數(shù)分選講講稿第19講.

1、第十九講3學(xué)時3、在不等式兩端取變限積分證明新的不等式例15證明：當x 0時,3Xx sin x63Xx 65X120證已知cosx 1，( X 0，只有2n時,等號成立)在此式兩端同時取0,x上的積分,Xcostdt0X0dt,sin x(x0)再次取0, x上的積分，得2X cosx 第三次取0,x上的積分，得sin x所以3XX 一6sin x上式再在0, x上的積分，得4 X24cosxcosx2x_24x_24再在0, x上的積分，得 sin x3X X +6 120X50.例16設(shè)f (x)是a,b上連續(xù)的凸函數(shù).試證:a,b,為X2，有幾何解釋:X1X221X2X1X2x f(t

2、)dtx1f(X1) f(X2)X1(X2 %),(0,1)，則X2X2X-IX11f (t)dt 0 f(x1(X2 xj)d(1)同理，令 tX2(X2 Xi),(0,1)，則1X21Lxif(t)dt0f(X2(X2 Sd從而X2,f(t)dtf (X1(X2 X1) f (X2(X2 X1) d注意到X1(X2 X1)與X2(X2 xJ關(guān)于中點 勺一X2對稱，f (x)又2為凸函數(shù)，所以1 f (X1(X2 X1)f (X22(X2 X1)XiX2另一方面，由(1X2X f(t)dtX2X1 X11)式及f (X)的凸性X-IX2).t X1 (X2 X1)1f (t)dt0 f( X

3、2(1)xjdf(X2) (1)f (X1) d1 .2f(X2) 2f(X1)f (X1) f (X2)2例17設(shè)函數(shù)g(X)在a,b上遞增.試證：x (a, b)X函數(shù)f (x) g(t)dt為凸函數(shù).c證 Q g(x)在a,b上遞增,X1, X2,X3(a,b), X1X2X3f (X2) f (X1)1X2X2X1X2X1g(t)dtX1cg(t)dt1X21X3x g(t)dtg(X2) x g(t)dtX? X1 X1X3 X2X3X2c g(t)dt c g(t)dtf(X3)f(X2)X3 X2X3 X2所以，f(x)為凸函數(shù).例 18 設(shè) f(x) , p(x)在a,b上連續(xù)

4、，p(x) 0, bpgdxa且m f (x) M ,(x)在m,M 上有定義，并且有二階導(dǎo)數(shù)，(x)0.試證:p(x)f (x)dxbaPgf (x) dxp(x) dxp(x)dx令n ，得b a P fiPi (fi) b aPiPibp(x) f (x)dx abbp(x) f (x) dx abp(x)dxp(x)dx證II (利用Taylor公式)記x0p(x)f (x)dxxia_(b na), PP(xJ,fif(Xi), (i1,2,L ,n)Q (x)0，(x)為凸函數(shù).由詹禁定理，取Piin,(j1,2丄n,n)，i1Pji 1j 1nni 1ifiii 1(fi)P1

5、f1p2 f2LPn fnP1(fjP2LPn (fn)(利用積分和)將區(qū)間a,b n等分，記piP2LPnPiP2L Pnp(x)dx則(y)(x0)(x0)(y x0) 2 ()(y x0)2.注意 （）0,(y)(xo)(x)(y X。).在上式中，令y f（x），然后兩邊乘以P(X)得b，得p(x)dxap(x) f(x)ba p(x)dx(Xo) bp(x)a p(x)dxgpm f(x) xoba P(x)dx在a,b上取積分b p(x)aba p(x)dx（X。）:bp(x) dxa p(x)dxbaP(x)其中aP(x)（X。）b p(x) f(x) x。abp(x)dx ad

6、xf(x) dxbp(x)dxaf(x)x0 dx（x。）baP(x)b(Xo) a p(x) f(x) X0 dxbp(x)dxaf(x)ba p(x)f (x)dxr dxp(x)dxa1ba P(x)dxba P(x)dxba p(x)f (x)dxabba P(x)dx a p(x)f (x)dxaap(x)f (x) dx(Xo)p(x) f (x)dxp(x)dxp(x)dx 4.5 不等式、Cauchy不等式及Schwarz不等式1. Cauchy不等式設(shè)（i 1,2丄，n）為任意實數(shù)，則nna b2.i 1i 1（Cauchy不等式）其中等號當且僅當ai與b成比例時成立.證1

7、（判別式法）n20 aix bi 1n22caix 2i 1nahi 1nb2i 1上式是關(guān)于x的二次二項式，保持非負，故判別式nabi 1n2aii 1nb2o.i 1證II （配方法）nQ a2i 1nb2i 1nahi 12ainb2j 1naibii 1najbjj 1n2 2ai bjj 1nabajbjj 11 n nj嚴 ajbi0.因此，Cauchy不等式成立.等號成立當且僅當a ajbi, （i1,2,L ,n）.證III （利用二次型）n2o a/ byi 1n2aii 1n2 a bii 1xynb2i 1即關(guān)于x, y的二次型，非負定,因此n2 aii 1n2aii 1

8、naibii 1na bii 1nb2i 1b2i 1方法iii可推廣.2. Schwarz不等式設(shè)f(x),g(x)在a,b上可積，則f (x)g(x)dx2(x)dxg2(x)dx.Cauchy不等式的積分形 式稱為Schwarz 不若f(x), g(x)在a,b上連續(xù)，其中等號當且僅當存在常數(shù),使得f (x)g(x)時成立.(，不同時為零)3. Schwarz不等式的應(yīng)用例 1 已知 f(x) 0，在a,b上連續(xù)，bf(x)dx 1. ak為任意實數(shù).求證：b2b2a f (x)cos kxdx a f (x)sin kxdx 1.證第一項應(yīng)用Schwarz不等式:f (x)cos kx

9、dxf(x) f(x)coskxdxbb2a f(x)dx & f(x)cos kxdx同理f (x)sin(1) + (2):2f (x)cos kxdx2kxdxbf (x)sin kxdx(1)f (x)cos kxdxbf (x)sin kxdxa1.例2設(shè)f (x)在a,b上有連續(xù)的導(dǎo)數(shù)，f(a) 0試證:f (x)f (x) dx2f (x) dx.證 令 g(x)：|f(t)|dt,則 g(x) |f(x)|,a x b.由 f (a)0，知If (x) |f (x) f (a)|xf (t)dtaX f (t) dtag(x)因此，f (x) f (x)|dx g(x)g (x

10、)dxaabag(X)dg(X)；g2(x)g2(b) g2(a)ba f (x)血1 b12 a2f (x) dxSchwarz2b2b12dxaf (x)fdx,2f (x) dx.例3設(shè)f (x)在a,b(b a)上有連續(xù)的n階導(dǎo)數(shù)f(n) (x),且 f(k)(a)0,(k0,1,2,L ,n 1).求證:其中0下證b f(k)(x)a2dxm k2(bm ka)(m)(x)12 2dxk m n.先證明n 1的情況.此時1,k(x)在a,b上有連續(xù)的導(dǎo)數(shù),(a)0.2(x)dx(x)由Schwarz不等式:2(x)(x兩端同時積分(b a)(x)2dx(t)dt,a,ba)(t)dt

11、X12dta2(t) dt(t)2dt1(b(x)dxa(x2(t)dtd(x,X2a(t)dtb2(x)dx.abaxaxxa)2a)2a)2b1a2xa) a(t) 2dt dxx a)2(x) 2dxa第二項積分:值大于零.兩邊同時開方:2(x)dxba) a(X)12 2dx對一般情況,(X)f(k)(x),a,bb f(k)(x) 2dx(ba)(k、(x)2dx例4設(shè)f (x)正的下界.dn試證:limn只需證明limn10先證bdna22(ba)2(k 2)(x)2dxm k2(bm ka)1f(m)(x) dx 2.(m)g(x)在a,b上連續(xù),bna f(x)| g(x)dx

12、,f (x)不恒為零，g(x)有(n 1,2,L).dn 1dnlim dmax f (x).x bddnn1存在dn 1單調(diào)增dnmax f (x),lim dn 1n dnlimndnf(x)g(x)dx n 1n 1.g(x)|f(x)W . g(x) f(x)|2dxdxg(x)|f(x)|n1dx1 1dn 1 dn21.Q dn0,dnd n 1dr即dn 1dndn 1Zdndn1 dn20再證dn 1有界.dnbag(x)|f(x)|n112因為f(x)在a,b上連續(xù)，所以M 0,使得f(x) M, x a,bd n 1dnbn 1a|f(X)| g(x)dxbna|f(x)| g(x)dxbnf(x)| g(x)dxMng(x)dx30既然dn 1dn單調(diào)有界,存在極限.lim dn 1n dn、平均值不等式limndnlimnba f(x)ng(x)dxmaxa x bf(x).基本形式：對任意n個實數(shù)ai 0,(i1,2丄,n),恒有Ln(即幾何平均值 算術(shù)平均值)其中等號成立當且僅當QqazL a. a-aana1a2Lan .設(shè)正值函數(shù)f (x)在0,1上連