It is often useful to approximate complicated functions using simpler ones. We consider the task of approximating a function by a polynomial.

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1 Taylor Polyomials ad Taylor Series It is ofte useful to approximate complicated fuctios usig simpler oes We cosider the task of approximatig a fuctio by a polyomial If f is at least -times differetiable at a (that is, if f () (a) exists), we defie the th degree Taylor polyomial of f at a by f (k) (a) T (x) = T,a (x) = (x a) k Remark If = 1, the T,a (x) = f(a) + f (a)(x a) is the liear approximatio of f at a Example Fid the degree 4 Taylor polyomial for cos x at 0 We compute Therefore f(x) = f (0) (x) = cos x f(0) = cos(0) = 1 f (x) = si x f (0) = si(0) = 0 f (x) = cos x f (0) = cos(0) = 1 f (3) (x) = si x f (3) (0) = si(0) = 0 f (4) (x) = cos x f (4) (0) = cos(0) = 1 T 4,0 (x) = 4 k=0 f (k) (0) (x 0) k = 1 0! x ! x ! x ! x ! x4 = 1 x2 2! + x4 4! The followig result ca be used to estimate the error i approximatig a fuctio f by the th degree Taylor polyomial for f at a Taylor s Iequality (Error Estimate) If f (+1) (x) K i a closed, bouded iterval I cotaiig a, the K x a +1 f(x) T,a (x) i I ( + 1)! Remark You will ot be tested o Taylor s Iequality If f is ifiitely differetiable at a (that is, if f () (a) exists for all ), the Taylor series of f at a is f () (a) (x a), where x is a variable The Taylor series of f at 0 is sometimes called the Maclauri series of f 1

2 2 Example Fid the Taylor series of e x at 0 We have f(x) = e x f(0) = e 0 = 1 f (x) = e x f (0) = 1 f () (x) = e x f () (0) = 1 Therefore the Taylor series of e x at 0 is f () (0) (x 0) = Example Fid the Taylor series of cos x at 0 x We have f(x) = cos x f(0) = cos(0) = 1 f (x) = si x f (0) = si(0) = 0 f (x) = cos x f (0) = cos(0) = 1 f (3) (x) = si x f (3) (0) = si(0) = 0 f (4) (x) = cos x f (4) (0) = cos(0) = 1 f (5) (x) = si x f (5) (0) = si(0) = 0 f (6) (x) = cos x f (6) (0) = cos(0) = 1 f (7) (x) = si x f (7) (0) = si(0) = 0 f (4k) (x) = cos x f (4k) (0) = cos(0) = 1 f (4k+1) (x) = si x f (4k+1) (0) = si(0) = 0 f (4k+2) (x) = cos x f (4k+2) (0) = cos(0) = 1 f (4k+3) (x) = si x f (4k+3) (0) = si(0) = 0 Therefore the Taylor series of cos x at 0 is f () (0) (x 0) = 1 x2 2! + x4 4! x6 6! + x8 8! ( 1) x 2 = (2)!

3 There is a atural questio to ask about Taylor series: Does the Taylor series of a fuctio coverge to the fuctio? The Taylor series of e x at 0 coverges to e x That is, for every fixed value of x x N = x N = ex Likewise, the Taylor series of si x at 0 coverges to si x, ad the Taylor series of cos x at 0 coverges to cos x We will prove these assertios i the optioal sectio below It ca happe that the Taylor series of f at a does ot coverge to f for ay value of x except x = a For example, if { e 1/x2 if x 0 f(x) = 0 if x = 0 the Taylor series of f at 0 oly coverges to f whe x = 0 It will be useful to memorize the followig 3 e x = si x = cos x = x ( 1) x 2+1 (2 + 1)! ( 1) x 2 (2)!

4 4 Optioal: Covergece of Taylor Series Theorem For ay a the Taylor series of e x at a coverges to e x For ay a the Taylor series of si x at a coverges to si x For ay a the Taylor series of cos x at a coverges to cos x The theorem is a cosequece of the followig lemma Lemma Suppose f has the property that for ay real umbers c ad d with c < d there is a costat K c,d > 0 such that f () (x) K c,d for every x i [c, d] ad every The for ay a the Taylor series of f at a coverges to f Proof of the theorem If f(x) = e x, the f () (x) = e x e d for every x i [c, d] If f(x) = si x, the f () (x) is either si x, cos x, si x, or cos x, hece f () (x) 1 i [c, d] Likewise, if f(x) = cos x, the f () (x) 1 i [c, d] The theorem ow follows immediately from the lemma To prove the lemma, we will eed the followig it formula = 0 for ay real umber r Proof of the it formula Let r be ay give real umber Choose a iteger m such that r m The for all > m we have = r 1 r 2 r m r m + 1 r 1 r ( ) = r m r m! m + 1 r r 1 Sice the squeeze theorem implies Therefore r m m! = r m+1 m! 1 r 1 r m+1 1 m! = r m+1 1 m! = 0, = 0 = 0 Now we are ready to prove the lemma

5 Proof of the lemma Let a be ay real umber We eed to show that for each fixed value of x the f () (a) sequece of partial sums of (x a) coverges to f(x) Fix x The th partial sum is T (x) = k k=0 f (k) (a) (x a) k (That is, the th partial sum of the Taylor series of f at a is the th degree Taylor polyomial of f at a) We write T (x) = (T (x) f(x) + f(x)) = (T (x) f(x)) + f(x) So we just eed to show that (T (x) f(x)) = 0 Choose c ad d such that [c, d] cotais a ad x By assumptio, there is a costat K c,d > 0 such that f (+1) (y) K c,d for every y i [c, d] ad every Taylor s iequality gives I particular, By the it formula, So the squeeze theorem implies T (y) f(y) K c,d y a +1 ( + 1)! T (x) f(x) K c,d x a +1 ( + 1)! K c,d x a +1 ( + 1)! for all y i [c, d] = 0 (T (x) f(x)) = 0 5