Continued Fractions continued. 3. Best rational approximations

Transcription

1 Cotiued Fractios cotiued 3. Best ratioal approximatios We hear so much about π beig approximated by 22/7 because o other ratioal umber with deomiator < 7 is closer to π. Evetually 22/7 is defeated by 333/06 ad o ratioal umber with deomiator < 06 is closer to π. The 355/3 is eve closer to π with ot much bigger deomiator ad that s why the aciets used 355/3 for more precise calculatios. These fractios arise from the Cotiued Fractio Algorithm. For α = π the algorithm gives B 0 D 0 = 3 B = 22 D 7 B 2 = 333 D 2 06 B 3 = 355 D For ay irratioal umber α we have see that the Cotiued Fractio Algorithm produces a sequece of ratioal umbers B / which coverges to α. We wat to prove that these ratioal umbers are best approximatios i the followig sese: Best Approximatio Theorem Let α be a positive irratioal umber ad let N. Suppose that p q < B () for some p q N. The q >. Proof: First I claim that B < B (2) i other words that α is closer to B / tha to B /. To prove this claim we will show that the ratio R = α B / α B / has absolute value R <. We have R = α B B = α B C α A = B C α + A

2 2 where i the ext-to-last step we have used the recursio relatios A = B C = from the previous chapter. It follows that D [ ] D B R = (α). (3) C A This matrix is almost the iverse of the matrix f. I fact sice det f = ( ) + we have [ ] [ ] f A B = = ( ) + D B. C C A Hece equatio (3) ca be writte as But recall that Pluggig this ito (4) we get Hece ( ) D [ ] A B R = (α). (4) C [ ] A B α = f (r + ) = (r + ). C ( ) R = r +. Sice 0 < < ad 0 < r + < we have R = ( ) r +. (5) R < as claimed. Now we ca prove the theorem. Let I be the ope iterval betwee B / ad B /. We kow that α I ad that α is closer to B / tha to B / so that p q < B < B. (6)

3 It follows that p/q is also cotaied i I. I particular sice I is a ope iterval we have B / p/q B /. The width of I is give by our crucial equality B B =. Sice p/q I we have > p q B = p qb. (7) q q To see the last iequality ote that the iteger p qb is ozero because p/q B /. The iequalities (7) imply that which proves the theorem. q > There is aother more visual way i which the umbers B / are best approximatios. Cosider the poits (p q) i the xy plae where p q N. We call these lattice poits. Each lattice poit (p q) determies a ratioal umber p/q. Now draw the lie L with equatio y = αx. Sice α is irratioal this lie L misses all the lattice poits (p q). Start your car at the origi (0 0) ad travel up the lie L observig earby lattice poits as they pass by. Every time you see a lattice poit (p q) that gets closer to you tha ay previously see lattice poit write dow the ratioal umber p/q. The remarkable fact is that the list you make will be oe other tha B D B 2 D 2 B 3 D So you ca compute the best ratioal approximatios to α without ever gettig out of your car. The distace from (p q) to L is equal to p qα + α 2. The precise form of this assertio is : If B/D is a ratioal umber such that B Dα < p qα 3

4 4 for all ratioal umbers p/q with 0 < q D the B/D = B / for some. This ca be proved i a similar way as the Best Approximatio Theorem. Exercise CF.9 Fid the ratioal umber p/q closest to π with q 25. Exercise CF.0 I the proof of the Best Approximatio Theorem the ratio / appeared (see Equatio (5)). Prove by iductio that for we have = a + a + a a = [0 a a... a ]. Hit: Use the recursio formula for. The ext exercise has othig to do with this sectio but it has to go somewhere. Exercise CF. Fid the umber α = [ ]. The ext two exercises are a trasitio from Cotiued Fractios to our ext topic: Ifiite Series. Exercise CF.2 Let (s ) be a sequece satisfyig the two coditios. s 2k 2 s 2k s 2k+ s 2k for all k N; 2. lim k (s 2k+ s 2k ) = 0. Prove that (s ) coverges ad that the limit s satisfies for all k N. s 2k s s 2k+ Hit: First show that the eve sequece is bouded above.

5 Exercise CF.3 I class we saw that the sequece s = B / satisfies coditios ad 2 of Exercise CF.2. Here is aother example: Let (a k ) be a oicreasig sequece which coverges to 0. Show that the sequece 5 s = a a ( ) a satisfies coditios ad 2 i Exercise CF.2 ad therefore coverges. This result is called the Alteratig Series Test. It is Theorem 5.3 i the text but the proof you will give here is shorter.