Ratioal Bouds for the Logarithm Fuctio with Applicatios Robert Bosch Abstract We fid ratioal bouds for the logarithm fuctio ad we show applicatios to problem-solvig. Itroductio Let a = + solvig the problem U385 from the joural Mathematical Reflectios I realized that I eeded good lower ad upper bouds for a depedig o. The classical a < e was ot eough. The problem U385 was proposed by Ágel Plaza Uiversidad de Las Palmas de Gra Caaria Spai: Evaluate More precisely I cojectured that + lim. a a +. To prove this cojecture I used the double iequality + e < a < + e + that I foud i []. Exactly problem 7 page 38 with solutio i page 6. The source of this problem is old from 87 i Nouvelles Aales de Mathématiques [] the proposer is ukow ad was solved by C. Moreau i [3]. Here we preset a ew proof. Later we will show how this result helps to fid the limit. Our proof of is based o o-stadard bouds for the logarithm fuctio these bouds are ratioal fuctios quotiet of two polyomials ad the seed iequalities the proof is iterative are the well-kow Mai Theorem Let us begi with the proof of ad 3. e x x + for all real x. Equality holds if ad oly if x =. e x x + for x R l x > for x >. 3 x Proof: Let fx = e x x. Note that f x = e x for x therefore f is a Mathematical Reflectios 7
icreasig fuctio for x. So fx f or equivaletly e x x +. If x the x + is egative ad e x is positive. Whe < x < the fuctio fx is also positive because f = e > ad f =. Suppose by cotradictio that exist x such that fx < the sice fx is clearly cotiuous by Bolzao s theorem there is a value x x with fx = cotradictio sice fx = oly for x =. To prove this last result suppose that fx = ad x. By Rolle s theorem exist x x such that f x = or e x = hece x = cotradictio. l x > x for x >. Proof: Now we are ready to fid the ratioal bouds. Theorem: x t dt > t dt. Proof: I : l + x x x > xx + I : l + x x + x I 3 : l + x I 4 : l x > xx + 6 x + 3 x x x + x >. I is a direct cosequece of. Now we itegrate I to obtai I Itegratig I the result is I 3 l + tdt tdt. Fially I 4 follows from itegrate 3 l + tdt l tdt > tt + t + dt. dt. t Notice that i each step the ew iequalities are refiemets of the previous oes. Holds xx + 6 x + 3 xx + x + x for x. Also x x x + x for x >. The result of cotiuig this algorithm are ot ratioal bouds. To see how to fid ew refiemets that are ratioal bouds look at [4]. Mathematical Reflectios 7
3 Applicatios We show a proof of the Arithmetic-Geometric Mea iequality usig iequality. Let α = x +x + +x. By e x α e x α e x α x α x α x α. After simple trasformatios this iequality is equivalet to AM-GM: α x x x. Let us see the proof of the double iequality. Takig logarithms o both sides we eed to prove l + < l + + < l +. + + The left had iequality is l + + + l >. The lower boud provided by 3 is ot effective here but I 4 it is The right had iequality becomes l + I ad I are ot good eough but I 3 it is + + 4 + > >. 6 + 3 + + + 7 6 + 5 + A ice cosequece of the double iequality is + l + <. + lim a = e. The problem U373 from Mathematical Reflectios is < 3 + >. Prove the followig iequality holds for all positive itegers + + + + + + 3 + + 3 + + Proposed by Nguye Viet Hug Haoi Uiversity of Sciece Vietam. < 3. Mathematical Reflectios 7 3
The published solutio is due to Albert Stadler Herrliberg Switzerlad usig the iequality I. k= + + + + k = k= = exp exp = exp + kk + l + k= k= kk + kk + k k + e < 3. k= = exp + Now we wat to show how applyig I we obtai a refiemet. We eed to show that Or equivaletly By I it is eough to prove that k= k= + < 6. kk + l + < l 6. kk + k= kk + + k + k + k= l 6. The first series telescopes to ad the secod oe ca be foud from the series for z cotπz for a appropriate z the value usig Wolfram Alpha is 7π π tah k + k + = 7.68687. k= Fially.68687 <.79759 = l 6. Notice that e.68687 5.43 < 5.4365 e. To fiish let us see our solutio to problem U385 the before metioed limit. The value of the limit is e. Deotig a = + e Mathematical Reflectios 7 4
the expressio to fid the limit is a a. a It oly remais to show that the expressio iside parethesis ted to. We shall use the double iequality a + a. The lower boud is because the sequece a is icreasig the upper boud is hard ad is true because the double iequality We obtai To fiish ote that ad doig x = / we have By L Hopital s rule. + e < a < + + e. a a < 4 4 4 < +. lim / = lim / x lim x x = lim x x =. Refereces [] G. Pólya - G. Szegő. Problems ad Theorems i Aalysis. Vol. I. Spriger 998. [] Problem 98. Nouv. Als. Math. Ser. Vol. p. 48. 87 [3] C. Moreau. Nouv. Als. Math. Ser. Vol. 3 p. 6. 874 [4] Flemmig Topsoe. Some bouds for the logarithmic fuctio. Uiversity of Copehage. available o the Iteret. Robert Bosch USA. bobbydrg@gmail.com Mathematical Reflectios 7 5