Evaluation of Def 留数计算 2=b点是f(2)的m阶极点 在b点的邻域内 f(2)=a-m(z-b)-m+…+a-1(2-b) +a0+a1(z-b)+a2(2-b)2+… 两端同乘以(z-b) (z-b)f(2)=a-m(2-b) a0(z-b)+a1(z-b) 考题:两端是否同乘以(b)
Residue theorem Evaluation of Definite Integrals Residue theorem Some Applications of Residue Theorem Residue at Infinity 3êO z = b:´f(z)m 4: 3b:S f(z) = a−m(z − b) −m + · · · + a−1(z − b) −1 + a0 + a1(z − b) + a2(z − b) 2 + · · · üàÓ¦±(z − b) k (z−b) k f(z)=a−m(z−b) k−m+· · ·+a−1(z−b) k−1 + a0(z−b) k+a1(z−b) k+1+· · · · · · · · · gKµüà´ÄÓ¦±(z − b) k , k > mº C. S. Wu 1ù 3ê½n9ÙA^()
Evaluation of Def 留数计算 2=b点是f(2)的m阶极点 在b点的邻域内 f(2)=a-m(z-b)-m+…+a-1(2-b) +a0+a1(z-b)+a2(z 两端同乘以(z-b) (z-b)f(2)=a-m(2-b) a0(z-b)+a1(z-b) 思考题:两端是否同乘以(z-b),k>m?
Residue theorem Evaluation of Definite Integrals Residue theorem Some Applications of Residue Theorem Residue at Infinity 3êO z = b:´f(z)m 4: 3b:S f(z) = a−m(z − b) −m + · · · + a−1(z − b) −1 + a0 + a1(z − b) + a2(z − b) 2 + · · · üàÓ¦±(z − b) k (z−b) k f(z)=a−m(z−b) k−m+· · ·+a−1(z−b) k−1 + a0(z−b) k+a1(z−b) k+1+· · · · · · · · · gKµüà´ÄÓ¦±(z − b) k , k > mº C. S. Wu 1ù 3ê½n9ÙA^()
Evaluation of Def 留数计算 特殊情形:z=b点是f(z)的一阶 在b点的邻域内
Residue theorem Evaluation of Definite Integrals Residue theorem Some Applications of Residue Theorem Residue at Infinity 3êO AÏ/µz = b:´f(z) 4: 3b:S f(z) = a−1(z − b) −1 + a0 + a1(z − b) + a2(z − b) 2 + · · · a−1 = lim z→b (z − b)f(z) C. S. Wu 1ù 3ê½n9ÙA^()
Evaluation of Def 留数计算 特殊情形:z=b点是f(z)的一阶 在b点的邻域内 f(2)=a-1(z-b)-1+ao +a1(z-b)+a2(x-b)2 im(2-b)f(2)
Residue theorem Evaluation of Definite Integrals Residue theorem Some Applications of Residue Theorem Residue at Infinity 3êO AÏ/µz = b:´f(z) 4: 3b:S f(z) = a−1(z − b) −1 + a0 + a1(z − b) + a2(z − b) 2 + · · · a−1 = lim z→b (z − b)f(z) C. S. Wu 1ù 3ê½n9ÙA^()
Evaluation of Def 留数计算 特殊情形:z=b点是f(z)的一阶 在b点的邻域内 f(2)=a-1(z-b)-1+ao +a1(z-b)+a2(x-b)2 a-1=lim(z-b)f(z)
Residue theorem Evaluation of Definite Integrals Residue theorem Some Applications of Residue Theorem Residue at Infinity 3êO AÏ/µz = b:´f(z) 4: 3b:S f(z) = a−1(z − b) −1 + a0 + a1(z − b) + a2(z − b) 2 + · · · a−1 = lim z→b (z − b)f(z) C. S. Wu 1ù 3ê½n9ÙA^()