第五章微分中值定理和Taylor展开85.1函数极值85.2中值定理定理5.2.1(Rolle).设函数f在[a,b]上连续，在(a,b)内可微，且f(a)=f(b),则存在≤E(a,b),使得f"(E)=0.证明.连续函数f在闭区间[a,b]上可以取到最大值M和最小值m.如果M=m，则f恒为常数，从而f'=0;如果M>m，则由f(a)=f(6)知m与M中至少有一个是被f在内点εE(a,b)处所取得，由Fermat定理，f(s)=0.口定理5.2.2(Lagrange)。设函数于在[a,b]上连续，在（a,b)内可微，则存在EE(a,b),使得F() = =1(), 或 () (a) = f()( - a),b-a证明.令F(n)= (m) -[(a)+ 二(a-a) b-n则F(a)=F(b)=0, F满足定理1的条件.从而日EE(a,b),使得 F(s)=0. 此口即为满足定理要求的6.定理5.2.3(Cauchy).设函数f,g在[a,b]上连续，在（a,b)内可微，且g(r)≠0,VE(a,b).则存在E(a,b),使得f(b) - f(a) -f'(E)g(b) -g(a)g ()证明.由定理1和g ≠0知 g(b)≠g(a).令F(1) = f(a) -[9(a)+ ()-((g(g() - g(a))g(b) - g(a)则F(a)=F(b)=0,F满足定理1的条件,从而日EE(a,b),使得F()=0.E即口为满足要求的点注：令g(a)=，则Cauchy定理可以推出Lagrange定理。以上三个定理均称为微分中值定理定理5.2.4(积分第一中值定理).设f,g在[a,b]上Riemann可积，且g(a)不变号，则日，inf,f()≤μ≤ sup,f(r),rE[a,b]re[a,b]1
1ÊÙ ©¥½nÚ Taylor Ðm §5.1 ¼ê4 §5.2 ¥½n ½n 5.2.1 (Rolle). ¼ê f 3 [a, b] þëY§3 (a, b) S§ f(a) = f(b). K3 ξ ∈ (a, b), ¦ f 0 (ξ) = 0. y². ëY¼ê f 34«m [a, b] þ± M Ú m. XJ M = m, K f ð~ê§l f 0 ≡ 0; XJ M > m, Kd f(a) = f(b) m M ¥k´ f 3S: ξ ∈ (a, b) ?¤§d Fermat ½n§f 0 (ξ) = 0. ½n 5.2.2 (Lagrange). ¼ê f 3 [a, b] þëY, 3 (a, b) S, K3 ξ ∈ (a, b), ¦ f 0 (ξ) = f(b) − f(a) b − a , ½ f(b) − f(a) = f 0 (ξ)(b − a). y². - F(x) = f(x) − [f(a) + f(b) − f(a) b − a (x − a)] K F(a) = F(b) = 0, F ÷v½n 1 ^. l ∃ ξ ∈ (a, b), ¦ F 0 (ξ) = 0. d ξ =÷v½n¦ ξ. ½n 5.2.3 (Cauchy). ¼ê f, g 3 [a, b] þëY, 3 (a, b) S, g 0 (x) 6= 0, ∀ x ∈ (a, b). K3 ξ ∈ (a, b), ¦ f(b) − f(a) g(b) − g(a) = f 0 (ξ) g 0(ξ) . y². d½n 1 Ú g 0 6= 0 g(b) 6= g(a). - F(x) = f(x) − [f(a) + f(b) − f(a) g(b) − g(a) (g(x) − g(a))] K F(a) = F(b) = 0, F ÷v½n 1 ^, l ∃ ξ ∈ (a, b), ¦ F 0 (ξ) = 0. ξ = ÷v¦:. 5: - g(x) = x, K Cauchy ½n±íÑ Lagrange ½n. ±þn½nþ¡ ©¥½n. ½n 5.2.4 (È©1¥½n). f, g 3 [a, b] þ Riemann È, g(x) ØCÒ, K ∃ µ, inf x∈[a,b] f(x) ≤ µ ≤ sup x∈[a,b] f(x), 1
2第五章微分中值定理和Taylor展开使得f(a)g(a)dz =g(a)dr证明.不失一般性，可设g(a)≥0.则inf f-g()≤f()g()≤supf-g(r)→inf f:g(a)d≤/f(a)g(r)da≤ sup f :g(a)da上式说明，如果Jg(a)da=0,则Jf(a)g(a)da=0,此时定理当然成立如果g(z)d>0,则令" f(a)g(a)drJ g(r)da则显然有inf fμ≤supf.定理得证口注： (1)当g()=1时，J f()=μ (b-a)(2)特别地，如果f连续，则由介值定理，日E[a,b]，使得f(s)=μ,此时f(r)g(a)da = f(s). / g(n)de.定理5.2.5（积分第二中值定理)．设于在[a,b]上Riemann可积(1)如果g在[a,上单调减，且g(a)≥0,VrE[a,b],则3E[a,b],使得f(r)g()dz = g(a).f(r)dr;(2)如果g在[a,b] 上单调增，且 g(α)≥0, V E[a,b],则日n E[a,bl],使得 (n)g(n)dr = g(6) / (n)dr;(3)一般地，如果g在[a,b]上单调函数，则日（E[a,l]，使得[~ f(n)g(n)da = g(a) - / f()d + g() /~ f(n)dr
2 1ÊÙ ©¥½nÚ Taylor Ðm ¦ Z b a f(x)g(x)dx = µ Z b a g(x)dx. y². Ø5, g(x) ≥ 0. K inf f · g(x) ≤ f(x)g(x) ≤ sup f · g(x) =⇒ inf f · Z b a g(x)dx ≤ Z b a f(x)g(x)dx ≤ sup f · Z b a g(x)dx þª`², XJ R b a g(x)dx = 0, K R b a f(x)g(x)dx = 0, d½n,¤á. XJ R b a g(x)dx > 0, K- µ = R b a f(x)g(x)dx R b a g(x)dx Kw,k inf f ≤ µ ≤ sup f. ½ny. 5: (1) g(x) ≡ 1 , R b a f(x) = µ · (b − a). (2) AO/, XJ f ëY, Kd0½n, ∃ ξ ∈ [a, b], ¦ f(ξ) = µ, d Z b a f(x)g(x)dx = f(ξ) · Z b a g(x)dx. ½n 5.2.5 (È©1¥½n). f 3 [a, b] þ Riemann È. (1) XJ g 3 [a, b] þüN~, g(x) ≥ 0, ∀ x ∈ [a, b], K ∃ ξ ∈ [a, b], ¦ Z b a f(x)g(x)dx = g(a) · Z ξ a f(x)dx; (2) XJ g 3 [a, b] þüNO, g(x) ≥ 0, ∀ x ∈ [a, b], K ∃ η ∈ [a, b], ¦ Z b a f(x)g(x)dx = g(b) · Z b η f(x)dx; (3) /, XJ g 3 [a, b] þüN¼ê, K ∃ ζ ∈ [a, b], ¦ Z b a f(x)g(x)dx = g(a) · Z ζ a f(x)dx + g(b) · Z b ζ f(x)dx.
3$5.3L'Hospital法则证明.我们只证明一个特殊情形：于连续，9连续可微，9’≤0.令F(）=f(t)dt,则F"=f(a).从而f(a)g()da =F (r)g(r)daFg'da= F(r)g(r)临 -= F(b) · g(b) - μ-g'da= F(b) · g(b) - μ(g(6) - g(a))这说明inf F-g(a)≤f(r)g(r)dr≤supF.g(a)因此,存在[a,b],使得f(a)g(a)dt = F()g(a) = g(a) - [f(r)drg5.3L'Hospital法则Motivation：设f.g为函数，求极限f(a)lim+10 g(r)困难的情形：(1) →o 时, f() →0, g() →0,(%) 型;(2)→0时, f()→80, g()→80,() 型定理5.3.1(L'Hospital)：设fg在(a,b)内可导，且g(a)≠0,VE(a,b).又设lim f(r) = 0 = lim g(z)如果极限f'(α)lim+a+ g'(r)存在(或为00),则f'(a)f(a)limlim+a+ g(r)r→a+ g'(r)证明.补充定义f(a)=g(a)=0,则于在[a,b)上连续.由 Cauchy中值定理V(a,b),(a,),使得f(a)f(r) - f(a)f'(6)g()g(z)g() - g(a)
§5.3 L’Hospital {K 3 y². ·y²AÏ/: f ëY, g ëY, g 0 ≤ 0. - F(x) = R x a f(t)dt, K F 0 = f(x). l Z b a f(x)g(x)dx = Z b a F 0 (x)g(x)dx = F(x)g(x)| b a − Z b a F g0 dx = F(b) · g(b) − µ · Z b a g 0 dx = F(b) · g(b) − µ(g(b) − g(a)). ù`² inf F · g(a) ≤ Z b a f(x)g(x)dx ≤ supF · g(a). Ïd, 3 ξ ∈ [a, b], ¦ Z b a f(x)g(x)dx = F(ξ)g(a) = g(a) · Z ξ a f(x)dx. §5.3 L’Hospital {K Motivation: f, g ¼ê, ¦4 limx→x0 f(x) g(x) (J/: (1) x → x0 , f(x) → 0, g(x) → 0, ( 0 0 ) .; (2) x → x0 , f(x) → ∞, g(x) → ∞, (∞ ∞) . ½n 5.3.1 (L’Hospital). f, g 3 (a, b) S, g(x) 6= 0, ∀ x ∈ (a, b). q lim x→a+ f(x) = 0 = lim x→a+ g(x) XJ4 lim x→a+ f 0 (x) g 0(x) 3 (½ ∞), K lim x→a+ f(x) g(x) = lim x→a+ f 0 (x) g 0(x) y². Ö¿½Â f(a) = g(a) = 0, K f 3 [a, b) þëY. d Cauchy ¥½n, ∀ x ∈ (a, b), ∃ ξ ∈ (a, x), ¦ f(x) g(x) = f(x) − f(a) g(x) − g(a) = f 0 (ξ) g 0(ξ) .
4第五章微分中值定理和Taylor展开当E-→a+时.→a+.从而f()→ limf'(a)g'(s)T-a+ g'()因此f(r)f'(r)limlimi+ g'(α)a+g()注：(1)如果仍有f(a+）=g(α+）=0,则可利用二次导数继续求：f(μ)f'(r)f"(r)limlimlimra+ g(a)=→α+ g (a)→a+ g"(r)高阶导数的情形类似(2)区间(a,b)换成(-0,b)或(a,80)时，有类似结论f(α)f'(a)limlim-00 g()0g()-f(r)f'(r)limlimF00g()+o0 g'(r)（作变量代换工=1即可）定理5.3.2(L'Hospital).设f,g在(a,b）内可导，g(r)≠0且lim g(a) = 00,-0如果极限f(r)lima+ g()存在（或为80),则f'(r)f(r)= limlimr-a+ g(r)r-a+ g'(μ)证明.我们对f'(a)I = lim20,>0,使得当(a,a+)时1-=<f(a)<I+E.g'(α)取c=a+，则由Cauchy中值公式，日E（a,c)，使得f'()f(r) f(a)1-<<l+e, rE(a,c)g()g(r) - g(a)
4 1ÊÙ ©¥½nÚ Taylor Ðm x → a + , ξ → a +, l f 0 (ξ) g 0(ξ) → lim x→a+ f 0 (x) g 0(x) , Ïd lim x→a+ f(x) g(x) = lim x→a+ f 0 (x) g 0(x) . 5: (1) XJEk f 0 (a +) = g 0 (a +) = 0, K|^gêUY¦: lim x→a+ f(x) g(x) = lim x→a+ f 0 (x) g 0(x) = lim x→a+ f 00(x) g 00(x) , p ê/aq. (2) «m (a, b) ¤ (−∞, b) ½ (a, ∞) , kaq(Ø: lim x→−∞ f(x) g(x) = lim x→−∞ f 0 (x) g 0(x) , lim x→+∞ f(x) g(x) = lim x→+∞ f 0 (x) g 0(x) . (Cþ x = 1 t =). ½n 5.3.2 (L’Hospital). f, g 3 (a, b) S, g(x) 6= 0 lim x→a+ g(x) = ∞, XJ4 lim x→a+ f(x) g(x) 3 ( ½ ∞), K lim x→a+ f(x) g(x) = lim x→a+ f 0 (x) g 0(x) . y². ·é l = lim x→a+ f 0 (x) g 0(x) 0, ∃ δ > 0, ¦ x ∈ (a, a+δ) l − ε < f 0 (x) g 0(x) < l + ε. c = a + δ 2 , Kd Cauchy ¥úª, ∃ ξ ∈ (a, c), ¦ l − ε < f(x) − f(a) g(x) − g(a) = f 0 (ξ) g 0(ξ) < l + ε, x ∈ (a, c).
5$5.3L'Hospital法则因为→a+时，g()→80,故在上式中令→a，得f'(r)1-es lim inf f<limsup<l+eg(r)g(r)-T0因为是任意取的，令→0,就得f'(a)lim inf f'()limsup=1g(r)g(r)t→a+z-a+注和定理1的注记一样，有类似的注记（略）例5.3.1.求极限r-sinlimr3T-0解sina11-cosra-sinr= limlim= lim6T33r22006元20例5.3.2.设f"(ro）存在，求极限f(ro +h)-f(zo)-f(ro)hlim :h2h-0解f(to+h)-f(ro)-f(ro)hh = lim (c + h) - (c) limf"(ro)h22hh-0h_0例5.3.3.设1E(0,1),an+1=n(1-n),n=1,2,...证明n.n→1.证明易见0<an<1,Vn≥1.从而Tn+1= Tn·(1 -In)<an设n→a,则在上式中令n→8,得a=a.(l-a)→a=0.即m→0.从而111lim (-)= lim=1001-InEnn=00n+1-因此1111(lim-)= 1lim1+nTn1n-0onann1-k=1这说明lim n.an =1
§5.3 L’Hospital {K 5 Ï x → a+ , g(x) → ∞, 3þª¥- x → a +, l − ε ≤ lim x→a+ inf f 0 (x) g 0(x) ≤ lim x→a+ sup f 0 (x) g 0 (x) ≤ l + ε. Ï ε ´?¿, - ε → 0, Ò lim x→a+ inf f 0 (x) g 0(x) = lim x→a+ sup f 0 (x) g 0(x) = l. 5 Ú½n 1 5P, kaq5P (Ñ). ~ 5.3.1. ¦4 limx→0 x − sin x x 3 . ) lim x→0 x − sin x x 3 = lim x→0 1 − cos x 3x 2 = lim x→0 sin x 6x = 1 6 . ~ 5.3.2. f 00(x0) 3, ¦4 lim h→0 f(x0 + h) − f(x0) − f 0 (x0)h h 2 . ) lim h→0 f(x0 + h) − f(x0) − f 0 (x0)h h 2 = lim h→0 f 0 (x0 + h) − f 0 (x0) 2h = 1 2 f 00(x0). ~ 5.3.3. x1 ∈ (0, 1), xn+1 = xn · (1 − xn), n = 1, 2, · · ·. y² n · xn → 1. y² ´ 0 < xn < 1, ∀ n ≥ 1. l xn+1 = xn · (1 − xn) < xn. xn → a, K3þª¥- n → ∞, a = a · (1 − a) ⇒ a = 0. = xn → 0. l limn→∞ ( 1 xn+1 − 1 xn ) = limn→∞ 1 1 − xn = 1 Ïd limn→∞ 1 n · ( 1 xn − 1 x1 ) = limn→∞ 1 n · nX−1 k=1 ( 1 xn − 1 xn ) = 1, ù`² limn→∞ n · xn = 1.
6第五章微分中值定理和Taylor展开例5.3.4.设于在（a,+o）中可微，则(1)如果lim·f'()=1,则limf(a)=+00.(2)如果α>0,使得lim (q·f(r) +· f'(r)) =β,则limg (a) = .ar证明.(1）→+o0时，log→+80，故f(a)f'(a)lim2 = lim a·f()=1lim→00 log-n-→00(2)当α>0时,→+o0(→+),故raofra-l.af+a".flim= lim二limf(a)raQ·2Q-1n-001lim(a·(a)+-)==00a$5.4Taylor展开Motivation：我们研究一元函数f的性态(1)如果f(a)在o处连续,则f()-f(ro) =o(1), (→ro)即，在o附近于可用常值函数f(ro）逼近(2)如果f()在o处可微，则f()-[f(ro) + f(ro)(r-o) =o(-o), (→ o),即，在To附近f可用线形函数L逼近L(r) = f(ro) + f'(ro) - (r - ro)(3)如果F"(ro）存在，则由85.2例2f() -[f(ro) + f(ro) (-) +"(ro) (-o)")=o( -o)"), (→ o),即f在ro附近可以用2次多项式逼近.一般地，我们有
6 1ÊÙ ©¥½nÚ Taylor Ðm ~ 5.3.4. f 3 (a, +∞) ¥. K (1) XJ lim n→+∞ x · f 0 (x) = 1, K limn→∞ f(x) = +∞. (2) XJ ∃ α > 0, ¦ limn→∞ (α · f(x) + x · f 0 (x)) = β, K limn→∞ f(x) = β α . y². (1) x → +∞ , log x → +∞, limn→∞ f(x) log x = limn→∞ f 0 (x) 1 x = limn→∞ x · f 0 (x) = 1 (2) α > 0 , x α → +∞ (x → +∞), limn→∞ f(x) = limn→∞ x α · f x α = limn→∞ x α−1 · αf + x α · f 0 α · x α−1 = limn→∞ 1 α (α · f(x) + x · f) = β α . §5.4 Taylor Ðm Motivation: ·ïļê f 5. (1) XJ f(x) 3 x0 ?ëY, K f(x) − f(x0) = o(1), (x → x0), =, 3 x0 NC f ^~¼ê f(x0) %C. (2) XJ f(x) 3 x0 ?, K f(x) − [f(x0) + f 0 (x0)(x − x0)] = o(x − x0), (x → x0), =, 3 x0 NC f ^/¼ê L %C, L(x) = f(x0) + f 0 (x0) · (x − x0) (3) XJ f 00(x0) 3, Kd §5.2 ~ 2, f(x) − [f(x0) + f 0 (x0) · (x − x0) + 1 2 f 00(x0) · (x − x0) 2 ] = o((x − x0) 2 ), (x → x0), = f 3 x0 NC±^ 2 gõª%C. /, ·k
7g5.4Taylor展开定理5.4.1（带Peano余项的Taylor公式).设f在zo处n次可导，则(*) f() =f(ro)+f(ro).(-o)+f"(ro) - (a- zo)? +..()(o) (o)"+o()"), (0)证明.用数学归纳法.n=1.2已经在前面证明.设n=k时（*）成立，则n=k+1时，由假设，fk+1(ro)存在，此时f(zo）在ao处k次可导，由归纳假设-f"(ro) (r - ro)? + ..f'(r) = f'(ro)+f"(ro)-(-ro)+ :21(+1)(ro) (0)+o(o)), (0),!从而由L'Hospital法则f() - [5(0) + f(c0) ( - 20) + .+ ( ( - 20)++)lim(r - ro)*+1) -I(c)+ "(c0) (α-20)++ ( (-0)lim(k + 1)(r- ro)k= 0.即f(a) = f(ro)+f (ro) (-ro)+f"(ro) -(-ro)?+...f(k+1)(ro) - ( - zo)h+1 + o((r - To)*+1), (a → ro).(k + 1)!口从而（*）对n=k+1也成立.由数学归纳法，（*）对任意的n都成立记f(n)(ro)Rn(a) = Rn(ro, z) = f(r) - [f(ro) + f'(ro) (r - ro) + .. +(α-To)"],n!称Rn为Taylor展开的余项.如果f在o处有n阶导数，则Rn(r)=o((r-ro)").(Peano余项)如果条件更强些，则有定理5.4.2(Taylor).设于在开区间(a,b)中有直到n+1阶导数，o,E(a,b)，则日EE(,ro)(或(r,ro))以及（E(ro,)(或(,ro),使得余项可表示为1(n+1)()-(-ro)n+1,(Lagrange余项）Rn(r) = 7(n+1)!以及Ra(1)=f(n+1)(C) (-C)" (μ- zo), (Cauchy 余项)一
§5.4 Taylor Ðm 7 ½n 5.4.1 ( Peano { Taylor úª). f 3 x0 ? n g, K (∗) f(x) = f(x0) + f 0 (x0) · (x − x0) + 1 2!f 00(x0) · (x − x0) 2 + · · · + 1 n! f (n) (x0) · (x − x0) n + o((x − x0) n ), (x → x0) y². ^êÆ8B{. n = 1, 2 ®²3c¡y². n = k (∗) ¤á, K n = k + 1 , db, f k+1(x0) 3, d f 0 (x0) 3 x0 ? k g, d8Bb, f 0 (x) = f 0 (x0) + f 00(x0) · (x − x0) + 1 2!f 000(x0) · (x − x0) 2 + · · · + 1 k! f (k+1)(x0) · (x − x0) k + o((x − x0) k ), (x → x0), l d L’Hospital {K limx→x0 f(x) − [f(x0) + f 0 (x0) · (x − x0) + · · · + f (k+1)(x0) (k+1)! · (x − x0) k+1] (x − x0) k+1 = limx→x0 f 0 (x) − [f 0 (x0) + f 00(x0) · (x − x0) + · · · + f (k+1)(x0) k! · (x − x0) k ] (k + 1)(x − x0) k = 0. = f(x) = f(x0) + f 0 (x0) · (x − x0) + 1 2!f 00(x0) · (x − x0) 2 + · · · + 1 (k + 1)!f (k+1)(x0) · (x − x0) k+1 + o((x − x0) k+1), (x → x0). l (∗) é n = k + 1 ¤á. dêÆ8B{, (∗) é?¿ n Ѥá. P Rn(x) = Rn(x0, x) = f(x) − [f(x0) + f 0 (x0) · (x − x0) + · · · + f (n) (x0) n! · (x − x0) n ], ¡ Rn Taylor Ðm{. XJ f 3 x0 ?k n ê, K Rn(x) = o((x − x0) n ). (Peano {) XJ^r , Kk ½n 5.4.2 (Taylor). f 3m«m (a, b) ¥k n+1 ê, x0, x ∈ (a, b). K ∃ ξ ∈ (x, x0)( ½ (x, x0)) ±9 ζ ∈ (x0, x)( ½ (x, x0)), ¦{L« Rn(x) = 1 (n + 1)!f (n+1)(ξ) · (x − x0) n+1 , (Lagrange {) ±9 Rn(x) = 1 n! f (n+1)(ζ) · (x − ζ) n · (x − x0), (Cauchy {)
8第五章微分中值定理和Taylor展开证明.令"f(k)-(r - t)k,F(t) =f(t) +/对t求导，得- f(n+1)(t) (r - t)n(**) F'(t) =n!由F的构造，有F(r) - F(ro) = Rn(μ)由微分中值公式,3（=+(-0),0<1,使得Rn(μ) = F(C)·(r-ro)f(n+)(C) (r-C)" (μ-zo)，(Cauchy余项)n!再由微分中值公式，日s,使得(取G(t)=(r-t)n+1)f(n+1)(6)Rn(r)F'()F(r) -F(ro)(r - ro)n+I 0 -(- ro)n+I (n + 1)!(n+1)(-)n即1f(n+1)($)-(-ro)n+1,(Lagrange余项)Rn(μ) =(n+1)!口这就得到了余项的两种表达式，如果关于的条件更强一些，例如f是n+1阶连续可微的，则由（**）知，F可积，由微积分基本公式，[" f(n+)() . (r -t)"dt.Rn(a) = F() - F(ro) = /F(t)dt =n!这是余项的一个精确积分表示。此时，由于（一t)”不变号，由第一积分中值定理，FE(ro,)(或(,o),使得fn+1($) f(n+1)(E) :(r-t)"dt =zo)n+1Rn(r) =n!(n + 1)!这也就是Lagrange余项.同理可得Cauchy余项如果f在o附近无穷次可微，则称形式和()(-0)"2n!为f在o处的Taylor展开.Taylor展开在ao=0的特殊情形也称Maclaurin展开公式.如果limRn(r)=0,则记E f(n+)(r0)f(r) =(r-ro)"(n+ 1)!2-0此时称于的Taylor展开收敛到自身
8 1ÊÙ ©¥½nÚ Taylor Ðm y². - F(t) = f(t) +Xn k=1 f (k) k! (x − t) k , é t ¦, (∗∗) F 0 (t) = 1 n! f (n+1)(t) · (x − t) n , d F E, k F(x) − F(x0) = Rn(x). d©¥úª, ∃ ζ = x0 + θ(x − x0), 0 < θ < 1, ¦ Rn(x) = F 0 (ζ) · (x − x0) = 1 n! f (n+1)(ζ) · (x − ζ) n · (x − x0), (Cauchy {) 2d©¥úª, ∃ ξ, ¦ ( G(t) = (x − t) n+1) Rn(x) (x − x0) n+1 = F(x) − F(x0) 0 − (x − x0) n+1 = F 0 (ξ) (n + 1)(x − ξ) n = f (n+1)(ξ) (n + 1)! , = Rn(x) = 1 (n + 1)!f (n+1)(ξ) · (x − x0) n+1 , (Lagrange {) ùÒ {ü«Lª. XJ'u f ^r , ~X f ´ n + 1 ëY, Kd (∗∗) , F 0 È, dÈ©Ä úª, Rn(x) = F(x) − F(x0) = Z x x0 F 0 (t)dt = Z x x0 f (n+1)(t) n! · (x − t) n dt. ù´{°(È©L«. d, du (x − t) n ØCÒ, d1È©¥½n, ∃ ξ ∈ (x0, x) (½ (x, x0), ¦ Rn(x) = 1 n! f (n+1)(ξ) · Z x x0 (x − t) n dt = f n+1(ξ) (n + 1)!(x − x0) n+1 . ùÒ´ Lagrange {. Ón Cauchy {. XJ f 3 x0 NCág, K¡/ªÚ X∞ n=0 f n (x0) n! (x − x0) n f 3 x0 ? Taylor Ðm. Taylor Ðm3 x0 = 0 AÏ/¡ Maclaurin Ð múª. XJ limn→∞ Rn(x) = 0, KP f(x) = X∞ n=0 f (n+1)(x0) (n + 1)! · (x − x0) n . d¡ f Taylor ÐmÂñg .
9g5.4Taylor展开例5.4.1.f(a）=一在（-1,1）中任意次可微.求o处f的Taylor展开解利用归纳法容易计算f的各阶导数为n!f(n)(r) =(1-a)n+i, n= 1,2, ..因此f(n)(O)=n!,故Taylor展开为0n=l+r+r?+...+rn+..n!n=o又因为余项Th1-(1 +#+?+...+2")=Rn(r) =→0,(n-00)1-T1-T故1=1+r+a?+...+r"+.., re(-l,1)1-r例5.4.2.求f(z）=e在a=0处的Taylor展开式解f(n)() =e, n≥1. 故 f(n)(0)=1 (n≥1), 从而 e在 =0 处的Taylor展开式为o0Sm=n=on!,n!n=0其余项为11eSan+1e0rn+1, 0 (0,1)Rn(r)(n + 1)!(n + 1)!因此有估计[Rn(a)/ ≤ el]/ /a|n+1→0 (n →8)(n + 1)!这说明r3r2rne"=1+r+: r E (-00,00)....++21 +3!n!特别地111e=1+1++...+21 +3n!由此容易说明e为无理数：首先11112<e <1+1+++3.4+234.51111111+1+?=3+4+22223
§5.4 Taylor Ðm 9 ~ 5.4.1. f(x) = 1 1−x 3 (−1, 1) ¥?¿g. ¦ x0 ? f Taylor Ðm. ) |^8B{N´O f ê f (n) (x) = n! (1 − x) n+1 , n = 1, 2, · · · Ïd f (n) (0) = n!, Taylor Ðm X∞ n=0 f (n)(0) n! x n = 1 + x + x 2 + · · · + x n + · · · qÏ{ Rn(x) = 1 1 − x − (1 + x + x 2 + · · · + x n ) = x n 1 − x → 0, (n → ∞) 1 1 − x = 1 + x + x 2 + · · · + x n + · · · , x ∈ (−1, 1) ~ 5.4.2. ¦ f(x) = e x 3 x = 0 ? Taylor Ðmª. ) f (n) (x) = e x , ∀ n ≥ 1. f (n) (0) = 1 (n ≥ 1), l e x 3 x = 0 ? Taylor Ðmª X∞ n=0 f (n) (0) n! · x n = X∞ n=0 x n n! , Ù{ Rn(x) = 1 (n + 1)! e ζx n+1 = 1 (n + 1)! e θxx n+1, θ ∈ (0, 1) ÏdkO |Rn(x)| ≤ e |x| |x| n+1 (n + 1)! → 0 (n → ∞) ù`² e x = 1 + x + x 2 2! + x 3 3! + · · · + x n n! · · · , x ∈ (−∞, ∞). AO/ e = 1 + 1 + 1 2! + 1 3! + · · · + 1 n! · · · ddN´`² e Ãnê: Äk 2 < e < 1 + 1 + 1 2 + 1 2 · 3 + 1 3 · 4 + 1 4 · 5 + · · · = 1 + 1 + 1 2 + 1 2 − 1 3 + 1 3 − 1 4 + 1 4 − 1 5 + · · · = 3
10第五章微分中值定理和Taylor展开如果e=,(p,g)=1,则10<q!.e-ql(1+1+...+q!11 +...)q!(?(q + 1))(q + 2)!111<+.q+1(q + 2)(q + 3)(q + 1)(q + 2)2<1q+1但q！!·e-q(1+1++）为整数，这就得到了矛盾！例5.4.3.求sin工，cos在工=0处的Taylor的展开解sinT三cosT.sin"三cosT=一sinT，因此sin(2k+1)(z) = (1)* cos T, sin(2k)() =(1) sin 2特别地sin(2k+1)(0) = (-1)k, sin(2k)(0) = 0因此2325(-1)nμ2n+1(-1)n+1±2n+3 cos Or.0<9<1,sinr=a-+..+3!+5!(2n + 1)!(2n + 3)!因为余项趋于零，故(-1)"α2n+102sinr=(2n + 1), E (-80, )n=0类似地，222426(-1)nz2n=1-COST=X+...,rE(-00,00)+-6!(2n)!n=0命题5.4.3.设f(z）在o=0处有Taylor展开n=0则oo(1)f(-r)有Taylor展开(-1)"an·r";an·zn，其中k为正整数；(2)f(rk）有Taylor展开n=0
10 1ÊÙ ©¥½nÚ Taylor Ðm XJ e = p q , (p, q) = 1, K 0 < q! · e − q!(1 + 1 + · · · + 1 q! ) = q!( 1 (q + 1)! + 1 (q + 2)! + · · ·) < 1 q + 1 + 1 (q + 1)(q + 2) + 1 (q + 2)(q + 3) + · · · = 2 q + 1 < 1 q! · e − q!(1 + 1 + · · · + 1 q! ) ê, ùÒ gñ! ~ 5.4.3. ¦ sin x, cos x 3 x = 0 ? Taylor Ðm. ) sin0 x = cos x, sin00 x = cos0 x = − sin x, Ïd sin(2k+1)(x) = (−1)k cos x, sin(2k) (x) = (−1)k sin x AO/ sin(2k+1)(0) = (−1)k , sin(2k) (0) = 0. Ïd sin x = x − x 3 3! + x 5 5! + · · · + (−1)nx 2n+1 (2n + 1)! + (−1)n+1x 2n+3 cos θx (2n + 3)! , 0 < θ < 1, Ï{ªu", sin x = X∞ n=0 (−1)nx 2n+1 (2n + 1)! , x ∈ (−∞, ∞) aq/, cos x = X∞ n=0 (−1)nx 2n (2n)! = 1 − x 2 2! + x 4 4! − x 6 6! + · · · , x ∈ (−∞, ∞) ·K 5.4.3. f(x) 3 x0 = 0 ?k Taylor Ðm X∞ n=0 an · x n , K (1) f(−x) k Taylor Ðm X∞ n=0 (−1)n an · x n ; (2) f(x k ) k Taylor Ðm X∞ n=0 an · x kn , Ù¥ k ê;