数学-外文翻译-外文文献-英文翻译-范德蒙行列式的相关应用.doc

范德蒙行列式的相关应用（一）范德蒙行列式在行列式计算中的应用范德蒙行列式的标准规范形式是：根据范德蒙行列式的特点，将所给行列式包括一些非范德蒙行列式利用各种方法将其化为范德蒙行列式，然后利用范德蒙行列式的结果，把它计算出来。常见的化法有以下几种：1.所给行列式各列（或各行）都是某元素的不同次幂，但其幂次数排列与范德蒙行列式不完全相同，需利用行列式的性质（如提取公因式，调换各行（或各列）的次序，拆项等）将行列式化为范德蒙行列式。例1 计算解 中各行元素都分别是一个数自左至右按递升顺序排列，但不是从0变到。而是由1递升至。如提取各行的公因数，则方幂次数便从0变到. 例2 计算解 本项中行列式的排列规律与范德蒙行列式的排列规律正好相反，为使中各列元素的方幂次数自上而下递升排列，将第列依次与上行交换直至第1行，第行依次与上行交换直至第2行第2行依次与上行交换直至第行，于是共经过次行的交换得到阶范德蒙行列式： 若的第行（列）由两个分行（列）所组成，其中任意相邻两行（列）均含相同分行（列）；且中含有由个分行（列）组成的范德蒙行列式，那么将的第行（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：例3 计算解 将的第一行乘以-1加到第二行得：再将上述行列式的第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：例4 计算 （1）解 先加边，那么再把第1行拆成两项之和，2.加行加列法各行（或列）元素均为某一元素的不同方幂，但都缺少同一方幂的行列式，可用此方法：例5 计算解 作阶行列式：=由所作行列式可知的系数为，而由上式可知的系数为：通过比较系数得：3.拉普拉斯展开法运用公式=来计算行列式的值：例6 计算 解 取第1，3，2行，第1，3，列展开得：=4.乘积变换法例7 设，计算行列式解 例8 计算行列式解 在此行列式中，每一个元素都可以利用二项式定理展开，从而变成乘积的和。根据行列式的乘法规则，其中 对进行例2中的行的变换，就得到范德蒙行列式，于是 = =5.升阶法例9 计算行列式解 将升阶为下面的阶行列式即插入一行与一列，使是关于的阶范德蒙行列式，此处是变数，于是故是一个关于的次多项式，它可以写成另一方面，将按其第行展开，即得比较中关于的系数，即得 (二) 范德蒙行列式在多项式理论中的应用例1 设若至少有个不同的根，则。证明 取为的个不同的根，则有齐次线形方程组 (2)其中看作未知量因为方程组（2）的系数行列式是Vander monde行列式，且所以方程组（2）只有零解，从而有即是零多项式。例2 设是数域F中互不相同的数，是数域F中任一组给定的不全为零的数，则存在唯一的数域F上次数小于n的多项式，使=，证明 设由条件，知 （3）因为互不相同，所以方程组（3）的系数行列式则方程组（3）有唯一解，即唯一的次数小于的多项式使得，例3设多项式， ，则不可能有非零且重数大于的根。证明 反设是的重数大于的根，则=0， 进而即（4）把（4）看成关于为未知量的齐次线形方程组则（4）的系数行列式 = 所以方程组（4）只有零解，从而，所以必有这与矛盾，故没有非零且重数大于的根。附件：(外文资料原文)New proof of the Vander monde determinant and some applications(A): a new method of proof: mathematical inductionWe on the n for that the inductive method.（1）When ， When the result is right.（2）The Vander monde determinant conclusion assumptions for the class, now look at the level of.in，Subtracting the rows times, the first rows by subtracting times, that is, a bottom-up sequentially subtracted from each row on row timeshare有The latter determinant is a Van Dear Mind determinant, according to 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, the proof is completed This result can be abbreviated as even the multiplication signImmediately by the results obtained necessary and sufficient condition for van der Mond determinant is zeroAt least two equal number nThe application of the Vander monde determinant（一）Vander monde determinant in the determinant calculationThe Vander monde determinant standards form：According to the characteristics of the Vander monde determinant given determinant using various methods, including some non-Vander monde determinant into the Vander monde determinant, and then use the results of the Vander monde determinant, it calculated. The common method of following：1. Given determinant of the columns (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, change each line (or column) order, the dissolution of items, etc.) as the determinant of the Vander monde determinant.Example 1.Solutions of elements of each row are a number from left to right in ascending order, but not from 0 to. But by a delivery ros. The common factor, such as extraction of each line number of a power from zero change to. Example 2Solution 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 frequency 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 rows, so after a total ofSub-line exchange Vander monde determinant of order ： If th row (column) consists of two branches (column), any two adjacent lines (columns) contain the 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 monde determinant：Example3Solution will be the first line of the d multiplied by -1 to the second line we have:Then the second row of the above determinant is multiplied by -1 added to line 3, the new determinant line 3 line 4 was multiplied by -1 added:2 plus line to add lawEach row (or column) the different 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 Solution for order determinant：= seen by made determinant coefficient, by the coefficients of the above equation :By comparing the coefficients obtained:3 Laplace expansion methodsApply the formula = to calculate the value of the determinant：Example 5Dereference lines 1，3，2, 1，3， series expansion：=4 product transformation methodExample 6Set up ，Compute the determinantSolution . Ascending OrderExample 7 compute the determinantThe solution l order for the following -order determinantInsert a row with a , order Vander monde determinant, where is the variable, so therefore G H polynomial, it can be written asOn the other hand, the started their first -line, that wasCompare factor that was (B) The Vander monde determinant polynomial theoryExample 1 set is at least type root, 。Proof of different root, homogeneous linear equations (2)Where as unknown amountVander monde Determinant is because the coefficients of the equations (2), and Equations (2) only the zero solution, thus is zero polynomial.Example 2 Let number different from each other in the number field 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 =，Proof Let, by the condition , know （3） different from each other, so the coefficients of the equations (3) determinantThe equations (3) has a unique solution, that is only the number of times less than a polynomial inMake ，Example 3 Let polynomial ， ， impossible to have a non-zero weight is greater than the root of .Proof Anti weight greater than root, =0， And then That（4）(4) as the unknown quantity in homogeneous system of linear equations (4) coefficient determinant = Equations (4) only the zero solution, thus must This is contradiction, it is non-zero root of multiplicity greater than no