Taylor Polynomials. The Tangent Line. (a, f (a)) and has the same slope as the curve y = f (x) at that point. It is the best

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1 Tylor Polyomils Let f () = e d let p() = Without usig clcultor, evlute f (1) d p(1) Ok, I m still witig With little effort it is possible to evlute p(1) = (144) (178) = 308 However f (1) = e 1, hmm, ot so much Whe we use clcultor we fid e 1 = How did your clcultor determie this vlue? Remember e is irrtiol umber whose vlue is ot kow How does your clcultor compute e π? This situtio is the whole poit of tryig to fid wys to pproimte fuctios tht re ot esily clculted by hd These iclude lot of the fmilir clculus fuctios: e, l, the trig fuctios d their iverses such s si, cos, rcsi, rct, d eve more fmilir fuctios such s d more geerlly It would be ice to be ble to pproimte such fuctios by polyomils, sice polyomils re simpler to clculte The good ews is tht this c be doe (rther simply) usig wht re clled Tylor polyomils Now we do t hve lot of time to sped doig this, so I will just preset few fcts d simple emple Your tet hs more emples d detils which you should look t The Tget Lie Recll tht if we hve fuctio y = f () tht is differetible t =, the the slope of the tget lie t is m = f () To obti the equtio of the tget lie to f t the poit (, f ()) we use this slope m = f () = y f () d solve for y: y = f () Geometriclly the tget lie is the uique lie tht psses through the poit (, f ()) d hs the sme slope s the curve y = f () t tht poit It is the best lier pproimtio to the curve Let s cll the equtio of the tget lie p 1 (), to idicte tht it is polyomil d our first pproimtio Figure 151 shows typicl emple of this situtio To mke the pproimtio eve better we could isist tht both the first d the secod derivtive of our pproimtio mtch those of the origil fuctio Let s cll this ew pproimtio p () Geometriclly this mes tht we wt the slope p 1 () Figure 151: The tget lie p 1 () is the uique lie tht psses through the poit (, f ()) d hs the sme slope s the curve y = f () t tht poit It is the best lier pproimtio to the curve

2 mth 131 tylor polyomils d cocvity of p to be the sme s f t = To crete such fuctio we eed to dd qudrtic (squre) term to our first pproimtio p () = f () + f () ( ) + c( ) }{{}}{{} p 1 () qudrtic term where c is some costt But which? We c figure out wht c hs to be usig the followig process Clculte the first d secod derivtives of p p () = () = If we evlute these derivtives t =, we get p () = () = So the first derivtive still mtches the first derivtive of f t We eed the secod derivtive to be f () I other words we eed f () = p () = c = So this mes p () = f () + f () ( ) + f () ( ) Figure 15 shows illustrtes this situtio Let s do oe further pproimtio p 3 tht requires the third derivtive (the jerk) to mtch tht of f t = This time we hve p 3 () = f () + f () ( ) + f () ( ) + d( ) 3 }{{ }}{{} cubic term p () f () p () Figure 15: The fuctio p () is the uique degree two polyomil through the poit (, f ()) tht hs the sme first d secod derivtives (slope d cocvity) s f t tht poit Notice tht the first three derivtives of p 3 re p 3 () = f () + f ()( ) + 3d( ) 3 () = f () + 3 d( ) 1 3 () = If we evlute these derivtives t =, we get 3 () = f () 3 () = f () 3 () = which we wt = f () The first two derivtives still mtch those of f t We eed the third derivtive to be f () I other words, we eed f () = d or d = f ()/ So this mes p 3 () = f () + f () ( ) + f () ( ) + f () ( ) 3 Figure 153 shows tht p () is eve better pproimtio of the origil curve f () p 3 () Figure 153: The fuctio p 3 () is the uique degree three polyomil through the poit (, f ()) tht hs the sme first three derivtives s f there

3 Do you see ptter? If we use tht fct tht! =, the we c write the lst pproimtio s p 3 () = f () + f () ( ) + f ()! ( ) + f () ( ) 3 We could keep goig with these pproimtios After doig this process times we would get degree polyomil whose first derivtives gree with those of f t = DEFINITION 151 (Tylor Polyoils) Assume f hs t lest derivtives The th order Tylor polyomil for f t = is p () = f () + f () ( ) + f ()! ( ) + f () ( ) f () () ( ) The poit is clled the ceter of the epsio p () hs the sme first derivtives s f () t = The th order Tylor polyomil c lso be writte s p () = [Remember tht 0! = 1 d tht f (0) () = f ()] f (k) () ( ) k k! Writig Tylor polyomil usig summtio ottio c t help but brig series to mid I fct the Tylor series for f cetered t is the ifiite degree polyomil p () = f (k) () ( ) k k! Now there re ll sorts of questios we c d will sk bout such epressios (covergece beig the most importt) But it turs out uder the right circumstces tht such series re ot just pproimtios to f, they re ctully equl to f i some itervl roud I other words, f () = p () d this gives us wy to clculte fuctios such s e or l tht we c t ordirily evlute Workig with Tylor Polyomils We wo t sped lot of time clcultig Tylor polyomils (or series) But here s fu emple tht illustrtes the ide EXAMPLE 1501 Let s tke the fuctio f () = e / d let s use = 0 s the ceter Determie the Tylor polyomils p 1 (), p (), d p 3 () Use this ptter to determie the geerl order Tylor polyomil p () for e / The use p 3 () to pproimte e 0 (Be creful, use = 04) SOLUTION By Defiitio 151 the formul for p () with ceter = 0 is give by p () = f (0) + f (0) ( 0) + f (0)! ( 0) + f (0) ( 0) f () (0) ( 0) So we eed to clculte the derivtives of f () = e / d the evlute them t 0 Well, f () = f (0) = f () = f (0) = f () = f (0) = f () = f (0) = f (k) () = f (k) (0) = 3

4 mth 131 tylor polyomils 4 I prticulr, we see tht the first three Tylor polyomils for e / cetered t 0 re d p 1 () = 1 + p () = p 3 () = p () = f (0) + f (0) ( 0) + f (0) = = k 1 1! = 1 + ( 0) + f (0) ( 0) f () (0) ( 0) The grphs of these fuctios re show i the figure o the et pge Notice how the pproimtio to e / improves s the order of the polyomil icreses Let s do the pproimtio of e 0 Sice our fuctio is e / we will eed to use = 04: e 0 = e 04/ p 3 (04) = ! + 04! = = Compre to the clcultor vlue: e 0 = 1140 So p 3 (4) is pretty good estimte I this cse it is esy to see tht the Tylor series for e / is the ifiite degree polyomil p () = 1 + or epressed s series 1 1! +! p () = =0 Ech umber c be plugged ito the series d we c determie whether p () eists (whether the series coverges for this vlue of ) or does ot eist (the series diverges) If we do this for ech d every, we c determie the domi of p () We will soo see tht this c be doe very efficietly Moreover, it is possible to show tht i this cse e / ctully equls p () This gives us mechicl wy to clculte e / The ultimte gol is to represet fmilir fuctios like si d e (which re doe s emples i your tet) s ifiite degree polyomils so tht the vlues of these fuctios c be estimted quickly This is how your clcultor works!

5 f () = e / p 3 () = ! +! p () = ! +! p 1 () = ! 1 p p 3 e / 5 p The first three Tylor polyomils for f () = e / cetered t 0 Notice how the pproimtio improves s icreses EXAMPLE 150 Let s tke the fuctio f () = si d let s use = 0 s the ceter Determie the Tylor polyomils p 1 (),, p 7 () EXAMPLE 1503 Let s tke the fuctio f () = l d let s use = 1 s the ceter Determie the Tylor polyomils p 1 (), p (), p 3 ()