cosx=[e^(ix)+e^(-ix)]/2
e^x cosx=[e^(1+i)x+e^(1-i)x]/2=1+a1x+a2x^2/2!+..anx^n/n!+....
an=[(1+i)^n+(1-i)^n]/2=[(√2)^n(cosnπ/4+isinnπ/4)+(√2)^n(cos-nπ/4+isin(-nπ/4)]/2
=2^(n/2)cos(nπ/4)
本文地址: http://www.goggeous.com/i/1/452837
文章来源:天狐定制
版权声明:除非特别标注,否则均为本站原创文章,转载时请以链接形式注明文章出处。
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2024-12-08 20:59:36职业培训
2024-12-08 20:59:32职业培训
2024-12-08 20:59:29职业培训
2024-12-08 20:59:22职业培训
2024-12-08 20:59:20职业培训
2024-12-08 20:59:13职业培训
2024-12-08 20:59:09职业培训
2024-12-08 20:59:04职业培训
2024-12-08 20:59:03职业培训
2024-12-08 20:59:03职业培训
2024-12-21 22:13职业培训
2024-12-01 19:05职业培训
2024-11-29 01:21职业培训
2025-01-07 17:09职业培训
2024-12-18 01:37职业培训
2025-01-03 09:49职业培训
2024-12-27 10:57职业培训
2025-01-02 01:16职业培训
2024-11-26 11:30职业培训
2024-12-27 22:33职业培训
扫码二维码
获取最新动态