MTH 146 Class 16 Notes

Transcription

1 MTH 46 Clss 6 Notes 0.4- Cotiued Motivtio: We ow cosider the rc legth of polr curve. Suppose we wish to fid the legth of polr curve curve i terms of prmetric equtios s: r f where b. We c view the cos si x f y f. We kow from sectio 0.2 tht the desired legth will be give by: b L We c simplify this formul quite bit by doig the followig clcultio: 2 2 dx dy d. d d dx dy dr cos r si dr si r cos r dr d d d d d. This leds to the followig formul. Formul: The legth of curve with polr equtio r f where b, is b 2 2 dr L r d d Grded Exmple: Fid the legth of the crdioid give by r si Solutio: First, we ote tht d obti: 2 0. where 0 2. You my use clcultor to pproximte the vlue of the defiite itegrl. dr cos d Now, we pply the formul tht we derived bove si cos 2si si cos L d d 2 2si d 8. 0

2 .- Sequeces Termiology: A sequece c be thought of s list of severl umbers writte i defiite order:,,,,, 2 3 We refer to s the th term of the sequece. We will del exclusively with ifiite sequeces which mes tht every sequece we will cosider will hve ifiite umber of terms. Note: A mthemticl wy to defie ifiite sequece would be: A ifiite sequece is fuctio whose domi is the turl umbers. Note: The book dopts the covetio tht sequeces do ot eed to begi with first term. For exmple, it is possible for sequece to begi with 3. The sequece, 2, 3, c be deoted by: f ( ) where f( ) is some fuctio or. Note: Whe we c deote sequece by f ( ) d we hve formul for f( ) we sy we hve explicit formul for the sequece. Exmple: () Write the first 5 terms of the sequece give by From the explicit formul we see tht, 2, 3, 4, d As side ote, two other wys to represet this sequece re: ,,,,,,, (b) Fid the terms: 3, 4, d 5 of the sequece give by 3 where 3 From the explicit formul we see tht 3 0, 4, d 5 2. Note: There is clerly some mbiguity here o wht we should cll the first term. We will igore this for ow. d

3 (c) Fid the first five terms of the followig sequece which is give recursively s: f f2 f f f2 where 3. This sequece is kow s the Fibocci sequece. First, we kow f d f2. The, by the recursive formul we hve: f3 f2 f 2, f4 2 3, d f Note: Oe big disdvtge to recursively defied sequeces is tht oe must fid certi previous terms i order to fid more terms. For exmple, i the bove exmple, oe must fid the 99 th d 00 th term before fidig the 0 st term. Note: There is wesome explicit formul for the Fibocci sequece (usully 5 5 prove i discrete mth or combitorics). It is: f Grded Exmple: Fid explicit formul for the sequece whose first few terms re: (), 8, 27, 64, 25, Solutio: Oe possibility is 3. (b), 3, 7, 5, 3, 63, 27, Solutio: Oe possibility is 2. (c),,,,,, Solutio: Oe possibility is. Note: I ech prt of this grded exmple oe c esily check whether or ot his/her solutio works by fidig the first few terms of the sequece tht his/her explicit formul defies. Questio to the Clss: If, wht do you thik lim should equl? Aswer: Oe possible swer is lim sice the terms of the sequece get rbitrrily close to.

4 Note: If time permits we will grph this sequece o the clcultor. Defiitio: A sequece hs the limit L d we write lim L or L s if for every 0 there is correspodig iteger N such tht if N the L. Note: If L exists s fiite umber we sy tht the sequece is coverget. Otherwise, the sequece is sid to be diverget. Note: Ituitively, this defiitio is syig tht sequece coverges to L if we c lwys swer the questio: How fr out must oe go to gurtee tht the remiig terms re withi epsilo of L? Exmple: Use the bove defiitio to prove tht if, the lim. We begi with some scrtch work d tryig to uderstd the problem ituitively. Let s begi by skig: How fr out i the sequece must we go to gurtee tht the remiig terms re withi of. To swer this we solve: This shows tht if we go beyod the 999 th term i the sequece, y term will be withi thousdth of oe. Now, we sk the more geerl questio: How fr out i the sequece must we go to gurtee tht the remiig terms re withi of (where 0 ). To swer this we solve:. This shows tht if we go beyod terms, y term will be withi of. Note: x deotes the smllest iteger greter th or equl to x. Now, we re prepred to begi the proof. Proof: Suppose is rbitrry positive umber d if N, the N. We see tht

5 . Thus, N we hve tht lim by defiitio. Defiitio: lim mes tht for every positive umber M there is iteger N such tht if N the M. Note: lim c be defied similrly. Theorem: If lim f ( x) x L d f whe is positive iteger, the lim L. Proof: The proof follows from the defiitio of limit of fuctio. Exmple: Use the bove theorem to show tht if, the lim. x Cosider the fuctio f( x). By l Hospitl s rule we kow x x lim f( x) lim lim. Thus, by the theorem lim. x x x x l Grded Exmple: Fid with full justifictio: lim. l( x) Solutio: Cosider the fuctio f( x). By l Hospitl s rule we kow x l( ) x x l( ) lim f( x) lim lim 0. Thus, by the theorem lim 0. x x x x The ext theorem tells us tht logues of the limit lws i sectio 2.3 hold for sequeces. They c be prove similrly to the limit lws preseted i chpter 2. Theorem: If d. lim b lim lim b 2. lim c c lim b re coverget sequeces d c is costt, the

6 lim b lim limb lim b lim limb p if lim b 0 p 5. lim lim if p 0 d 0. Theorem: If b c for 0 d lim lim c L, the lim b L. Note: This is the logue of the squeeze theorem d its proof is similr to the proof of the squeeze theorem. Recommeded Homework: odd