x), g(x) where f nd g re polyonomils, 1st find its (PDF) If [degree of f] [degree of g], then one must first does If [degree of f] < [degree of g] ie, hve strictly bigger bottoms then first fctor y = g(x) into: fctors px + q nd irreducible you need ) fctors x 2 + bx + c (to be sure it s irreducible, Next, collect up like terms nd follow the following rules Rule 1: For ech fctor of the form (px+q) m where m 1, the decomposition of y = f(x) contins g(x) sum of prtil frctions of the form, where ech A i is rel number, Rule 2: For ech fctor of the form (x 2 + bx + c) n where n 1, the decomposition of y = f(x) g(x) contins sum of prtil frctions of the form, where the A i s nd B i s re rel number, Trig Substitution (Recll tht the integrnd is the function you re integrting) Here, is constnt nd > 0 if the integrnd involves 2 u 2, then one mkes the substitution u = if the integrnd involves 2 +u 2, then one mkes the substitution u = if the integrnd involves u 2 2, then one mkes the substitution u = Improper Integrls 0 Fill-in-the boxes Below,, b, c R with < c < b If f : [0, ) R is continuous, then we define the improper integrl 0 0 If f : (, 0] R is continuous, then we define the improper integrl 0 0 Prof Girrdi Pge 2 of 11 Mth 142
If f : (, ) R is continuous, then we define the improper integrl Review of Bsic Concepts If f : (, b] R is continuous, then we define the improper integrl If f : [, b) R is continuous, then we define the improper integrl If f : [, c) (c, b] R is continuous, then we define the improper integrl An improper integrl s bove converges precisely when An improper integrl s bove diverges precisely when An improper integrl s bove diverges to precisely when An improper integrl s bove diverges to precisely when Prof Girrdi Pge 3 of 11 Mth 142
Sequences Let { n } n=1 be sequence of rel numbers Complete the below sentences The limit of { n } n=1 is the rel number L provided for ech ɛ > 0 there exists nturl number N so tht if the nturl number n stisfies > then < If the limit of { n } n=1 is L R, then we denote this by { n } n=1 converges provided { n } n=1 diverges provided { n } n=1 Prctice tking bsic limits (Importnt, eg, for Rtio nd Root Tests) 5n 17 + 6n 2 + 1 lim 7n 18 + 9n 3 + 5 = lim 36n 17 6n 2 1 4n 17 + 9n 3 + 5 = -5n 18 + 6n 2 + 1 36n lim 7n 17 + 9n 3 + 5 = lim 17 6n 2 1 4n 17 + 9n 3 + 5 = Cn you do similr problems? Commonly Occurring Limits Thoms Book 101, Theorem 5 pge 578 ln n (1) lim n = (2) lim n n = (3) lim c 1/n = (c > 0) (4) lim c n = ( c < 1) (5) lim ( 1 + c n) n = (c R) x n (6) lim n! = (c R) Let < r < (Needed for Geometric Series Wrning, don t confuse sequences with series) If r < 1, then lim r n = If r = 1, then lim r n = If r > 1, then lim r n = If r = -1, then lim r n = If r < -1, then lim r n = Prof Girrdi Pge 4 of 11 Mth 142
Series In this section, ll series re understood to be n=1, unless otherwise indicted Review of Bsic Concepts For forml series n=1 n, where ech n R, consider the corresponding sequence {s N } N=1 of prtil sums, so s N = N n=1 n Then the forml series n : converges if nd only if converges to L R if nd only if diverges if nd only if Now ssume, furthermore, tht n 0 for ech n Then the sequence {s N } N=1 of prtil sums either is bounded bove (by some finite number), in which cse the series n or is not bounded bove (by some finite number), in which cse the series n Stte the n th -term test for n rbitrry series n Fix r R For N 17, let s N = N n=17 rn (Note the sum strts t 17) Then, for N > 17, s N = r s N = (1 r) s N = nd if r 1, then s N = (your nswer cn hve s but not sign) (your nswer cn hve s but not sign) (your nswer should hve neither s nor sign) (your nswer should hve neither s nor sign) Geometric Series where < r < The series r n (hint: look t the previous questions): converges if nd only if diverges if nd only if p-series where 0 < p < The series 1 n p converges if nd only if diverges if nd only if This cn be shown by using the (the hrd to compute series) 1 n p to (the esy to compute improper integrl) here, nme the test one uses nd compring dx x=1 Prof Girrdi Pge 5 of 11 Mth 142
Tests for Positive-Termed Series (so for n where n 0) 01 Stte the Integrl Test with Reminder Estimte for positive-termed series n Let f : [1, ) R be so tht Then (1) n = f (n) for ech n N (2) f is function (3) f is function (4) f is function n converges if nd only if nd if n converges, then ( ) ( N ) 0 k k k=1 k=1 converges 02 Stte the Direct Comprison Test for positive-termed series n If when n 17 nd, then n converges If when n 17 nd, then n diverges Hint: sing the song to yourself 03 Stte the Limit Comprison Test for positive-termed series n Let b n > 0 nd L = lim n bn If 0 < L <, then If L = 0, then If L =, then Gol: cleverly pick positive b n s so tht you know wht b n does (converges or diverges) nd the sequence { n bn }n converges 04 Helpful Intuition Fill in the 3 boxes using: e x, ln x, x q Use ech once, nd only once Consider positive power q > 0 There is (some big number) N q > 0 so tht if x N q then Prof Girrdi Pge 6 of 11 Mth 142
Tests for Arbitrry-Termed Series (so for n where < n < ) 05 By definition, for n rbitrry series n, (fill in these 3 boxes with convergent or divergent) n is bsolutely convergent if nd only if n is n is conditionlly convergent if nd only if n is nd n is n is divergent if nd only if n is divergent 06 Stte the Rtio nd Root Tests for rbitrry-termed series n with < n < Let ρ = lim n+1 n or ρ = lim n 1 n If then n converges bsolutely If If then n diverges then the test is inconclusive 07 Stte the Alternting Series Test (AST) & Alternting Series Estimtion Theorem Let Then (1) u n 0 for ech n N (2) lim u n = (3) u n u n+1 for ech n N nd we cn estimte (ie, pproximte) the infinite sum n=1 ( 1)n u n by the finite sum N k=1 ( 1)k u k nd the error (ie reminder) stisfies N ( 1) k u k ( 1) k u k k=1 k=1 Prof Girrdi Pge 7 of 11 Mth 142
Condsider (forml) power series with rdius of convergence R [0, ] h(x) = Power Series (Here x 0 R is fixed nd { n } n=0 is fixed sequence of rel numbers) n (x x 0 ) n, (11) Without ny other further informtion on { n } n=0, nswer the following questions The choices for the next 4 boxes re: AC, CC, DIVG, nything Here, n=0 AC stnds for: lwys bsolutely convergent CC stnds for: lwys conditionlly convergent DIVG stnds for: is lwys divergent nything stnds for: cn do nything, ie, there re exmples showing tht it cn be AC, CC, or DIVG (1) At the center x = x 0, the power series in (11) (2) For x R such tht x x 0 < R, the power series in (11) (3) For x R such tht x x 0 > R, the power series in (11) (4) If R > 0, then for the endpoints x = x 0 ±R, the power series in (11) For this prt, fill in the 7 boxes Let R > 0 nd consider the function y = h(x) defined by the power series in (11) (1) The function y = h(x) is lwys differentible on the intervl (mke this intervl s lrge s it cn be, but still keeping the sttement true) Furthermore, if x is in this intervl, then h (x) = (12) n= Wht cn you sy bout the rdius of convergence of the power series in (12)? (2) The function y = h(x) lwys hs n ntiderivtive on the intervl (mke this intervl s lrge s it cn be, but still keeping the sttement true) Futhermore, if α nd β re in this intervl, then x=β x=α h(x) dx = n= x=β x=α Prof Girrdi Pge 8 of 11 Mth 142
Tylor/Mclurin Polynomils nd Series Review of Bsic Concepts Let y = f(x) be function with derivtives of ll orders in n intervl I contining x 0 Let y = P N (x) be the N th -order Tylor polynomil of y = f(x) bout x 0 Let y = R N (x) be the N th -order Tylor reminder of y = f(x) bout x 0 Let y = P (x) be the Tylor series of y = f(x) bout x 0 Let c n be the n th Tylor coefficient of y = f(x) bout x 0 The formul for c n is c n = b In open form (ie, with nd without -sign) P N (x) = c In closed form (ie, with -sign nd without ) P N (x) = d In open form (ie, with nd without -sign) P (x) = e In closed form (ie, with -sign nd without ) P (x) = f We know tht f(x) = P N (x) + R N (x) Tylor s BIG Theorem tells us tht, for ech x I, R N (x) = for some c between nd g A Mclurin series is Tylor series with the center specificlly specified s x 0 = Prof Girrdi Pge 9 of 11 Mth 142
Commonly Used Tylor Series Here, expnsion refers to the power series expnsion tht is the Mclurin series An expnsion for y = e x is, which is vlid precisely when x An expnsion for y = cos x is, which is vlid precisely when x An expnsion for y = sin x is, which is vlid precisely when x An expnsion for y = 1 1 x is, which is vlid precisely when x An expnsion for y = ln(1+x) is, which is vlid precisely when x An expnsion for y = rctn x is, which is vlid precisely when x Prof Girrdi Pge 10 of 11 Mth 142
Prmetric Curves In this prt, fill in the 4 boxes Consider the curve C prmeterized by x = x (t) for t b y = y (t) 1) Express dy dx in terms of derivtives with respect to t Answer: dy dx = 2) The tngent line to C when t = t 0 is y = mx + b where m is evluted t t = t 0 3) Express d2 y dx 2 using derivtives with respect to t Answer: d 2 y dx 2 = 4) The rc length of C, expressed s on integrl with respect to t, is Arc Length = Polr Coordintes Here, CC stnds for Crtresin coordintes ( while PC) stnds for polr ( coordintes ) A point with PC (r, θ) lso hs PC, θ + 2π s well s, θ + π A point P R 2 with CC (x, y) nd PC (r, θ) stisfies the following x = & y = & r 2 = & = { y x if x 0 DNE if x = 0 The period of f(θ) = cos(kθ) nd of f(θ) = sin(kθ) is To sketch these grphs, we divide the period by nd mke the chrt, in order to detect the Now consider sufficiently nice function r = f(θ) which determines curve in the plne The the re bounded by polr curves r = f(θ) nd the rys θ = α nd θ = β is Are = θ=β θ=α The rc length of the polr curves r = f(θ) is dθ Arc Length = θ=β θ=α dθ Prof Girrdi Pge 11 of 11 Mth 142