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<P align=center><B >微分中值定理证明中辅助函数的构造<o:p></o:p></B></P>
<P ><o:p> </o:p></P>
<P ><B>摘要:</B>本文总结了证明微分中值命题时常用的六种构造辅助函数的方法,并给出了具体的应用.<o:p></o:p></P>
<P ><B>关键词:</B>辅助函数;原函数;不定积分;罗尔定理<o:p></o:p></P>
<P ><B><o:p><FONT face="Times New Roman"> </FONT></o:p></B></P>
<P align=center><B ><FONT face="Times New Roman">Construction of auxiliary function in proving differential mean value theorem<o:p></o:p></FONT></B></P>
<P align=center><FONT face="Times New Roman">LIANG Hui<o:p></o:p></FONT></P>
<P ><FONT face="Times New Roman">(Class 2 Grade 2001, Major of Math. and applied Math., Dept. of Math., <st1:place w:st="on"><st1:PlaceName w:st="on">Huanggang</st1:PlaceName> <st1:PlaceName w:st="on">Normal</st1:PlaceName> <st1:PlaceType w:st="on">University</st1:PlaceType></st1:place>, Huangzhou 438000 Hubei)</FONT></P>
<P ><B><FONT face="Times New Roman">Abstract</FONT></B>:<FONT face="Times New Roman">In this paper, we summarized six methods about construction of auxiliary function in proving differential mean value theorem</FONT>,<FONT face="Times New Roman">and discussed material application of them</FONT>.</P>
<P ><B><FONT face="Times New Roman">Key words</FONT></B><B>:</B><FONT face="Times New Roman">auxiliary function; original function; indefinite integral; Rolle theorem</FONT></P>
<P ><o:p><FONT face="Times New Roman"> </FONT></o:p></P>
<P >利用微分中值定理解决问题时,通常需要构造一个辅助函数,由这个辅助函数满足某个中值定理的条件而得到要证明的结论,而构造性方法是高等数学中一个重要的分析技巧,这往往成为解题的难点所在.本文主要介绍证明微分中值命题时常用的六种构造辅助函数的方法.<o:p></o:p></P>
<P ><B ><FONT face="Times New Roman">1 </FONT></B><B >原函数法</B><B ><o:p></o:p></B></P>
<P >此法是将结论变形并向罗尔定理的结论靠拢,凑出适当的原函数作为辅助函数,主要思想分为四点:<FONT face="Times New Roman">(1)</FONT>将要证的结论中的<v:shapetype><FONT face="Times New Roman"> <v:stroke joinstyle="miter"></v:stroke><v:formulas><v:f eqn="if lineDrawn pixelLineWidth 0"></v:f><v:f eqn="sum @0 1 0"></v:f><v:f eqn="sum 0 0 @1"></v:f><v:f eqn="prod @2 1 2"></v:f><v:f eqn="prod @3 21600 pixelWidth"></v:f><v:f eqn="prod @3 21600 pixelHeight"></v:f><v:f eqn="sum @0 0 1"></v:f><v:f eqn="prod @6 1 2"></v:f><v:f eqn="prod @7 21600 pixelWidth"></v:f><v:f eqn="sum @8 21600 0"></v:f><v:f eqn="prod @7 21600 pixelHeight"></v:f><v:f eqn="sum @10 21600 0"></v:f></v:formulas><v:path o:connecttype="rect" gradientshapeok="t" o:extrusionok="f"></v:path><o:lock aspectratio="t" v:ext="edit"></o:lock></FONT></v:shapetype><v:shape><v:imagedata></v:imagedata></v:shape>换成<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>;<FONT face="Times New Roman">(2)</FONT>通过恒等变形将结论化为易消除导数符号的形式;<FONT face="Times New Roman">(3)</FONT>用观察法或积分法求出原函数(等式中不含导数符号),并取积分常数为零;<FONT face="Times New Roman">(4)</FONT>移项使等式一边为零,另一边即为所求辅助函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P >例<FONT face="Times New Roman">1</FONT>:证明柯西中值定理.<o:p></o:p></P>
<P >分析:在柯西中值定理的结论<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>中令<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,得<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,先变形为<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>再两边同时积分得<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,令<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,有<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>故<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>为所求辅助函数.<o:p></o:p></P>
<P >例<FONT face="Times New Roman">2</FONT>:若<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">,</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">,</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">,</FONT>…<FONT face="Times New Roman">,</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>是使得<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>的实数.证明方程<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在(<FONT face="Times New Roman">0</FONT>,<FONT face="Times New Roman">1</FONT>)内至少有一实根.<o:p></o:p></P>
<P >证:由于<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><o:p></o:p></P>
<P >并且这一积分结果与题设条件和要证明的结论有联系,所以设<o:p></o:p></P>
<P ><v:shape><v:imagedata></v:imagedata></v:shape>(取<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>),则<o:p></o:p></P>
<P ><FONT face="Times New Roman">1</FONT>)<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在<FONT face="Times New Roman">[0</FONT>,<FONT face="Times New Roman">1]</FONT>上连续<o:p></o:p></P>
<P ><FONT face="Times New Roman">2</FONT>)<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在(<FONT face="Times New Roman">0</FONT>,<FONT face="Times New Roman">1</FONT>)内可导<o:p></o:p></P>
<P ><FONT face="Times New Roman">3</FONT>)<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">=0</FONT>,<FONT face="Times New Roman"> </FONT><v:shape><v:imagedata><FONT face="Times New Roman"></FONT></v:imagedata></v:shape><o:p></o:p></P>
<P >故<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>满足罗尔定理的条件,由罗尔定理,存在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>使<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,即<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>亦即<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<FONT face="Times New Roman"> <o:p></o:p></FONT></P>
<P >这说明方程<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在(<FONT face="Times New Roman">0</FONT>,<FONT face="Times New Roman">1</FONT>)内至少有实根<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P ><B ><FONT face="Times New Roman"> 2 </FONT></B><B >积分法</B><B ><o:p></o:p></B></P>
<P >对一些不易凑出原函数的问题,可用积分法找相应的辅助函数.<o:p></o:p></P>
<P ><FONT face="Times New Roman"> </FONT>例<FONT face="Times New Roman">3</FONT>:设<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在<FONT face="Times New Roman">[1</FONT>,<FONT face="Times New Roman">2]</FONT>上连续,在(<FONT face="Times New Roman">1</FONT>,<FONT face="Times New Roman">2</FONT>)内可导,<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.证明存在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>使<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P >分析:结论变形为<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,不易凑成<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.我们将<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>换为<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,结论变形为<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,积分得:<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,即<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,从而可设辅助函数为<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,有<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.本题获证.<o:p></o:p></P>
<P >例<FONT face="Times New Roman">4</FONT>:设函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>上连续,在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>内可微,<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.证明存在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,使得:<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P >证:将<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>变形为<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><v:shape><v:imagedata><FONT face="Times New Roman"></FONT></v:imagedata></v:shape><v:shape><v:imagedata><FONT face="Times New Roman"></FONT></v:imagedata></v:shape>,将<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>换为<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,则<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,两边关于<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>积分,得:<FONT face="Times New Roman"> </FONT><B ><v:shape><v:imagedata><FONT face="Times New Roman"></FONT></v:imagedata></v:shape><v:shape><v:imagedata><FONT face="Times New Roman"></FONT></v:imagedata></v:shape></B>,所以<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><v:shape><v:imagedata><FONT face="Times New Roman"></FONT></v:imagedata></v:shape>,其中<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,由<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>可得<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.由上面积分的推导可知,<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>为一常数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,故其导数必为零,从整个变形过程知,满足这样结论的<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>的存在是不成问题的.因而令<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,易验证其满足罗尔定理的条件,原题得证.<o:p></o:p></P>
<P ><B ><FONT face="Times New Roman">3 </FONT></B><B >几何直观法</B><B ><o:p></o:p></B></P>
<P >此法是通过几何图形考查两函数在区间端点处函数值的关系,从而建立适当的辅助函数.<o:p></o:p></P>
<P >例<FONT face="Times New Roman">5</FONT>:证明拉格朗日中值定理.<o:p></o:p></P>
<P ><v:shape><v:imagedata></v:imagedata><w:wrap type="tight"></w:wrap></v:shape>分析:通过弦<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>两个端点的直线方程为<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,则函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>与直线<FONT face="Times New Roman">AB</FONT>的方程之差即函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在两个端点处的函数值均为零,从而满足罗尔定理的条件故上式即为要做辅助函数.<o:p></o:p></P>
<P >例<FONT face="Times New Roman">6</FONT>:若<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>上连续且<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.试证在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>内至少有一点<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,使<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P >分析:由图可看出,此题的几何意义是说,连续函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>的图形曲线必跨越<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>这一条直线,而两者的交点的横坐标<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,恰满足<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.进而还可由图知道,<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata><w:wrap type="tight"></w:wrap></FONT></v:shape>对<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>上的同一自变量值<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,这两条曲线纵坐标之差<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>构成一个新的函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,它满足<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman"><0,</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">>0,</FONT>因而符合介值定理的条件.当<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>为<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>的一个零点时,<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>恰等价于<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.因此即知证明的关键是构造辅助函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P ><B ><FONT face="Times New Roman">4 </FONT></B><B >常数</B><B ><FONT face="Times New Roman">k</FONT></B><B >值法</B><B ><FONT face="Times New Roman"> <o:p></o:p></FONT></B></P>
<P >此方法构造辅助函数的步骤分为以下四点:<o:p></o:p></P>
<P ><FONT face="Times New Roman">1)</FONT>将结论变形,使常数部分分离出来并令为<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P ><FONT face="Times New Roman">2)</FONT>恒等变形使等式一端为<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>及<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>构成的代数式,另一端为<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>及<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>构成的代数式.<o:p></o:p></P>
<P ><FONT face="Times New Roman">3</FONT>)观察分析关于端点的表达式是否为对称式.若是,则把其中一个端点设为<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,相应的函数值改为<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P ><FONT face="Times New Roman">4</FONT>)端点换变量<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>的表达式即为辅助函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P >例<FONT face="Times New Roman">7</FONT>:设<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>上连续,在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>内可导,<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,试证存在一点<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,使等式<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>成立.<o:p></o:p></P>
<P >分析:将结论变形为<B ><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape></B>,令<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,则有<B ><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape></B>,令<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,可得辅助函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P >例<FONT face="Times New Roman">8</FONT>:设<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>上存在,在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,试证明存在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,使得<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P >分析:令<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,于是有<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,上式为关于<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>三点的轮换对称式,令<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>(<FONT face="Times New Roman">or</FONT>:<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,<FONT face="Times New Roman">or</FONT>:<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>),则得辅助函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P ><B ><FONT face="Times New Roman">5 </FONT></B><B >分析法</B><B ><o:p></o:p></B></P>
<P >分析法又叫倒推法,就是从欲证的结论出发借助于逻辑关系导出已知的条件和结论.<o:p></o:p></P>
<P >例<FONT face="Times New Roman">9</FONT>:设函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在<FONT face="Times New Roman">[0</FONT>,<FONT face="Times New Roman">1]</FONT>上连续,在(<FONT face="Times New Roman">0</FONT>,<FONT face="Times New Roman">1</FONT>)内可导,证明在(<FONT face="Times New Roman">0</FONT>,<FONT face="Times New Roman">1</FONT>)内存在一点<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,使得<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P >分析:所要证的结论可变形为:<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,即<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,因此可构造函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,则对<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>与<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在<FONT face="Times New Roman">[0</FONT>,<FONT face="Times New Roman">1]</FONT>上应用柯西中值定理即可得到证明.<o:p></o:p></P>
<P >例<FONT face="Times New Roman">10</FONT>:设函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在<FONT face="Times New Roman">[0,1]</FONT>上连续,在(<FONT face="Times New Roman">0</FONT>,<FONT face="Times New Roman">1</FONT>)内可导,且<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">=0</FONT>,对任意<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>有<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.证明存在一点<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>使<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>(<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>为自然数)成立.<o:p></o:p></P>
<P >分析:欲证其成立,只需证<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>由于对任意<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>有<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,故只需证:<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>即<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,于是引入辅助函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>(<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>为自然数).<o:p></o:p></P>
<P >例<FONT face="Times New Roman">11</FONT>:设函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在区间[<FONT face="Times New Roman">0</FONT>,<FONT face="Times New Roman">+</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>]上可导,且有<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>个不同零点:<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.试证<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在<FONT face="Times New Roman">[0</FONT>,<FONT face="Times New Roman">+</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">]</FONT>内至少有<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>个不同零点.(其中,<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>为任意实数)<o:p></o:p></P>
<P >证明:欲证<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在<FONT face="Times New Roman">[0</FONT>,<FONT face="Times New Roman">+</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>)内至少有<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>个不同零点,只需证方程<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">=0</FONT>在<FONT face="Times New Roman">[0</FONT>,<FONT face="Times New Roman">+</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">]</FONT>内至少有<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>个不同实根.<o:p></o:p></P>
<P >因为,<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,故只需证方程<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>内至少有<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>个不同实根.<o:p></o:p></P>
<P >引入辅助函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,易验证<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在区间<FONT face="Times New Roman">[</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">]</FONT>,<FONT face="Times New Roman">[</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">]</FONT>,…,<FONT face="Times New Roman">[</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">]</FONT>上满足罗尔定理的条件,所以,分别在这<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>个区间上应用罗尔定理,得<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,其中<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>且<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><o:p></o:p></P>
<P >以上说明方程<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在<FONT face="Times New Roman">[</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">]</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">[</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">]</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>…<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">[</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">]</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">[0</FONT>,<FONT face="Times New Roman">+</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">]</FONT>内至少有<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>个不同实根,从而证明了方程<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">=0</FONT>在<FONT face="Times New Roman">[0</FONT>,<FONT face="Times New Roman">+</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">]</FONT>内至少有<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>个不同实根.<o:p></o:p></P>
<P ><B><FONT face="Times New Roman">6 </FONT></B><B>待定系数法</B><B><o:p></o:p></B></P>
<P >在用待定系数法时,一般选取所证等式中含<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>的部分为<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,再将等式中一个端点的值<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>换成变量<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,使其成为函数关系,等式两端做差构造辅助函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,这样首先可以保证<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">=0</FONT>,而由等式关系<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">=0</FONT>自然满足,从而保证<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>满足罗尔定理条件,再应用罗尔定理最终得到待定常数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>与<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>之间的关系.<o:p></o:p></P>
<P >例<FONT face="Times New Roman">12</FONT>:设<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>是<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>上的正值可微函数,试证存在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,使<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P >证明:设<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,令<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>容易验证<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman"> </FONT>上满足罗尔定理条件,由罗尔定理,存在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>使<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,解得<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,故<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P >例<FONT face="Times New Roman">13</FONT>:设函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>上连续,在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>内可导,则在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>内至少存在一点<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>使<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P >证明:将所证等式看作<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,设<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,令<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,则<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>满足罗尔定理条件,由罗尔定理得,存在一点<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,使<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,即<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,若<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman">=0</FONT>,则<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,结论成立;若<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,则<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,从而有<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P >例<FONT face="Times New Roman">14</FONT>:设<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,则存在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>使<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P >分析:对于此题设<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>作函数<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><v:shape><v:imagedata><FONT face="Times New Roman"></FONT></v:imagedata></v:shape>.应用罗尔定理可得存在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,使<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,即<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,从而<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,这样并不能证明原结论,遇到这种情况,说明所作的辅助函数不合适,则需要将所证明的等式变形,重新构造辅助函数.<o:p></o:p></P>
<P >证明:将所证等式变形为<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,设<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><v:shape><v:imagedata><FONT face="Times New Roman"></FONT></v:imagedata></v:shape>,令<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><v:shape><v:imagedata><FONT face="Times New Roman"></FONT></v:imagedata></v:shape>,则<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>满足罗尔定理条件,用罗尔定理可得存在<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,使<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,即<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,于是<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>,故<v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape>.<o:p></o:p></P>
<P >总之,证明微分中值命题的技巧在于:一是要仔细观察,适当变换待证式子;二是要认真分析,巧妙构造辅助函数.抓住这两点,即可顺利完成证明.<o:p></o:p></P>
<P ><B >参考文献</B><o:p></o:p></P>
<P >[1] 张家秀.关于构造辅助函数的几种方法[J].高等理科教育.2003(4).<o:p></o:p></P>
<P >[2] 瞿美玲,等.微分中值定理的研究[J].河南教育学院学报.2003-6.<o:p></o:p></P>
<P >[3] 张小莉.构造辅助函数的两种方法[J].湖北工学院学报.2002-9.<o:p></o:p></P>
<P >[4] 聂洪珍,张翠萍.关于构造辅助函数证明微分中值定理的进一步探讨[J].鞍山师范学院学报.2003-8.<o:p></o:p></P>
<P >[5] 王跃恒.关于中值定理证明中辅助函数的构造[J].数学理论与应用.2003-12.<o:p></o:p></P>
<P >[6] 李山.微分中值定理及辅助函数[J].宿州教育学院学报.2001(2).<o:p></o:p></P>
<P ><B>致谢: </B>在论文的撰写过程中,衷心的感谢吴卫兵老师对我的精心指导!<o:p></o:p></P> |
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