数学专业英语－Mathematical Discovery

yunjiang · 发表于 2004-5-6 09:57:53

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>To give the flavor of Polya’s thinking and writing in a very beautiful but subtle case , a case that involve a change in the conceptual mode , I shall quote at length from his Mathematical Discovery (vol.II , pp.54 ff): 
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> EXAMPLE I take the liberty a little experiment with the reader , I shall state a simple but not too commonplace theorem of geometry , and then I shall try to reconstruct the sequence of idoas that led to its proof . I shall proceed slowly , very slowly , revealing one clue after the other , and revealing each gradually . I think that before I have finished the whole story , the reader will seize the main idea (unless there is some special hampering circumstance ) . But this main idea is rather unexpected , and so the reader may experience the pleasure of a little discovery .
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> A.If three circles having the same radius pass through a point , the circle through their other three points of intersection also has the same radius .
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Fig.1 Three circles through one point.
 
 This is the theorem that we have to prove . The statement is short and clear , but does not show the details distinctly enough . If we draw a figure (Fig .1) and introduce suitable notation , we arrive at the following more explicit restatement :
 B . Three circles k , l , m have the same radius r and pass through the same point O . Moreover , l and m intersect in the point A , m and k in B , k and l in C . Then the circle e through A , B , C has also the radius 
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Fig .2 too crowded .
 Fig .1 exhibits the four circles k , l , m , and e and their four points of intersection A, B , C , and O . The figure apt to be unsatisfactory , however , for it is not simple , and it is still incomplete ; something seems to be missing ; we failed to take into account something essential , it seems .
 We are dialing with circles . What is a circle ? A circle is determined by center and radius ; all its points have the same distance , measured by the length of the radius , from the center . We failed to introduce the common radius r , and so we failed to take into account an essential part of the hypothesis . Let us , therefore , introduce the centers , K of k , L of l , and M of m . Where should we exhibit the radius r ? there seems to be no reason to treat any one of the three given circles k ; l , and m or any one of the three points of intersection A , B , and C better than the others . We are prompted to connect all three centers with all the points of intersection of the respective circle ; K with B , C , and O , and so forth .
 The resulting figure (Fig . 2) is disconcertingly crowded . There are so many lines , straight and circular , that we have much trouble old－fashioned magazines . The drawing is ambiguous on purpose ; it presents a certain figure if you look t it in the usual way , but if you turn it to a certain position and look at it in a certain peculiar way , suddenly another figure flashes on you , suggesting some more or less witty comment on the first . Can you recognize in our puzzling figure , overladen with straight and circles , a second figure that makes sense ?
 We may hit in a flash on the right figure hidden in our overladen drawing , or we may recognize it gradually . We may be led to it by the effort to solve the proposed problem , or by some secondary , unessential circumstance . For instance , when we are about to redraw our unsatisfactory figure , we may observe that the whole figure is determined by its rectilinear part (Fig . 3) .
 This observation seems to be significant . It certainly simplifies the geometric picture , and it possibly improves the logical situation . It leads us to restate our theorem in the following form .
 C . If the nine segments 
KO , KC , KB ,
LC , LO , LA ,
MB , MA , MO ,
 
are all equal to r , there exists a point E such that the three segments 
EA , EB , EC ,
 
are also equal to r .
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Fig . 3 It reminds you －of what ?
 This statement directs our attention to Fig . 3 . This figure is attractive ; it reminds us of something familiar . (Of what ?)
 Of course , certain quadrilaterals in Fig .3 . such as OLAM have , by hypothesis , four equal sided , they are rhombi , A rhombus I a familiar object ; having recognized it , we can “see “ the figure better . (Of what does the whole figure remind us ?)
 Oppositc sides of a rhombus are parallel . Insisting on this remark , we realize that the 9 segments of Fig . 3 . are of three kinds ; segments of the same kind , such as AL , MO , and BK , are parallel to each other . (Of what does the figure remind us now ?)
 We should not forget the conclusion that we are required to attain . Let us assume that the conclusion is true . Introducing into the figure the center E or the circle e , and its three radii ending in A , B , and C , we obtain (supposedly ) still more rhombi , still more parallel segments ; see Fig . 4 . (Of what does the whole figure remind us now ?)
 Of course , Fig . 4 . is the projection of the 12 edges of a parallele piped having the particularity that the projection of all edges are of equal length . 
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Fig . 4 of course !
 
 Fig . 3 . is the projection of a “nontransparent “ parallelepiped ; we see only 3 faces , 7 vertices , and 9 edges ; 3 faces , 1 vertex , and 3 edges are invisible in this figure . Fig . 3 is just a part of Fig . 4 . but this part defines the whole figure . If the parallelepiped and the direction of projection are so chosen that the projections of the 9 edges represented in Fig . 3 are all equal to r (as they should be , by hypothesis ) , the projections of the 3 remaining edges must be equal to r . These 3 lines of length r are issued from the projection of the 8th , the invisible vertex , and this projection E is the center of a circle passing through the points A , B , and C , the radius of which is r .
 Our theorem is proved , and proved by a surprising , artistic conception of a plane figure as the projection of a solid . (The proof uses notions of solid geometry . I hope that this is not a treat wrong , but if so it is easily redressed . Now that we can characterize the situation of the center E so simply , it is easy to examine the lengths EA , EB , and EC independently of any solid geometry . Yet we shall not insist on this point here .)
 This is very beautiful , but one wonders . Is this the “ light that breaks forth like the morning . “ the flash in which desire is fulfilled ? Or is it merely the wisdom of the Monday morning quarterback ? Do these ideas work out in the classroom ? Followups of attempts to reduce Polya’s program to practical pedagogics are difficult to interpret . There is more to teaching , apparently , than a good idea from a master .
——From Mathematical Experience

yunjiang · 发表于 2004-5-6 09:58:09

<DIV class=Section1 style="LAYOUT-GRID: 15.6pt none"><

0cm 0cm 0pt; TEXT-ALIGN: center" align=center>Vocabulary<

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0cm 0cm 0pt">subtle 巧妙的，精细的clue 线索，端倪hamper 束缚，妨碍disconcert 使混乱，使狼狈ambiguous 含糊的，双关的witty 多智的，有启发的rhombi 菱形（复数）rhombus 菱形parallelepiped 平行六面体projection 射影solid geometry 立体几何pedagogics 教育学，教授法commonplace 老生常谈；平凡的</DIV> <DIV class=Section3 style="LAYOUT-GRID: 15.6pt none"></DIV>

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