既然你只是要知道爲什麽degeneration law长那样的话... --------------------------------------------------------------- Let F be a field of characteristic 0, a_1,...,a_n be n different values in F, m_1+1,...,m_n+1 be n positive integers. Given c_ij values in F where 0 <= j <= m_i. There is a unique polynomial P(x) with degree < N where N = (m_1+1) + ... + (m_n+1). We claim that P(x) is the one given by Hermite's algorithm. The algorithm constructs a table T the values of which are given by values in n small rectangles following degeneration law (DL) and inductively filling in difference quotients. On the other hand, the values on the top row are those appearing in the intepolation formula and once given, the whole table is determined by inductive formulas (IF) and only addition and multiplication are involved in (IF). 1. Let P(x) be the desired polynomial. We start with N distinct values b_1,...,b_N and P(b_1),...,P(b_N) to form the table. The accuracy of the interpolation formula in this case is guaranteed by Newton form. The value b_1,...,b_N are given by a_1, a_1+y_11,..., a_1+y_(1m_1), a_2, a_2+y_21,..., a_2+y_{2m_2},..., a_n, a_n+y_n1,..., a_n+y_(nm_n), regarded as values in the polynomial ring F[{y_ij}]. The table thus written is denoted by T({y_ij}) with values in the function field F({y_ij}). 2. We claim that T({y_ij}) has values in F[{y_ij}]. Suppose the top row is d_0,...,d_(N-1). The interpolation formula is d_0+d_1(x-b_1)+...+d_(N-1)(x-b_1)...(x-b_(N-1))=P(x). Suppose 0 <= m <= N-1 is the least one such that d_m is not in F[{y_ij}]. We separate P(x) into the tow parts d_0+...+d_m(x-b_1)...(x-b_m) + higher terms. This is the only way to decompose P(x) into a multiple of (x-b_1)...(x-b_(m+1)) and a polynomial of degree atmost m in F({y_ij})[x]. Since P(x) is actually in F[{y_ij}][x], such decomposition is made within F[{y_ij}]. However, the leading coefficient of the first part is not in F[{y_ij}] and we get a contraction. 3. We now evaluate at y_ij = 0 for all possible (i,j). The interpolate formula is d_0(0)+...+d_(N-1)(0)(x-a_1)^(m_1)...(x-a_n)^(m_n)=P(x). The table T(0) is given by having d_0(0),...,d_(N-1)(0) on the top row and (IF) because a ring homomorphism preserves (IF). It suffices to check the values in n small rectangles follow (DL). We call the rth small rectangle T({y_ij})_r and we are looking at T(0)_r. 4. Consider the problem where we are considering the interpolation with only conditions on values and higher derivatives at a_r, the table being T'({y_rj}), and the polynomial solving this problem denoted by P'. T'(0) follows (DL) because we have Taylor expansion. Since P-P' has vanishing order at least m_n+1, T({y_ij})_r and T'({y_rj}) differ from terms of degree at least m_n+1-s > 0 on each entry in the s+1th column. As there are m_n+1 column in that rectangle, T(0)_r=T'(0). -- 第01话 似乎在课堂上听过的样子 第02话 那真是太令人绝望了 第03话 已经没什麽好期望了 第04话 被当、21都是存在的 第05话 怎麽可能会all pass 第06话 这考卷绝对有问题啊 第07话 你能面对真正的分数吗 第08话 我，真是个笨蛋 第09话 这样成绩，教授绝不会让我过的 第10话 再也不依靠考古题 第11话 最後留下的补考 第12话 我最爱的学分 --

※ 发信站: 批踢踢实业坊(ptt.cc), 来自: 160.39.29.33 (美国) ※ 文章网址: https://webptt.com/cn.aspx?n=bbs/Math/M.1691362921.A.E78.html ※ 编辑: cmrafsts (160.39.29.33 美国), 08/07/2023 07:18:06