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yearover银虫 (正式写手)
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[求助]
一个偏于数学上的翻译，请求指正
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a. The PWE method converges to the true value downward with the increasing of the numbers of plane waves. One hundred plane waves are needed to guarantee the relative error less than 1.0%. PWE方法随着the numbers of plane waves的增加，向下收敛接近到精确解。100个plane waves可以保证相对误差小于1%。 b. Results show that a very low convergence happens with the large n due to the well-known Gibbs oscillations at the interfaces while the computational time is increasing quickly. The reason is that the order of matrix in Eq. (11) is 4n+2, which leads to a large amount of matrix operation with a growth of power series manner. 当n值很大时，收敛速度变慢，其原因是Gibbs oscillations at the interfaces ；同时需要大量的计算时间。原因是方程11中的矩阵的阶数为4n+1，因此，随着n的增大，矩阵计算过程（步骤）呈幂级数级增长。 如有不足之处请指教，谢谢 |
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aixiaoxi123
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yearover: 金币+20, ★★★★★最佳答案 2012-06-01 11:32:37
yearover: 金币+20, ★★★★★最佳答案 2012-06-01 11:32:37
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a. The PWE method converges downward to the exact solution as the number of plane waves increases .When the the number of plane waves reaches 100,the relative error can be keept less than 1.0%. b. Results show that when the value of n is very large,the speed of convergence will be low due to the well-known Gibbs oscillations at the interfaces and at the same time,there needs a mount of time to caculate because the order of matrix of Eq. (11) is 4n+2,that is to say when n increases,the matrix caculation processes will increase by power series. |
9楼2012-05-31 21:44:18
jiujian1984
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2楼2012-05-29 12:43:54
8814402
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爱与雨下: 金币+1 2012-05-29 19:36:53
yearover: 金币+5, 翻译EPI+1, ★有帮助, 谢谢，不过我把表达的意思说错了，应该是按照中文写的英文，请看英文写法是否正确。 2012-05-29 22:54:19
爱与雨下: 金币+1 2012-05-29 19:36:53
yearover: 金币+5, 翻译EPI+1, ★有帮助, 谢谢，不过我把表达的意思说错了，应该是按照中文写的英文，请看英文写法是否正确。 2012-05-29 22:54:19
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a. The PWE method converges to the true value downward with the increasing of the numbers of plane waves. One hundred plane waves are needed to guarantee the relative error less than 1.0%. 随着平面波数量的增加， PWE方法向下收敛接近到精确解。为保证相对误差小于1%，需要100个平面波。 b. Results show that a very low convergence happens with the large n due to the well-known Gibbs oscillations at the interfaces while the computational time is increasing quickly. The reason is that the order of matrix in Eq. (11) is 4n+2, which leads to a large amount of matrix operation with a growth of power series manner. 结果显示，当n值很大时，收敛速度变慢，其原因是众所周知的界面Gibbs振动，而同时需要的计算时间快速增加。原因是方程11中的矩阵的阶数为4n+1，造成大量的矩阵计算过程（步骤）呈幂级数增长。 |
3楼2012-05-29 17:41:11
aixiaoxi123
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4楼2012-05-29 21:16:15
