Foolig Newto s Method You might thik that if the Newto sequece of a fuctio coverges to a umber, that the umber must be a zero of the fuctio. Let s look at the Newto iteratio ad see what might go wrog: f = + f ( ) ( ) Most tetbooks give eamples where the Newto sequece gets stuck(oscillates), hits a horizotal taget ad fails, or simply coverges to a differet zero tha the oe iteded, but I do t see tetbooks give eamples of Newto sequeces covergig to ozeros. Normally the Newto sequece { } coverges to a umber L ad the fuctio ad its derivative are cotiuous, so we ca let i the Newto formula to coclude that f ( L) L= L f L Assumig that f ( L) ( L) ( L) ( ) f f =., we coclude that the Newto sequece coverges to a zero of f. If we ca get a Newto sequece { } to coverge to a umber L with the property that ( ) diverges to ±, the L might ot be a zero of f. { f } 3
Cosider the fuctio si ( ) ; f ( ) =, ad let s start the ; = sequece with =. 3 a) Fid a formula for the Newto sequece, ad verify that it coverges to a ozero of f. b) Fid a formula for f ( ) ad determie its behavior as.
A Stirlig-like Iequality Stirlig s asymptotic approimatio! e comes from the iequality <! < + e e. Let s use some elemetary calculus to derive a weaker iequality: = i= I. l (! ) l( i) From the two graphs, you ca deduce the followig double iequality: + l d < l (! ) < l d. 3 + Itegrate the left ad right sides, epoetiate, ad complete the iequality: e <! < e +.
II. Fid the iterval of covergece of the power series = edpoits. (Most tetbooks just ask for the radius of covergece!)!. Use part I. for the III. a) If k is a positive iteger, fid the radius of covergece of the power series k (! ) ( k)! =. b) If k = check the edpoits. c) If k, use the result of I. to check the edpoits.
Evaluatig Proper/Improper Itegrals with little or o Itegratio. l l I. For the improper itegral d +, let s look at the two cases: d + ad l For d +, l < < = 3, so it s coverget by compariso. + + l d. + For l l l d, + + + = l absolutely coverget by compariso., but l l d is coverget. So d + is Use the substitutio u = to fid its value. II. Evaluate l d usig the substitutio ( + )( + + ) u =. {Hit: z z = z.} For l l d, < ( + )( + + ) ( )( ) ( )( ) + + + + + + For l l l d, ( + )( + + ) ( )( ) ( )( + + + + + + ) = l
III. If you use the substitutio u = i the itegral d, you arrive at u u d = du = ( ) du = du. Is it okay to coclude that u u u u d =? Eplai. IV. a) Use the substitutio u = alog with the idetities si = cos cos = si to evaluate the defiite itegral si d. cos + si ad b) Evaluate the defiite itegral ( si ) d for a positive iteger. ( cos ) + ( si ) V. Evaluate si d usig the substitutio u + cos. ad cos( ) = cos. = ad the idetities si( ) = si
f d f d by showig that VI. Show that if f is cotiuous the ( si ) = ( si ) f ( si ) d= usig the substitutio symmetry. u =, si + = cos, ad
Limit Problems si I. What happes if you try L Hopital s Rule o lim +? si Fid lim + >. si by cosiderig the iequality + + + which is valid for II. + si lim.{see problem I.} + c III. Fid the value of c so that lim = 9. c IV. Fid a simple formula for lim b b b b b, for b >. V. Fid lim si. L Hopital s Rule wo t work, so try somethig else. ta VI. Fid the followig limits: e l a) lim b) e l lim
VII. Fid lim ( + )( + ) ( + ) by observig the followig: ( + )( + ) ( + ) l = l ( + ) + l( + ) + + l( + ) l l ( ( ) ) l ( ( ) ) l ( ( = ) ) l + + + + + + = l( ) l ( ) l( ) l l l l + + + + + + + + + terms = l ( ) l ( ) l ( ) + + + + + + The last epressio is a Riema sum of some defiite itegral. VIII. The alteratig series does it coverge to? = ( ) + coverges by the Alteratig Series Test, but what Let s look at the eve partial sums: S = + + 3 4 + + + + + + = + + + + 3 3 4 3 S Solvig the previous equatio for S, we get Fid S = + + + + + + + + 3 3 = + + + + + + ( ) So = lim S = lim + + +. + + = lim + + +, ad you ll kow the sum of the series. + +
Method : Calculate lim + + + + + + by rewritig it as lim + + + ad idetifyig it as a defiite itegral. + + + Method : + + + + From the pictures you get IX. Telescopers = ( ) + a) ( ) + + + d < + + + < d + + + b) + {Hit: ( ) + = + = +.} c) ta {Hit: + + = y ta ta y= ta + y, choose ad y carefully.}
I. = ( l ) l {Hit: For Assorted Series e l > e, ( ) ( ) l l > e.} II. {Try somethig like I.} ( l( l )) l = 3 III. a) Show that + + + + =. { } b) Show that if { a } is a sequece of positive umbers, the if l( ) { a } is decreasig. I other words, show that if l( a ) l( a) a is decreasig, the +, the a + a. c) For >, show that l( + ). {Hit: l( ) + = dt + t, ad + t.} + d) Show that a = l is a decreasig sequece by showig that f ( ) = l + l has a egative derivative. {Hit:Use part c).}
e) Determie whether the alteratig series ( ) + + + is absolutely = coverget, coditioally coverget, or diverget usig the previous results. IV. a) Startig with = ad get e e d =, you get that! = = = ad fid the sum of a series. +! d e + =. Now itegrate from = to! =. Evaluate the itegrals o both sides of the equatio b) You ca verity the sum you foud i part a) by oticig that. So fid the sum ( ) ( ) ( ) ( ) ( ) ( ) + + = = = ( + )! +! +! +! +! = = = = of this telescopic series ad verify the previous result.
The Goat/Cow Grazig i the Grass/Seaweed Problem I. Suppose that after a strig is woud clockwise aroud a circle of radius a, its free ed is at the poit A( a,). Now the strig is uwoud, always stretched tight so the uwoud portio TP is taget to the circle at T. The set of poits traced out by the free ed of the strig is called the ivolute of the circle. T( acos t, asit ) t t t P at at sit (?,?) at cost T t P A a, ( ) Fid the parametric equatios of the ivolute of the circle. = y =
II. Suppose the circle i the previous problem represets the cross-sectio of a cylidrical water tak of radius a, ad the strig is a rope of legth a. The rope is achored at the poit B opposite poit A. If the other ed of the rope is tied to a cow, let s eamie the regio that ca be grazed by the cow. Here is a diagram showig the rope i various positios: Q P B A S The boudary of the grazig regio ca be broke dow ito three pieces: APQ is a portio of the ivolute of the circle, QR is a semicircle, ad RSA is the reflectio across the -ais of a portio of the ivolute. Fid the legth of the boudary of the grazig regio. β L = () t + y () t dt α R
III. Fid the area of the grazig regio. b β yd = y() t () t dt ; curve ad -ais a α Area = b β dy = () t y () t dt ; curve ad y-ais a α
IV. Now suppose that a sea cow(maatee) is tied to a poit o the surface of a sphere of radius a by a rope of legth a. Try to fid the surface area ad the volume of the grazig regio of the sea cow. Volume b β = = a b () () yd yt t dt discs α β ; () () () ydy= t y t y t dt; shells a α β () () () Surface Area = y t t + y t dt α
V. Now suppose that the rope i the previous problem has legth a ad is achored at the poit A before beig woud completely aroud the tak. R S P A APQ is a portio of the ivolute, QR is a semicircle, ad RSA is a reflectio of a portio of the ivolute. Q Attempt all the previous calculatios: Legth, area, surface area, ad volume.
Iteratio ad More Grazig I. Aalyze the followig recursively defied sequeces usig a cobweb diagram: a) a = 6, a+ = 3a b) a = 8, a+ = 3a lima = lima = c) + a = 5, a = 6 + a d) = 6, + = 5 a a a lima = lima =
( ) a + e) a =, a + = f) 4 ( a ) + a = 4, a + = 4 lima = lima = g) Determie the covergece or divergece of the series ( ) 3 a 5 recursively by the followig: ; a a = a + = {Hit: Determie lima by cob-web aalysis.} = a whose terms are defied
II. A farmer has a feced circular pasture of radius a ad wats to tie a cow to the fece with a rope of legth b so as to allow the cow to graze half the pasture. How log should the rope be to accomplish this? a b a The legth of the rope, b, must be loger tha a ad shorter tha a, i.e. a< b< a. To fid the area of the grazig regio, we ca use polar coordiates: si b ( ) a The grazig area b si b ( ) si a ( ) a = 4a si θdθ + b dθ 4a si d b d = θ θ + θ. b si ( ) b si a ( a )
a We wat this to equal half the pasture area which is, so we get the equatio b si ( a ) a 4a si θdθ + b dθ =. If we multiply both sides by ad perform the a b si ( ) a itegratios, we arrive at the equatio a) Verify the previous equatio. b b b b b 4 si 4 + =. a a a a a b If we let =, we get the simplified equatio ( 4 ) si 4 + =, ad a we re lookig for the solutio, with < <. Here s a plot of the leftside ad rightside of the equatio: ( 4 ) si 4 + solutio If we rearrage the equatio, we ca produce a sequece that will coverge to the solutio: = + 4 ( 4 ) si
+ 4 ( 4 ) si = Let =, ad + + 4 ( 4 ) si =. + 4 ( 4 ) si =. b) Complete the cobweb diagram for the recursive sequece.
Here are the first 4 terms of the sequece geerated by Ecel:.363 3.3795 4.568 5.5558 6.5749 7.584 8.5854 9.5865.587.587.587 3.5873 4.5873 So the rope legth, b, should be approimately.5873a.