5 Short Proofs of Simplified Stirling s Approximation

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1 5 Short Proofs of Smplfed Strlg s Approxmato Ofr Gorodetsky, drtymaths.wordpress.com Jue, 20 0 Itroducto Strlg s approxmato s the followg (somewhat surprsg) approxmato of the factoral,, usg elemetary fuctos: ( e ) 2π () (f() g() meas: the rato f() g() mathematca James Strlg. teds to as goes to fty.) It was proved 730 by the Scottsh Ths approxmato has may applcatos, amog them - estmato of bomal ad multomal coeffcets. It has a wde varety of proofs, for example: Applyg the Euler-Maclaur formula to the tegral l x dx. Usg Cauchy s formula from complex aalyss to extract the coeffcets of e x : = 2π z = e z z + dz. Those proofs are ot complcated at all, but they are ot too elemetary ether. I ths short ote I wll preset, usg guded exercses, 5 dfferet proofs for weaker versos of Strlg s approxmato. I most of the real-lfe applcatos, the weaker versos are more tha eough. The proofs are self-cotaed ad completely elemetary - othg more tha basc calculus (tegrato ad dfferetato, Taylor seres, deftos of e ad e x ) s requred. I am partcularly fod of proof o. 2 ad the secod part of proof o., sce I came up wth them o my ow (although t s hghly probable that they were dscovered years before, as t happes the harsh world of Mathematcs). Let s beg Frst proof: The Rectagle Method We ll prove the followg double equalty: e( e ) e( e ) (2) where equalty holds oly for =. Ulke Strlg s approxmato, whch s asymptotc ature, ths equalty s true for all - a bg advatage. The basc des to compare the sum l to the tegral l x dx.

2 . Wrte ths equalty logarthmcally: l + l l + + l (3) = 2. Show that: l xdx < l < + Ths s the geometrc step of the proof ad the oe that gves t ts ame. 3. Prove that l x dx = x l x x + C. l x dx (). By summg the left equalty of over = 2,, ad the rght oe over =,,, deduce the equalty. Challege: Ths method, of approxmatg tegrals usg rectagles, ca be refed. Prove the followg 2 dettes ad deduce Strlg s approxmato (except for the costat beg 2π, though you ca fd lower ad upper bouds for t). It s assumed that f s a twce dfferetable fucto. t+ 2 t 2 f(x) dx = f(t) + f (ξ) 2 for some ξ [t 2, t + 2 ] (Mdpot Rule) t+ f(x) dx = f(t)+f(t+) t 2 f (ξ) 2 for some ξ [t, t + ] (Trapezod Rule) 2 Secod proof: Start at the Cocluso We re gog to prove that C( e ), where C s some postve costat. The full approxmato states that C = 2π , ad after the proof I challege you to boud t from above by e There s somethg aoyg about the proof - t uses a pror kowledge about. It begs by approxmatg the rato ( e ), so we had to kow Strlg s approxmato beforehad to eve thk about ths rato.. Let a = ( e ). We eed to prove that a coverges to some postve fte lmt (whch wll be the verse of our costat C). 2. As the prevous proof, wrte everythg logarthmcally, for smplcty s sake: We eed to verfy that the seres = l(a ) = l( + ) + l(a ) (5) = a+ l( ) coverges to a fte lmt. 3. Verfy that a+ a = (+ )+ 2 e ad that ths rato coverges to. Use the secod-order Taylor approxmato of l( + x) aroud x = 0 to show that l( a+ ) = O( 2 ) (Ths s the key step).. Prove that 2 coverges ad deduce that so does a+ l( ). Coclude C( e ). 2

3 Challege: Show that {a } creases. Deduce e( e ). Gve C = 2π, deduce that Strlg s Approxmato s actually a lower boud: ( e ) 2π. Combg, we have the followg equalty: 3 Thrd Proof: Mootoc Covergece Ths proof gves 2 wthout tegrals or geometrcal sghts. ( e ) 2π e( e ) (6). (Iducto step) Show that f we attempt to prove the equalty by ducto o, the ducto step s equvalet to the followg equalty: Ths equalty s equvalet to. (Do you see why?) ( + ) ( < e < + ) + (7) 2. (Lmt) Show that both ( + ), ( + )+ ted to e. Thus, t s eough to show that ( + ) s creasg ad ( + )+ s decreasg. 3. (Icreasg sequece) Show that a = ( + ) s mootoe creasg by expadg t wth Newto s Bomal Theorem ad comparg a, a + term by term. (Ht: wrte the k th term as k! 0 <k ( )). (Decreasg sequece) Do the same wth b = ( + )+. Fourth Proof: The Power of the Power-Seres for e x As e x s aalytc ad s ts ow dervatve, t has the followg well-kow power-seres expaso aroud 0: e x = x k k! (8) We ll use the fact that the RHS cotas factorals to prove > ( e ), ad wth some more work we ll be able to sharpe t to > C ( e ) for some C > 0. It ca actually gve the full approxmato - ad I challege you to try.. Plug x = the power-seres of e x. Boud e from below usg the th term: e >. Deduce > ( e ). That s eat - but we ca (ad wll) do better. 2. Exame the sequece {a k = k k! } k 0 (summads of e ) more carefully ad verfy the followg: a k s umodal, ad acheves ts maxmum at both k = ad k = (Ht: Cosder a k+ a k ). Ths mmedately mproves the equalty by a multplcatve costat of 2: > 2( e ). It creases up to some dex, ad the starts to decrease deftely: r : a 0 a a r a r+ a r+2. 3

4 3. Mapulate e = k k! ad get the equalty = ( e ) Splt the seres to two sums (each s actually asymptotc to 2 2π): k! k = k! k (9) k =0 ( ) (0) We re gog to dscard. k= k! k = k ( + ) () k= =0. I 0, cosder oly the last terms to get the followg boud: > ( e ) ( ) > ( e ) ( ) (2) =0 Verfy that ( ) teds to e. Deduce that for ay C less tha e we have > ( e ) C for large eough. I thak T. Kalvar for a major smplfcato of ths step. Challege: By dog a more careful aalyss, mprove the costat e. Try ad get 2e. about π 2? Extreme Challege: Show that both 0, are asymptotc to for x 0) e x 2 0 e s harder. What dx = π 2. (Ht: e x +x + x e x 5 Ffth Proof: Cetral Bomal Coeffcet Uder Examato A commo usage combatorcs ad probablty for Strlg s Approxmato s the estmato of ( ), the cetral bomal coeffcet, as: ()! 2 ( e ) π (( e ) 2π) = (3) 2 π We wll prove very close estmates wth almost o aalyss. Credt where credt s due: Ths proof s mostly based o B. Schmulad s excellet post o MSE.. (Baby steps) Prove > ( ) > + for. 2. (Mapulato) Wrte ( ) as = ( 2 ).

5 3. (Aalyss) Prove the equalty < 2 < + for 2. Although t ca be proved drectly by squarg, there s also a elegat proof wth o computatos. Ca you fd t? Deduce + ( ), wth equalty oly for =. (Less elegat aalyss shows 2 3( )+ 3+, whch tur gves 3+ ( ).). (Mootoe Sequece) Use step 2 to deduce that ( ( ) ) 2 creases, ( + )( ( ) ) 2 decreases, ad so ( ) coverges to a costat C. Combe those observatos to get the followg equalty: + 2 C 2 ( ) C 2 () Assumg C = 2π (.e. assumg Strlg s approxmato), we obta the followg sharp result: ( ) π( (5) + 2 ) π 5. (Formula for Costat) Wrte ( ) as j= ( + ) j= ( ). Deduce that C = j ( )( + ). 6 Strlg s Full Approxmato j= (+ ). Together wth step 2, show that (( ) ) 2 = Although t was ot the teto of these otes, we ca recover Strlg s Approxmato wth just a lttle bt more work. If we combe the d ad the 5th proof, we fd that 2 j ( + ) ( e ). The fte product sde the square root s kow as Walls product. I wll break dow the ad hoc calculato of ths product, whch gves the fal value of π 2. Ths wll gve us Strlg s Approxmato.. Let I = π 0 s (x) dx. Note that I s a decreasg sequece that starts wth I 0 = π, I = Use tegrato by parts to show that for 2, I = ( )(I 2 I ), or: I I 2 =. 3. Use the last equalty together wth the fact that I s decreasg to show that lm I I + =.. Express the lmt as a fte product: I 0 I j +. Deduce that j ( + ) = π 2. 5

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