Mth 07H Topics since the second exm Note: The finl exm will cover everything from the first two topics sheets, s well. Absolute convergence nd lternting series A series n converges bsolutely if n converges. If n converges then n converges. A series which converges but does not converge bsolutely is clled conditionlly convergent. An lternting series hs the form ( ) n n with n 0 for ll n. If the sequence n is decresing nd hs limit 0, then the lternting series test sttes tht ( ) n n converges. For exmple, ( )n /(n + ) converges, but not bsolutely, so it is conditionlly convergent. Even more, if the lternting series test implies tht ( ) n n converges, then the N-th prtil sum, s N = N ( )n n, is within n+ of the sum of the series (since ll of the lter prtil sums lie between s N nd s N+ ). So,forexmple, n= ( )n+ /n 2 converges, nd 99 n= ( )n+ /n 2 iswithin/(00) 2 = /0000 of the infinite sum. For the series n= /n2, on the other hnd, the integrl test cn only conclude tht its til, n=00 /n2, is t most /00. The rtio test Previous tests hve you compre your series with something else (nother series, n improper integrl); this test compres series with itself (sort of) Rtio Test: n, n 0 ll n; lim n+ = L n n If L < then n converges bsolutely If L >, then n diverges If L =, then try something else! The bsic ide: We re relly (limit) compring to the series L n Ex: 4 n converges by the rtio test n n diverges by the rtio test n5 Power series Ide: turn series into function, by mking the terms n depend on x replce n with n x n ; series of powers n x n = power series centered t 0 n (x ) n = power series centered t Big question: for wht x does it converge? Solution from the Rtio Test lim n+ = L, or lim n n = L, set R = n L then n (x ) n converges bsolutely for x < R Ex.: x n = x diverges for x > R ; ; conv. for x < Why cre bout power series? n Ide: prtil sums k x k re polynomils; R = rdius of convergence
if f(x)= n x n, then the poly s mke good pproximtions for f Differentition nd integrtion of power series Ide: if you differentite or integrte ech term of power series, you get power series which is the derivtive or integrl of the originl one. If f(x) = n (x ) n hs rdius of conv R, then so does g(x) = n n (x ) n, nd g(x) = f (x) n= n nd so does g(x) = n+ (x )n+, nd g (x) = f(x) x n Ex: f(x) =, then f (x) = f(x), so (since f(0) = ) f(x) = e x = Ex.: x = x n x n+, so ln( x) = (for x < ), so n+ ( ) n x n+ (replcing x with x) ln(x+) =, so n+ ( ) n (x ) n+ (replcing x with x ) ln(x) = n+ Ex:. rctnx = ( x 2 ) dx = ( x 2 ) n dx = ( ) n x 2n+ Tylor series Ide: strt with function f(x), find power series for it. If f(x) = n (x ) n, then (term by term diff.) 2n+ f (n) () = n ; So n = f(n) () f (n) () Strting with f, define P(x) = (x ) n, the Tylor series for f, centered t. n f (k) () P n (x) = (x ) k, the n-th Tylor polynomil for f. k! ( ) n Ex.: f(x) = sinx, then P(x) = (2n+)! x2n+ x n (for x < Big questions: Is f(x) = P(x)? (I.e., does f(x) P n (x) tend to 0?) If so, how well do the P n s pproximte f? (I.e., how smll is f(x) P n (x)?) Error estimtes f (n) () f(x) = (x ) n 2
mens tht the vlue of f t point x (fr from ) cn be determined just from the behvior of f ner (i.e., from the derivs. of f t ). This is very powerful property, one tht we wouldn t ordinrily expect to be true. The mzing thing is tht it often is: T f, (x) = f (n) () (x ) n ; T n,f, (x) = n f (k) () (x ) n ; k! R n (x)= f(x) T n,f, (x) = n-th reminder term = error in using T n,f, to pproximte f Tylor s reminder theorem : estimtes the size of R n (x) If f(x) nd ll of its derivtives (up to n+) re continuous on [,b], then f (n+) (t) f(b) = T n,f, (b) + (n+)! (b t)n. How? By strting from f(b) = f()+ the wrong wy! This in turn implies: f(b) = T n,f, (b) + f(n+) (c) (n+)! (b )n+, for some c in [,b] every x in [,b] f (t) dt, nd repetedly integrting by prts i.e., for ech x, R n (x) = f(n+) (c) (n+)! (x )n+, for some c between nd x so if f (n+) (x) M for every x in [,b], then R n (x,) M (n+)! (x )n+ for Ex.: f(x)=sinx, then f (n+) (x) for ll x, so R n (x,0) x n+ (n+)! 0 s n ( ) n ( ) n so sinx = (2n+)! x2n+ Similrly, cosx = (2n)! x2n Using Tylor s reminder to estimte vlues of functions: e x (x) n = (n)!, so e=e = (n)! e c R n (,0) = f(n+) (c) = (n+)! (n+)! e (n+)! 4 (n+)! since e < 4 (since ln(4) > (/2)()+(/4)(2) = ) (Riemnn sum for integrl of /x) 4 so since (3+)! = 4.58 0, e = ++ 2 + 6 + 24 + 20 + +, to 0 deciml plces. 3! Other uses: if you know the Tylor series, it tells you the vlues of the derivtives t the center. Ex.: e x = (x) n (n)!, so xex = (x) n+, so (n)! 5th deriv of xe x, t 0, is 5!(coeff of x 5 ) = 5! 4! = 5 Substitutions: new Tylor series out of old ones Ex. sin 2 x = cos(2x) = 2 2 ( ( ) n (2x) 2n (2n)! = (2x)2 ( ( 2 2! + (2x)4 4! (2x)6 6! 3 +
= 2x2 23 x 4 + 25 x 6 27 x 8 + 2! 4! 6! 8! Integrte functions we cn t hndle ny other wy: (x) 2 n Ex.: e x2 = (n)!, so (x) 2n+ e x2 dx = (2n+) Fourier series Ide: different wy to express function s sum of nicer functions. The nice functions this time, though, re trig functions (insted of powers)! The other ide: Tylor series of f is built using informtion from only round the center x = of the series. For mny functions, though, this tells us nothing (i.e., we get no good pproximtion) further from. For exmple, it cn tell us nothing pst point of discontinuity of f. A different pproch uses integrtion to cpture informtion verged over n entire intervl. Strting with periodic function f, with period (for the ske of illustrtion, ny number cn be used) 2π, so f(x+2π) = f(x) for every x, the ide is to express f s n (infinite) sum of nice functions, lso hving period 2π. One nturl choice to mke is the functions sin(nx) nd cos(nx), so we will ttempt to write f(x) = n sin(nx)+b n cos(nx) Two immedite questions: does such series converge, nd cn we ctully do this?! Usully, yes! Just s with Tylor series, the right question to sk is: wht vlues must n nd b n hve? The nswer is obtined by integrtio Since nd sin(mx)cos(nx) dx = 0 for ll m nd n (the integrnds re odd functions!), sin(mx)sin(nx) dx = 0 nd while π cos(mx)cos(nx) dx = 0 for m n, sin(nx) sin(nx) dx = (these cn be verified by integrtion by prts!), this mens tht f(x)sin(mx) dx cos(nx)cos(nx) dx = π 2 = n sin(nx)sin(mx) dx +b n cos(nx)sin(mx) dx = π 2 m nd f(x)cos(mx) dx = n sin(nx)cos(mx) dx +b n cos(nx)cos(mx) dx = π 2 b m nd so n = 2 f(x)sin(nx) dx nd b n = 2 f(x)cos(nx) dx. π π So if we cn express function s n infinite sum of trig functions, this is wht the coefficients must be equl to! For exmple, if we compute this for the squre wve, the function f with f(x) = for x [,0) nd f(x) = for x [0,π) (nd which then repets this pttern in both directions), some computtion gives us tht n = 2( ( )n ) nd b n = 0 (since f is n odd function). Grphing the sums nπ N 2( ( ) n ) sin(nx) for incresingly lrge vlues of N does give sequence of nπ 4
functions which give good pproximtions to f! It is somewht beyond the scope of our course to verify this, but the theory behind ll of this is tht the coefficients we hve computed succeed in giving the smllest possible vlue for the integrl N [f(x) ( n sin(nx)+b n cos(nx))] 2 dx for every N, tht these integrls decrese with N, nd (usully!) converge to 0, implying tht over most of the intervl [,π] the error N f(x) [ n sin(nx)+b n cos(nx)] must be smll! Polr coordintes Ide: describe points in the plne in terms of (distnce,direction). r = (x 2 +y 2 ) /2 = distnce, θ = rctn(y/x) = ngle with the positive x-xis. x = rcosθ, y = rsinθ The sme point in the plne cn hve mny representtions in polr coordintes: (,0) rect = (,0) pol = (,2π) pol = (,6π) pol =... A negtive distnce is interpreted s positive distnce in the opposite direction (dd π to the ngle): ( 2,π/2) pol = (2,π/2+π) pol = (0, 2) rect An eqution in polr coordintes cn(in principl) be converted to rectngulr coords, nd vice vers: E.g., r = sin(2θ) = 2sinθcosθ cn be expressed s r 3 = (x 2 +y 2 ) 3/2 = 2(rsinθ)(rcosθ) = 2yx, i.e., (x 2 +y 2 ) 3 = 4x 2 y 2 Grphing in polr coordintes: grph r = f(θ) s if it were Crtesin; this llows us to identify the vlues of θ (= sectors of the circle) where r is positive/negtive nd incresing/decresing (i.e., moving wy from/towrds the origin). Now wrp the Crtesin grph round the origin, using the vlues of θ where f = 0 nd f = 0 s guide. Given n eqution in polr coordintes r = f(θ), i.e., the curve (f(θ),θ) pol,θ θ θ 2 we cn compute the slope of its tngent line, by thinking in rectngulr coords: x = f(θ)cosθ,y = f(θ)sinθ, so dy dx = dy/dθ dx/dθ = f (θ)sinθ +f(θ)cosθ f (θ)cosθ f(θ)sinθ Arclength: the polr curve r = f(θ) is relly the (rectngulr) prmetrized curve x = f(θ)cosθ,y = f(θ)sinθ, nd (x (θ)) 2 +(y (θ)) 2 ) /2 = (f (θ)) 2 +(f(θ)) 2 ) /2, so the rclength for θ b is (f (θ)) 2 +(f(θ)) 2 ) /2 dθ Are: if r = f(θ), θ b describes closed curve (f() = f(b) = 0), then we cn compute the re inside the curve s sum of res of sectors of circle, ech with re pproximtely πr 2 ( θ/2π) = (f(θ))2 θ 2 so the re cn be computed by the integrl 2 (f(θ))2 dθ For the re between two polr curves: if f(θ) g(θ) for α θ β, then β Are = 2 (f(θ))2 2 (g(θ))2 dθ α 5