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数学专业英语-Differential Calculus

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发表于 2004-5-6 09:29:10 | 显示全部楼层 |阅读模式
< ><B><FONT face="Times New Roman">Historical Introduction</FONT></B></P>
< ><FONT face="Times New Roman" size=3>Newton and Leibniz,quite independently of one another,were largely responsible for developing the ideas of integral calculus to the point where hitherto insurmountable problems could be solved by more or less routine methods.The successful accomplishments of these men were primarily due to the fact that they were able to fuse together the integral calculus with the second main branch of calculus,differential calculus.</FONT></P>
< ><FONT face="Times New Roman" size=3>The central idea of differential calculus is the notion of derivative.Like the integral,the derivative originated from a problem in geometry—the problem of finding the tangent line at a point of a curve.Unlile the integral,however,the derivative evolved very late in the history of mathematics.The concept was not formulated until early in the 17<SUP>th</SUP> century when the French mathematician Pierre de Fermat,attempted to determine the maxima and minima of certain special functions.</FONT></P>
<P ><FONT face="Times New Roman" size=3>Fermat’s idea,basically very simple,can be understood if we refer to a curve and assume that at each of its points this curve has a definite direction that can be described by a tangent line.Fermat noticed that at certain points where the curve has a maximum or minimum,the tangent line must be horizontal.Thus the problem of locating such extreme values is seen to depend on the solution of another problem,that of locating the horizontal tangents.</FONT></P>
<P ><FONT face="Times New Roman" size=3>This raises the more general question of determining the direction of the tangent line at an arbitrary point of the curve.It was the attempt to solve this general problem that led Fermat to discover some of the rudimentary ideas underlying the notion of derivative.</FONT></P>
<P ><FONT face="Times New Roman" size=3>At first sight there seems to be no connection whatever between the problem of finding the area of a region lying under a curve and the problem of finding the tangent line at a point of a curve.The first person to realize that these two seemingly remote ideas are,in fact, rather intimately related appears to have been Newton’s teacher,Isaac Barrow(1630-1677).However,Newton and Leibniz were the first to understand the real importance of this relation and they exploited it to the fullest,thus inaugurating an unprecedented era in the development of mathematics.</FONT></P>
<P ><FONT face="Times New Roman" size=3>Although the derivative was originally formulated to study the problem of tangents,it was soon found that it also provides a way to calculate velocity and,more generally,the rate of change of a function.In the next section we shall consider a special problem involving the calculation of a velocity.The solution of this problem contains all the essential fcatures of the derivative concept and may help to motivate the general definition of derivative which is given below.</FONT></P>
<P ><B><FONT face="Times New Roman">A  Problem Involving Velocity</FONT></B></P>
<P ><FONT face="Times New Roman" size=3>Suppose a projectile is fired straight up from the ground with initial velocity of 144 feet persecond.Neglect friction,and assume the projectile is influenced only by gravity so that it moves up and back along a straight line.Let f(t) denote the height in feet that the projectile attains t seconds after firing.If the force of gravity were not acting on it,the projectile would continue to move upward with a constant velocity,traveling a distance of 144 feet every second,and at time t we woule have f(t)=144 t.In actual practice,gravity causes the projectile to slow down until its velocity decreases to zero and then it drops back to earth.Physical experiments suggest that as the projectile is aloft,its height f(t) is given by the formula</FONT></P>
<P ><FONT face="Times New Roman"><FONT size=3>(1)</FONT>       <FONT size=3>f(t)=144t –16 t<SUP>2</SUP></FONT></FONT></P>
<P ><FONT face="Times New Roman" size=3>The term –16t<SUP>2</SUP> is due to the influence of gravity.Note that f(t)=0 when t=0 and when t=9.This means that the projectile returns to earth after 9 seconds and it is to be understood that formula (1) is valid only for 0&lt;t&lt;9.</FONT></P>
<P ><FONT face="Times New Roman" size=3>The problem we wish to consider is this:To determine the velocity of the projectile at each instant of its motion.Before we can understand this problem,we must decide on what is meant by the velocity at each instant.To do this,we introduce first the notion of average velocity during a time interval,say from time t to time t+h.This is defined to be the quotient.</FONT></P>
<P ><FONT face="Times New Roman" size=3>Change in distance during time interval   =f(t+h)-f(t)/h</FONT></P>
<TABLE  cellSpacing=0 cellPadding=0 border=1>

<TR >
<TD  vAlign=top width=276>
<P ><FONT face="Times New Roman">Length of time interval</FONT></P></TD></TR></TABLE>
<P ><FONT face="Times New Roman" size=3>This quotient,called a difference quotient,is a number which may be calculated whenever both t and t+h are in the interval[0,9].The number h may be positive or negative,but not zero.We shall keep t fixed and see what happens to the difference quotient as we take values of h with smaller and smaller absolute value.</FONT></P>
<P ><FONT face="Times New Roman" size=3>The limit process by which v(t) is obtained from the difference quotient is written symbolically as follows:</FONT></P>
<P ><FONT size=3><FONT face="Times New Roman">V(t)=lim<SUB>(h</SUB></FONT><SUB>→<FONT face="Times New Roman">0)</FONT></SUB><FONT face="Times New Roman"> [f(t+h)-f(t)]/h</FONT></FONT></P>
<P ><FONT face="Times New Roman" size=3>The equation is used to define velocity not only for this particular example but,more generally,for any particle moving along a straight line,provided the position function f is such that the differerce quotient tends to a definite limit as h approaches zero.</FONT></P>
<P ><FONT face="Times New Roman" size=3>The example describe in the foregoing section points the way to the introduction of the concept of derivative.We begin with a function f defined at least on some open interval(a,b) on the x axis.Then we choose a fixed point in this interval and introduce the difference quotient</FONT></P>
<P ><FONT face="Times New Roman" size=3>[f(x+h)-f(x)]/h</FONT></P>
<P ><FONT face="Times New Roman" size=3>where the number h,which may be positive or negative(but not zero),is such that x+h also lies in(a,b).The numerator of this quotient measures the change in the function when x changes from x to x+h.The quotient itself is referred to as the average rate of change of f in the interval joining x to x+h.</FONT></P>
<P ><FONT face="Times New Roman" size=3>Now we let h approach zero and see what happens to this quotient.If the quotient.If the quotient approaches some definite values as a limit(which implies that the limit is the same whether h approaches zero through positive values or through negative values),then this limit is called the derivative of f at x and is denoted by the symbol f’(x) (read as “f prime of x”).Thus the formal definition of f’(x) may be stated as follows:</FONT></P>
<P ><FONT face="Times New Roman" size=3>Definition of derivative.The derivative f’(x)is defined by the equation</FONT></P>
<P ><FONT size=3><FONT face="Times New Roman">f’(x)=lim<SUB>(h</SUB></FONT><SUB>→<FONT face="Times New Roman">o)</FONT></SUB><FONT face="Times New Roman">[f(x+h)-f(x)]/h</FONT></FONT></P>
<P ><FONT face="Times New Roman" size=3>provided the limit exists.The number f’(x) is also called the rate of change of f at x.</FONT></P>
<P ><FONT face="Times New Roman" size=3>In general,the limit process which produces f’(x) from f(x) gives a way of obtaining a new function f’ from a given function f.This process is called differentiation,and f’ is called the first derivative of f.If f’,in turn,is defined on an interval,we can try to compute its first derivative,denoted by f’’,and is called the second derivative of f.Similarly,the nth derivative of f denoted by f^(n),is defined to be the first derivative of f^(n-1).We make the convention that f^(0)=f,that is,the zeroth derivative is the function itself.</FONT></P>
 楼主| 发表于 2004-5-6 09:29:26 | 显示全部楼层
< 0cm 0cm 0pt; TEXT-ALIGN: center" align=center><A><B><FONT face="Times New Roman">Vocabulary</FONT></B></A></P>< 0cm 0cm 0pt"><FONT face="Times New Roman">differential calculus</FONT>微积分<FONT face="Times New Roman">         differentiable</FONT>可微的</P>< 0cm 0cm 0pt"><FONT face="Times New Roman">intergral calculus  </FONT>积分学<FONT face="Times New Roman">           differentiate </FONT>求微分</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">hither to          </FONT>迄今<FONT face="Times New Roman">              integration   </FONT>积分法</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">insurmountable     </FONT>不能超越<FONT face="Times New Roman">          integral     </FONT>积分</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">routine   </FONT>惯常的<FONT face="Times New Roman">                     integrable  </FONT>可积的</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">fuse      </FONT>融合<FONT face="Times New Roman">                       integrate  </FONT>求积分</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">originate </FONT>起源于<FONT face="Times New Roman">                     sign-preserving</FONT>保号</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">evolve     </FONT>发展,引出<FONT face="Times New Roman">                axis  </FONT>轴(单数)</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">tangent line </FONT>切线<FONT face="Times New Roman">                    axes  </FONT>轴(复数)</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">direction    </FONT>方向<FONT face="Times New Roman">                    contradict </FONT>矛盾</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">horizontal  </FONT>水平的<FONT face="Times New Roman">                   contradiction </FONT>矛盾</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">vertical   </FONT>垂直的<FONT face="Times New Roman">                    contrary    </FONT>相反的</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">rudimentary </FONT>初步的,未成熟的<FONT face="Times New Roman">         composite function </FONT>合成函数,复合函数</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">area        </FONT>面积<FONT face="Times New Roman">                     composition </FONT>复合函数</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">intimately  </FONT>紧密地<FONT face="Times New Roman">                   interior  </FONT>内部</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">exploit  </FONT>开拓,开发<FONT face="Times New Roman">                  interior point  </FONT>内点</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">inaugurate </FONT>开始<FONT face="Times New Roman">                      imply  </FONT>推出,蕴含</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">projectile </FONT>弹丸<FONT face="Times New Roman">                      aloft  </FONT>高入云霄</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">friction</FONT>摩擦<FONT face="Times New Roman">                         initial  </FONT>初始的</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">gravity </FONT>引力<FONT face="Times New Roman">                          instant  </FONT>瞬时</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">rate of change </FONT>变化率<FONT face="Times New Roman">                 integration by parts</FONT>分部积分</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">attain     </FONT>达到<FONT face="Times New Roman">                       definite integral   </FONT>定积分</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">defferential   </FONT>微分<FONT face="Times New Roman">                   indefinite integral  </FONT>不定积分</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">differentiation </FONT>微分法<FONT face="Times New Roman">                average   </FONT>平均</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">                    </FONT></P>
 楼主| 发表于 2004-5-6 09:29:43 | 显示全部楼层
< 0cm 0cm 0pt; TEXT-ALIGN: center" align=center><FONT face="Times New Roman"><B>Notes</B><p></p></FONT></P>< 0cm 0cm 0pt 18pt; TEXT-INDENT: -18pt; tab-stops: list 18.0pt; mso-list: l69 level1 lfo14"><FONT face="Times New Roman">1.       Newton and Leibniz,quite independently of one another,were largely responsible for developing…by more or less routine methods.</FONT></P>< 0cm 0cm 0pt 18pt">意思是:在很大程度上是牛顿和莱伯尼,他们相互独立地把积分学的思想发展到这样一种程度,使得迄今一些难于超越的问题可以或多或少地用通常的方法加以解决。</P><P 0cm 0cm 0pt 18pt">这里<FONT face="Times New Roman">responsible for</FONT>的基本意义是:“对<FONT face="Times New Roman">…</FONT>负责”,但也可作“应归功于”解,这里应理解为“归功于”。</P><P 0cm 0cm 0pt 21pt; TEXT-INDENT: -21pt; mso-char-indent-count: -2.0; mso-char-indent-size: 10.5pt"><FONT face="Times New Roman">2</FONT>.<FONT face="Times New Roman">The example described in the foregoing section points the way to the introduction of the concept of derivative.</FONT></P><P 0cm 0cm 0pt 21pt; TEXT-INDENT: -21pt; mso-char-indent-count: -2.0; mso-char-indent-size: 10.5pt"><FONT face="Times New Roman">   </FONT>意思是:前面一节所描述的例子指出了引进导数概念的方法。</P><P 0cm 0cm 0pt 21pt; TEXT-INDENT: -21pt; mso-char-indent-count: -2.0; mso-char-indent-size: 10.5pt"><FONT face="Times New Roman">   </FONT>这里<FONT face="Times New Roman">described</FONT>是过去分词,<FONT face="Times New Roman">foregoing</FONT>是现在分词,两者都用作定语,切不可误认<FONT face="Times New Roman">described</FONT>为过去式谓语。类似句子如:</P><P 0cm 0cm 0pt 21pt; TEXT-INDENT: -21pt; mso-char-indent-count: -2.0; mso-char-indent-size: 10.5pt"><FONT face="Times New Roman">     We begin with a function f defined on some interval(a,b);</FONT></P><P 0cm 0cm 0pt 18pt; TEXT-INDENT: -18pt; tab-stops: list 18.0pt; mso-list: l58 level1 lfo15"><FONT face="Times New Roman">3.       The quotient itself is referred to as the average of change of f in the interval joining x to x+h.</FONT></P><P 0cm 0cm 0pt 18pt">意思是:商本身是指区间<FONT face="Times New Roman">x</FONT>到<FONT face="Times New Roman">x+h</FONT>上<FONT face="Times New Roman">f</FONT>的平均变化率。这里<FONT face="Times New Roman">be referred to as</FONT>意思是:“把<FONT face="Times New Roman">…</FONT></P><P 0cm 0cm 0pt; TEXT-INDENT: 21.75pt">认为是“</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">4</FONT>.<FONT face="Times New Roman">We make the convention that f<SUP>0</SUP>=f</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">   </FONT>意思是:我们约定(按惯例)<FONT face="Times New Roman">f<SUP>0</SUP>=f</FONT></P>
 楼主| 发表于 2004-5-6 09:30:02 | 显示全部楼层
< 0cm 0cm 0pt; TEXT-ALIGN: center" align=center><FONT face="Times New Roman"><B>Exercise</B><p></p></FONT></P>< 0cm 0cm 0pt"><FONT face="Times New Roman">I.1.Fill in the missing words in column B such that the word in column B corresponds to the word in column A in the same sense as “integration”corresponds to “differentiation”</FONT></P>< 0cm 0cm 0pt"><FONT face="Times New Roman">                A                        B</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">        Differentiation                  integration</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">        Differential</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">        Differentiate</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">        Differentiable</FONT></P><P 0cm 0cm 0pt 18pt; TEXT-INDENT: -18pt; tab-stops: list 18.0pt; mso-list: l69 level1 lfo14"><FONT face="Times New Roman">1.       Then choose the correct word from either column A or column B and insert it in each of the blanks.</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman"> (i)The process of finding the derivative of a function is called(           )</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman"> (ii) If f(x) has a derivative at the point x<SUB>0 </SUB>then f(x) is said to be (       )at x<SUB>0.<p></p></SUB></FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman"> (iii) </FONT>∫<FONT face="Times New Roman">f(x)dx is called the indefinite(       ) of f(x).</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman"> <p></p></FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">II  Translate the following two examples into Chinese(pay attention to the phrases used):</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">   Example 1 Find the derivative of </FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">          f(x)=(3x+1)<SUP>4</SUP>/(x<SUP>2</SUP>+2)<SUP>3</SUP></FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">   solution:Taking the logarithms of both sides,we have</FONT></P><P 0cm 0cm 0pt; TEXT-INDENT: 27pt"><FONT face="Times New Roman">ln f(x)=4ln(3x+1)-3ln(x<SUP>2</SUP>+2)</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">   Differentiating both sides of the above equation,we obtain</FONT></P><P 0cm 0cm 0pt; TEXT-INDENT: 26.25pt"><FONT face="Times New Roman">f’(x)/f(x)=12/(3x+1)-6/(x<SUP>2</SUP>+2)</FONT></P><P 0cm 0cm 0pt 44.25pt; TEXT-INDENT: -18pt; tab-stops: list 44.25pt; mso-list: l25 level1 lfo16"><FONT face="Times New Roman">(1)    and (2)together yield </FONT></P><P 0cm 0cm 0pt 26.25pt"><FONT face="Times New Roman">f’(x)=[(3x+1)<SUP>4</SUP>/(x2+2)<SUP>3</SUP>][12/(3x+1)-6/(x<SUP>2</SUP>+2)]</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">   Example 2 Integrate </FONT>∫<FONT face="Times New Roman"> x<SUP>2</SUP> cosxdx</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">   Solutionet u=x<SUP>2</SUP>,v=sin x;</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">          Then du=2xdx,dv=cos x dx</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">          So we have I=</FONT>∫<FONT face="Times New Roman">x<SUP>2</SUP> cos x dx =</FONT>∫<FONT face="Times New Roman">u dv</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">          By applying integration by parts,we have</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">  I=</FONT>∫<FONT face="Times New Roman">udv=uv-</FONT>∫<FONT face="Times New Roman">vdu=x<SUP>2</SUP> sinx-2</FONT>∫<FONT face="Times New Roman">x sinx dx               (1)</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">          Applying integration by parts once again to the indefinite integral</FONT>∫<FONT face="Times New Roman">xsinxdx,we get</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">   </FONT>∫<FONT face="Times New Roman">x sinx dx = -x cos x +sin x +c                     (2)</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">          Substituting (2)into (1)yields</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">   </FONT>∫<FONT face="Times New Roman"> x<SUP>2</SUP> cos x dx=x<SUP>2</SUP> sin x+2x cos x-2 sin x +c </FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">          where c is an arbitrary constant.</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">III. Translate the following example into English:</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">   </FONT>求<FONT face="Times New Roman">y=ln(2x<SUP>2</SUP>-4)</FONT>的导数</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">   </FONT>【解】令<FONT face="Times New Roman">y=ln u,u=(2x2-4),</FONT>则<FONT face="Times New Roman">dy/du=1/u. du/dx=4x   (1)</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">         </FONT>据复合函数求导数的公式,我们有</P><P 0cm 0cm 0pt"><FONT face="Times New Roman">         dy/du=dy/du-du/dx                        (2)</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">         </FONT>把(<FONT face="Times New Roman">1</FONT>)代入(<FONT face="Times New Roman">2</FONT>)式得<FONT face="Times New Roman">dy/dx=(1/u)4x</FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman">         </FONT>把<FONT face="Times New Roman">u</FONT>换为<FONT face="Times New Roman">2x<SUP>2</SUP>-4,</FONT>最后得<FONT face="Times New Roman">dy/dx=4x/(2x<SUP>2</SUP>-4)       </FONT></P><P 0cm 0cm 0pt 36pt; TEXT-INDENT: -36pt; tab-stops: list 36.0pt; mso-list: l0 level1 lfo17"><FONT face="Times New Roman">IV.               Theorem Let f be defined on an open interval I,and assume that f has a relative maximum or a relative minimum at an interior point c of I.If the derivative f’(c )exists,then f’(c )=0</FONT></P><P 0cm 0cm 0pt 36pt"><FONT face="Times New Roman">The proof of this theorem is given in Chinese as follows.Turn it into English:</FONT></P><P 0cm 0cm 0pt 36pt">证明:</P><P 0cm 0cm 0pt 63pt; TEXT-INDENT: -36pt; tab-stops: list 63.0pt; mso-list: l49 level1 lfo18">(1)       在<FONT face="Times New Roman">I</FONT>上定义函数<FONT face="Times New Roman">Q</FONT>(<FONT face="Times New Roman">x</FONT>)</P><P 0cm 0cm 0pt 63pt"><FONT face="Times New Roman">Q(x)=[f(x)-f(c )]/(x-c)   x</FONT>≠<FONT face="Times New Roman">c</FONT></P><P 0cm 0cm 0pt 63pt; TEXT-INDENT: 26.25pt"><FONT face="Times New Roman">f’(c )                 x=c</FONT></P><P 0cm 0cm 0pt 63pt; TEXT-INDENT: -63pt; mso-char-indent-count: -6.0; mso-char-indent-size: 10.5pt"><FONT face="Times New Roman">      (2)   </FONT>因为<FONT face="Times New Roman">f’(c)</FONT>存在,故当<FONT face="Times New Roman">x</FONT>趋向<FONT face="Times New Roman">c</FONT>时,<FONT face="Times New Roman">Q</FONT>(<FONT face="Times New Roman">x</FONT>)趋向<FONT face="Times New Roman">Q</FONT>(<FONT face="Times New Roman">c</FONT>),也即<FONT face="Times New Roman">Q</FONT>(<FONT face="Times New Roman">x</FONT>)在点<FONT face="Times New Roman">x</FONT>=<FONT face="Times New Roman">c</FONT>连续<FONT face="Times New Roman"> </FONT></P><P 0cm 0cm 0pt 63pt; TEXT-INDENT: -36pt; tab-stops: list 63.0pt; mso-list: l49 level1 lfo18">(2)       我们将证明:若<FONT face="Times New Roman">f’</FONT>(<FONT face="Times New Roman">c</FONT>)=<FONT face="Times New Roman">Q</FONT>(<FONT face="Times New Roman">c</FONT>)≠<FONT face="Times New Roman">0</FONT>,则导致矛盾;</P><P 0cm 0cm 0pt 63pt; TEXT-INDENT: -36pt; tab-stops: list 63.0pt; mso-list: l49 level1 lfo18">(3)       设<FONT face="Times New Roman">Q</FONT>(<FONT face="Times New Roman">c</FONT>)<FONT face="Times New Roman">&gt;0</FONT>,根据连续函数的保号性质,存在<FONT face="Times New Roman">c</FONT>点的一个领域,在此领域里,<FONT face="Times New Roman">Q</FONT>(<FONT face="Times New Roman">x</FONT>)是正的;</P><P 0cm 0cm 0pt 63pt; TEXT-INDENT: -36pt; tab-stops: list 63.0pt; mso-list: l49 level1 lfo18">(4)       因此在此领域内,对所有<FONT face="Times New Roman">x</FONT>≠<FONT face="Times New Roman">c</FONT>,<FONT face="Times New Roman">Q</FONT>(<FONT face="Times New Roman">x</FONT>)的分子和分母同号;</P><P 0cm 0cm 0pt 63pt; TEXT-INDENT: -36pt; tab-stops: list 63.0pt; mso-list: l49 level1 lfo18">(5)       即是说,当<FONT face="Times New Roman">x&gt;c</FONT>时,<FONT face="Times New Roman">f</FONT>(<FONT face="Times New Roman">x</FONT>)<FONT face="Times New Roman">&gt;f(c ),</FONT>而当<FONT face="Times New Roman">x&lt;c</FONT>时,<FONT face="Times New Roman">f</FONT>(<FONT face="Times New Roman">x</FONT>)<FONT face="Times New Roman">&lt;f</FONT>(<FONT face="Times New Roman">c</FONT>)这与<FONT face="Times New Roman">f</FONT>在<FONT face="Times New Roman">x</FONT>=<FONT face="Times New Roman">c</FONT>处有一极值相矛盾;</P><P 0cm 0cm 0pt 63pt; TEXT-INDENT: -36pt; tab-stops: list 63.0pt; mso-list: l49 level1 lfo18">(6)       因此<FONT face="Times New Roman">Q</FONT>(<FONT face="Times New Roman">c</FONT>)<FONT face="Times New Roman">&gt;0</FONT>不可能,同理可证<FONT face="Times New Roman">Q</FONT>(<FONT face="Times New Roman">c</FONT>)<FONT face="Times New Roman">&lt;0</FONT>也不真。</P><P 0cm 0cm 0pt"><FONT face="Times New Roman"> <p></p></FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman"> <p></p></FONT></P><P 0cm 0cm 0pt"><FONT face="Times New Roman"> <p></p></FONT></P>
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