數(shù)學(xué)外文翻譯--范德蒙行列式的相關(guān)應(yīng)用

1、<p>　　分類號(hào)：__________ 學(xué)校代碼：11059</p><p>　　學(xué) 號(hào)：0907021021</p><p>　　畢業(yè)論文外文翻譯材料</p><p>　　學(xué)生姓名： 陳仁俊 </p><p

2、>　　學(xué) 號(hào)： 0907021021 </p><p>　　專業(yè)班級(jí)： 數(shù)學(xué)一班 </p><p>　　指導(dǎo)教師： 王敏秋 </p><p>　　正文：外文資料譯文 附 件：外文資料原文 </p><p&g

3、t;　　范德蒙行列式的相關(guān)應(yīng)用</p><p>　?。ㄒ唬┓兜旅尚辛惺皆谛辛惺接?jì)算中的應(yīng)用</p><p>　　范德蒙行列式的標(biāo)準(zhǔn)規(guī)范形式是：</p><p>　　根據(jù)范德蒙行列式的特點(diǎn)，將所給行列式包括一些非范德蒙行列式利用各種方法將其化為范德蒙行列式，然后利用范德蒙行列式的結(jié)果，把它計(jì)算出來。常見的化法有以下幾種：</p><p>　　1

4、.所給行列式各列（或各行）都是某元素的不同次冪，但其冪次數(shù)排列與范德蒙行列式不完全相同，需利用行列式的性質(zhì)（如提取公因式，調(diào)換各行（或各列）的次序，拆項(xiàng)等）將行列式化為范德蒙行列式。</p><p><b>　　例1 計(jì)算</b></p><p>　　解 中各行元素都分別是一個(gè)數(shù)自左至右按遞升順序排列，但不是從0變到。而是由1遞升至。如提取各行的公因數(shù)，則方冪次數(shù)便從

5、0變到.</p><p><b>　　例2 計(jì)算</b></p><p>　　解 本項(xiàng)中行列式的排列規(guī)律與范德蒙行列式的排列規(guī)律正好相反，為使中各列元素的方冪次數(shù)自上而下遞升排列，將第列依次與上行交換直至第1行，第行依次與上行交換直至第2行第2行依次與上行交換直至第行，于是共經(jīng)過</p><p>　　次行的交換得到階范德蒙行列式：</p&

6、gt;<p>　　若的第行（列）由兩個(gè)分行（列）所組成，其中任意相鄰兩行（列）均含相同分行（列）；且中含有由個(gè)分行（列）組成的范德蒙行列式，那么將的第行（列）乘以-1加到第行（列），消除一些分行（列）即可化成范德蒙行列式：</p><p><b>　　例3 計(jì)算</b></p><p>　　解 將的第一行乘以-1加到第二行得：</p>&l

7、t;p>　　再將上述行列式的第2行乘以-1加到第3行，再在新行列式中的第3行乘以-1加到第4行得：</p><p><b>　　例4 計(jì)算</b></p><p><b>　?。?）</b></p><p><b>　　解 先加邊，那么</b></p><p>　　再把第

8、1行拆成兩項(xiàng)之和，</p><p><b>　　2.加行加列法</b></p><p>　　各行（或列）元素均為某一元素的不同方冪，但都缺少同一方冪的行列式，可用此方法：</p><p><b>　　例5 計(jì)算</b></p><p><b>　　解 作階行列式：</b><

9、;/p><p><b>　　=</b></p><p>　　由所作行列式可知的系數(shù)為，而由上式可知的系數(shù)為：</p><p><b>　　通過比較系數(shù)得：</b></p><p><b>　　3.拉普拉斯展開法</b></p><p>　　運(yùn)用公式=來計(jì)算行

10、列式的值：</p><p><b>　　例6 計(jì)算 </b></p><p>　　解 取第1，3，2行，第1，3，列展開得：</p><p><b>　　=</b></p><p><b>　　4.乘積變換法</b></p><p>　　例7 設(shè)，計(jì)算行

11、列式</p><p><b>　　解 </b></p><p><b>　　例8 計(jì)算行列式</b></p><p>　　解 在此行列式中，每一個(gè)元素都可以利用二項(xiàng)式定理展開，從而變成乘積的和。根據(jù)行列式的乘法規(guī)則，，其中</p><p>　　對(duì)進(jìn)行例2中的行的變換，就得到范德蒙行列式，于是<

12、/p><p><b>　　=</b></p><p><b>　　=</b></p><p><b>　　5.升階法</b></p><p><b>　　例9 計(jì)算行列式</b></p><p>　　解 將升階為下面的階行列式</

13、p><p>　　即插入一行與一列，使是關(guān)于的階范德蒙行列式，此處是變數(shù)，于是故是一個(gè)關(guān)于的次多項(xiàng)式，它可以寫成</p><p>　　另一方面，將按其第行展開，即得</p><p>　　比較中關(guān)于的系數(shù)，即得</p><p>　　(二) 范德蒙行列式在多項(xiàng)式理論中的應(yīng)用</p><p>　　例1 設(shè)若至少有個(gè)不同的根，則。&

14、lt;/p><p>　　證明 取為的個(gè)不同的根，則有齊次線形方程組</p><p><b>　　(2)</b></p><p><b>　　其中看作未知量</b></p><p>　　因?yàn)榉匠探M（2）的系數(shù)行列式是Vander monde行列式，且</p><p>　　所以方程組

15、（2）只有零解，從而有即是零多項(xiàng)式。</p><p>　　例2 設(shè)是數(shù)域F中互不相同的數(shù)，是數(shù)域F中任一</p><p>　　組給定的不全為零的數(shù)，則存在唯一的數(shù)域F上次數(shù)小于n的多項(xiàng)式</p><p><b>　　，使=，</b></p><p><b>　　證明 設(shè)由條件，知</b></p

16、><p><b>　?。?）</b></p><p>　　因?yàn)榛ゲ幌嗤?，所以方程組（3）的系數(shù)行列式</p><p>　　則方程組（3）有唯一解，即唯一的次數(shù)小于的多項(xiàng)式</p><p><b>　　使得，</b></p><p><b>　　例3設(shè)多項(xiàng)式， </

17、b></p><p>　　，則不可能有非零且重?cái)?shù)大于的根。</p><p>　　證明 反設(shè)是的重?cái)?shù)大于的根，則=0，</p><p><b>　　進(jìn)而即</b></p><p><b>　?。?）</b></p><p>　　把（4）看成關(guān)于為未知量的齊次線形方程組則（

18、4）的系數(shù)行列式</p><p><b>　　= </b></p><p>　　所以方程組（4）只有零解，從而，所以必有</p><p>　　這與矛盾，故沒有非零且重?cái)?shù)大于的根。</p><p>　　附件：(外文資料原文)</p><p>　　New proof of the Vander mon

19、de determinant and some applications</p><p>　　(A): a new method of proof: mathematical induction</p><p>　　We on the n for that the inductive method.</p><p>　　（1）When ， When the

20、result is right.</p><p>　?。?）The Vander monde determinant conclusion assumptions for the class, now look at the level of.in</p><p>　　，Subtracting the rows times, the first rows by subtracting

21、 times, that is, a bottom-up sequentially subtracted from each row on row timeshare</p><p><b>　　有</b></p><p>　　The latter determinant is a Van Dear Mind determinant, according to

22、 the induction assumption, it is equal to all possible difference ；Contains difference all appear in front of the consequent conclusion van dear Mend determinant of the level n the establishment of mathematical induction

23、, the proof is completed </p><p>　　This result can be abbreviated as even the multiplication sign</p><p>　　Immediately by the results obtained necessary and sufficient condition for van der Mond

24、 determinant is zero</p><p>　　At least two equal number n</p><p>　　The application of the Vander monde determinant</p><p>　?。ㄒ唬￢ander monde determinant in the determinant calculatio

25、n</p><p>　　The Vander monde determinant standards form：</p><p>　　According to the characteristics of the Vander monde determinant given determinant using various methods, including some non-Vand

26、er monde determinant into the Vander monde determinant, and then use the results of the Vander monde determinant, it calculated. The common method of following：</p><p>　　1. Given determinant of the columns (

27、or rows) are different powers of an element, but the number of power arrangement with the Vander monde determinant is not exactly the same, the need to use the nature of the determinant (such as extraction common divisor

28、, change each line (or column) order, the dissolution of items, etc.) as the determinant of the Vander monde determinant.</p><p>　　Example 1.</p><p>　　Solutions of elements of each row are a nu

29、mber from left to right in ascending order, but not from 0 to. But by a delivery rosé. The common factor, such as extraction of each line number of a power from zero change to..</p><p><b>　　Exampl

30、e 2</b></p><p>　　Solution The law of the law of the determinant arranged with the arrangement of the Vander monde determinant on the contrary, to make the columns in the elements of a power of frequen

31、cy from top to bottom in ascending order, the columns uplink switch in turn until the first line, rows sequentially exchanged with the uplink until the second row 2nd row sequentially with uplink switch until the first

32、rows, so after a total of</p><p>　　Sub-line exchange Vander monde determinant of order ：</p><p>　　If th row (column) consists of two branches (column), any two adjacent lines (columns) contain t

33、he same branch (column); and contains the Vander monde ranks n branches (column)type, then the i-th row (column) multiplied by -1 added to the -row (column) to eliminate some of the branches (column) into the Vander mon

34、de determinant：</p><p><b>　　Example3</b></p><p>　　Solution will be the first line of the d multiplied by -1 to the second line we have:</p><p>　　Then the second row of t

35、he above determinant is multiplied by -1 added to line 3, the new determinant line 3 line 4 was multiplied by -1 added:</p><p>　　2 plus line to add law</p><p>　　Each row (or column) the differen

36、t square a power of the whose elements are all the of an element, but are the lack of the the determinant of the same party a power of, available this method:Example 4 </p><p>　　Solution for order determina

37、nt：</p><p><b>　　=</b></p><p>　　seen by made ??determinant coefficient, by the coefficients of the above equation :</p><p>　　By comparing the coefficients obtained:</p

38、><p>　　3 Laplace expansion methods</p><p>　　Apply the formula = to calculate the value of the determinant：</p><p><b>　　Example 5</b></p><p>　　Dereference lin

39、es 1，3，2, 1，3， series expansion：</p><p><b>　　=</b></p><p>　　4 product transformation method</p><p><b>　　Example 6</b></p><p>　　Set up ，Compute t

40、he determinant</p><p><b>　　Solution </b></p><p>　　. Ascending Order</p><p>　　Example 7 compute the determinant</p><p>　　The solution l order for the follow

41、ing -order determinant</p><p>　　Insert a row with a ,, order Vander monde determinant, where is the variable, so therefore G H polynomial, it can be written as</p><p>　　On the other hand, the

42、 started their first -line, that was</p><p>　　Compare factor that was</p><p>　　(B) The Vander monde determinant polynomial theory</p><p>　　Example 1 set is at least type root

43、, 。</p><p>　　Proof of different root, homogeneous linear equations (2)</p><p>　　Where as unknown amount</p><p>　　Vander monde Determinant is becau

44、se the coefficients of the equations (2), and</p><p>　　Equations (2) only the zero solution, thus is zero polynomial.</p><p>　　Example 2 Let number different from each other in the number fi

45、eld F, is any number field F Group given not all zero number, then there exists a unique number field F number of less than polynomial in n ，Make =，</p><p>　　Proof Let, by the condition , know</p>

46、<p><b>　　（3）</b></p><p>　　different from each other, so the coefficients of the equations (3) determinant</p><p>　　The equations (3) has a unique solution, that is only the numb

47、er of times less than a polynomial in</p><p><b>　　Make ，</b></p><p>　　Example 3 Let polynomial ， </p><p>　　， impossible to have a non-zero weight is greater than the roo

48、t of .</p><p>　　Proof Anti weight greater than root, =0，</p><p>　　And then That</p><p><b>　　（4）</b></p><p>　　(4) as the unknown quantity in homogene

49、ous system of linear equations (4) coefficient determinant</p><p><b>　　= </b></p><p>　　Equations (4) only the zero solution, thus must </p><p>　　This is contradiction