ORDERS OF GROWTH KEITH CONRAD

Transcription

1 ORDERS OF GROWTH KEITH CONRAD Itroductio Gaiig a ituitive feel for the relative growth of fuctios is importat if you really wat to uderstad their behavior It also helps you better grasp topics i calculus such as covergece of improper itegrals ad ifiite series We wat to compare the growth of three differet kids of fuctios of x, as x : power fuctios for r > (such as x 2 or x = x /2 ), expoetial fuctios a x for a >, logarithmic fuctios log b x for b > Examples are plotted i Figure The relative sizes are differet for x ear ad for large x f(x) x Figure Graph of y = x 2, y = x, y = e x, y = l(x) All power fuctios, expoetial fuctios, ad logarithmic fuctios (as defied above) ted to as x But these three classes of fuctios ted to at differet rates The mai result we wat to focus o is the followig oe It says e x grows faster tha ay power fuctio while log x grows slower tha ay power fuctio (A power fuctio meas with r >, so /x 2 = x 2 does t cout)

2 2 KEITH CONRAD log x Theorem For each r >, lim = ad lim x ex x = This is illustrated i Figure The fuctios icrease at first, but ted to for larger x After we prove Theorem ad look at some cosequeces of it i Sectio 2, we will compare power, expoetial, ad log fuctios with the sequeces! ad i Sectio 3 At the ed we will show that betwee ay two sequeces with differet orders of growth we ca isert ifiitely may sequeces with differet orders of growth i betwee them 2 Proof of Theorem ad some corollaries Proof (of Theorem ) First we focus o the limit /e x Whe r = this says x (2) as x ex This result follows from L Hopital s rule To derive the geeral case from this special case, write ( ) x/r r (22) e x = rr e x/r With r stayig fixed, as x also x/r, so (x/r)/e x/r by (2) with x/r i place of x The the right side of (22) teds to as x, so we re doe Now we show (log x)/ as x Writig y for log( ) = r log x, log x = y/r e y = r y e y As x, also y, so (/r)(y/e y ) by (2) with y i place of x Thus (log x)/ p(x) Corollary 2 For ay polyomial p(x), lim x e x = Proof By Theorem, for ay k > we have x k /e x as x This is also true whe k = Writig p(x) = a d x d + a d x d + + a x + a, we have p(x) x d e x = a d e x + a x d x d e x + + a e x + a e x Each x k /e x appearig here teds to as x, so p(x)/e x teds to as x (log x) k Corollary 22 For ay r > ad k >, lim x = Proof Let y = log( ) = r log x, so (log x) k = yk /r k e y = r k yk e y As x, also y Therefore (/r k )(y k /e y ) by Theorem (sice r k > ) We derived Corollary 22 from Theorem, but the argumet ca be reversed too Take k = i Corollary 22 to get the log part of Theorem ad use the chage of variables y = e x i /e x to get the expoetial part of Theorem from Corollary 22 Specifically, whe y = e x (log y)r =, ex y ad as x we have y = e x, so by Corollary 22 we get (log y) r /y /e x as x Therefore

3 ORDERS OF GROWTH 3 (log x) k Corollary 23 For ay o-costat polyomial p(x) ad positive umber k, lim x p(x) Proof For large x, p(x) sice o-zero polyomials have oly a fiite umber of roots Write p(x) = a d x d + a d x d + + a x + a, where d > ad a d The (log x) k p(x) = (log x)k x d a d + a d /x + + a /x d As x, the first factor teds to by Corollary 22 while the secod factor teds to /a d, so the product teds to Corollary 24 As x, x /x Proof The logarithm of x /x is log(x /x ) = (log x)/x, which teds to as x Expoetiatig, x /x = e (log x)/x e = We coclude this sectio by givig a secod proof of Corollary 22 which does t rely o aythig we have doe so far Thus, we could cosider Corollary 22 as the mai result ad Theorem as a special case! Proof We will use estimates o the itegral for log x: log x = x dt t For r >, we have /t /t r whe t Therefore whe x > < log x = x dt t x dt t r = x t r dt = xr r r < xr r If we ru through these estimates with r/2 i place of r (which is fie sice r/2 > too) the we get (23) < log x xr/2 r/2 = 2xr/2 r The reaso we use r/2 is because ow whe we divide by we get a decayig term o the right side: < log x 2 r/2 As x, the right side teds to, so (log x)/ But this is t complete: we wat (log x) k / for ay k >, ot just (log x)/ To get this, let s ru through the iequalities agai usig r/(2k) i place of r This amouts to substitutig r/2 with r/(2k) i (23), ad the result is = Now raise to the k-th power: < log x xr/(2k) r/(2k) = 2kxr/(2k) r < (log x) k ( ) 2k k /2 r

4 4 KEITH CONRAD Dividig by as we did before, < (log x)k ( ) 2k k r /2 As x, the right side teds to, so we have show (log x) k / Replacig e x with a x for ay a > ad log x with log b x for ay b > leads to completely aalogous results Theorem 25 Fieal umbers a > ad b > For ay r > ad iteger k >, lim x a x =, For ay o-costat polyomial p(x), lim x lim (log b x) k x = p(x) (log =, lim b x) k ax x p(x) Proof To deduce this theorem from earlier results, write a x = e (log a)x ad log b x = (log x)/(log b) The umbers log a ad log b are positive The, for istace, if we set y = (log a)x, a x = xr e (log a)x = y r (log a) r e y Whe x, also y sice log a >, so the behavior of /a x follows from that of y r /e y usig Theorem Sice log b x = (log x)/(log b) is a costat multiple of log x, carryig over the results o log x to log b x is just a matter of rescalig For istace, if we set y = log x, so log b x = y/ log b, the (log b x) k = yk /(log b) k e ry = (log b) k = y k (e r ) y As x, also y, so the expoetial fuctio (e r ) y domiates over the power fuctio y k : y k /(e r ) y Therefore (log b x) k / as x 3 Growth of basic sequeces We wat to compare the growth of five kids of sequeces: power sequeces r for r > :, 2 r, 3 r, 4 r, 5 r, expoetial sequeces a for a > : a, a 2, a 3, a 4, a 5, log sequeces log b for b > :, log b 2, log b 3, log b 4, log b 5,!:, 2, 6, 24, 2, :, 4, 27, 256, 325, The first three sequeces are just the fuctios we have already treated, except the real variable x has bee replaced by a iteger variable That is, we are lookig at those old fuctios at iteger values of x ow Some otatio to covey domiatig rates of growth will be coveiet For two sequeces x ad y, write x y to mea x /y as I other words, x grows substatially slower tha y (if it just grew at half the rate, for istace, the x /y would be aroud /2 rather tha ted to ) For istace, 2 ad

5 ORDERS OF GROWTH 5 Remark 3 The otatio x y does ot mea x < y for all Maybe some iitial terms i the x sequece are larger tha the correspodig oes i the y sequece, but this will evetually stop ad the log term growth of y domiates For istace, 2 eve though 2 < for all small Ideed, the ratio 2 = teds to as, but the ratio is ot small util gets quite large Theorem tells us that (3) log r e for ay r > By Theorem 25, we ca say more geerally that (32) log b r a for ay a > ad b > How do the sequeces! ad fit ito (32)? They belog o the right, as follows Theorem 32 For ay a >, a! Equivaletly, a! lim =, lim! = Proof To compare a ad!, we will use Euler s beautiful itegral formula for!: (33)! = x e x dx I case you re ot familiar with (33), let s recall how to prove it by iductio o Whe =, x e x dx = e x dx, which is by itegratio Assumig x e x dx =! for some, we will compute x + e x dx usig itegratio by parts with u = x + ad dv = e x dx: x + e x dx = x + e x + ( + )x e x dx = lim b b+ e b + ( + )x e x dx The limit is by Theorem 25, ad the itegral is ( + ) x e x dx, which by iductio is ( + )! = ( + )! To apply (33), we obtai a a lower boud for! by makig the itegral ru over [, ]:! > x e x dx O the iterval [, ], e x has its smallest value at the right ed: e x e Therefore x e x x e o [, ] Itegratig both sides of this iequality from x = to x = gives x e x dx = e x e dx x dx = + e + ( ) = e +

6 6 KEITH CONRAD ( ) Therefore! > e +, so a! < a ( ae (/e) (/( + )) = ) + This fial expressio is a upper boud o a /! How does it behave as? For large, ae/ /2, so (ae/) (/2) Therefore (ae/) Sice the other factor ( + )/ teds to, we see our upper boud o a /! teds to, so a /! as To show the other part of the theorem, that!/ as, we will get a upper boud o! ad divide the upper boud by Write e x as e x/2 e x/2 i Euler s factorial itegral:! = x e x dx = (x e x/2 )e x/2 dx The fuctio x e x/2 drops off to as x Where does it have its maximum value? The derivative is x e x/2 ( x/2) (check this), so x e x/2 vaishes at x = 2 Checkig the sigs of the derivative to the left ad right of x = 2, we see x e x/2 has a maximum value at x = 2, where the value is (2) e Therefore x e x/2 (2) e for all x >, so! = (x e x/2 )e x/2 dx (2) e e x/2 dx = (2) e e x/2 dx = (2) e 2 Dividig throughout by gives ( )! 2 2 e Sice 2 < e, the right side teds to, so!/ as The fact that a! is ituitively reasoable, for the followig reaso: each of these expressios (a,!, ad ) is a product of umbers, but the ature of these umbers is differet I a, all umbers are the same value a, which is idepedet of : a = a } a {{ a} times I!, the umbers are the itegers from to :! = 2 3 ( ) Sice the terms i this product keep growig, while the terms i a stay the same, it makes sese that! grows faster tha a (at least oce gets larger tha a) I, all umbers equal : = } {{ } times Sice all the terms i this product equal, while i! the terms are the umbers from to, it is plausible that grows a lot faster tha! To summarize our results o sequeces, we combie (32) ad Theorem 32: log b r a!

7 ORDERS OF GROWTH 7 Here a >, b >, ad r > (ot just r >!) All sequeces here ted to as, but the rates of growth are all differet: ay sequece which comes to the left of aother sequece o this list grows at a substatially smaller rate, i the sese that the ratio teds to For example, ca we fid a (atural) sequece whose growth is itermediate betwee ad r for every r >? That is, we wat to fid a sigle sequece of umbers x such that x r for every r > Oe choice is x = log Ideed, log = log, so log, ad for ay r > log r = log r, which teds to sice r > ad log grows slower tha ay power fuctio (with a positive expoet) by Theorem Usig powers of log, we ca write dow ifiitely may sequeces with differet rates of growth betwee ad every sequece r for r > : log (log ) 2 (log ) 3 (log ) k r, where k rus through the positive itegers Is it possible to isert ifiitely may sequeces with differet rates of growth betwee ay two sequeces with differet rates of growth? Theorem 33 If x y, there are sequeces {z () }, {z (2) }, {z (3) }, such that x z () z (2) z (3) y? Proof Sice x /y, for large the ratio x /y is small Specifically, < x /y < for large For small positive umbers, takig roots makes them larger but less tha : < a < = < a < a < 3 a < < k a < < Sice x /y < for large, this presets us with the iequalities < x x x x < < y y 3 < < y k < < y for large ad k =, 2, 3, Multiply through by y : (34) < x < x y < x /3 y 2/3 < < x /k y /k < < y For k < l, the ratio of the k-th root sequece to the l-th root sequece is x /k y /k ( ) /k /l x x /l y /l = y Sice /k /l >, this ratio teds to as Therefore (34) leads to ifiitely may sequeces with growth itermediate betwee {x } ad {y }, amely the sequeces z (k) = x /k y /k for k = 2, 3, 4, : (35) x x y x /3 y 2/3 x /k (If you wat to label the first sequece with k =, set z (k) y /k y = x /(k+) y /(k+) for k =, 2, 3, )

8 8 KEITH CONRAD The differece betwee (34) ad (35) is that (34) is a set of iequalities which is valid for large (amely large eough to have x /y < ), while (35) is a statemet about rates of growth betwee differet sequeces: it makes o sese to ask if (35) is true at a particular value of, ay more tha it would make sese to ask if the limit relatio + is true at = 45