逆计算

In the studies of EEG forward problems, primarily the EEG computation models are established, the analysis solution for concentric circle models and the boundary element method for piecewise humongous models are developed. These provide the basis for EEG inverse problems.

在EEG正问题的研究中，首先建立EEG计算模型，研究了针对同心球模型的解析解法和针对任意分层均匀介质的边界元法，为EEG逆问题研究奠定了基础。

Firstly, the approach formulates a cost functional to turn the inverse problem into a constrained minimization problem according to least squares criterion, then the resulting constrained minimization problem is transformed into an unconstrained minimization problem by using a penalty function technique, and then the closed Fr chet derivatives of the Lagrange function with respect to the properties are derived based on the calculus of variations, finally, one can solve the resulting problem by using any gradient-based algorithm and the finite-difference time-domain method.

该方法首先以最小二乘准则构造目标函数，将逆问题表示为约束最小化问题；接着应用罚函数法转化为无约束最小化问题；然后基于变分计算导出闭式的拉格朗日函数关于特征参数的Fr chet导数；最后借助梯度算法和时域有限差分法迭代反演德拜模型参数。

If the Landweber method and pre-iteration method were convergent, the residue sequence would be monotone decreasing. By analyzing the convergence, the fact is demonstrated that for the severe ill-conditioned matrix of sensitivity coefficient, because of the limitation of the precision of the computer, the residue sequence is not monotone decreasing; thereby the reconstruction images diverge from the true distribution of the conductivity. So the generalized inverse method with threshold is established and the selection way of the regularization parameter is presented, by which the better images are reconstructed.

Landweber法和预迭代法收敛，则其残差序列单调递减，通过对它们的收敛性进行分析，证实对于病态性严重的灵敏度系数矩阵，由于受计算精度的限制，迭代过程中的残差序列不再单调递减，从而导致重建图象偏离实际电导率分布，因此作者建立了带阈值的广义逆算法，提出了规则化参数选择方法，获得了较好的图象重建质量。

In this paper,we solve convolution equation based on Vandermonde Matrix and the method of image displacement elimination in space domain that we propose,so that we can reconstruct sectional image quickly.

本文利用Vandermonde矩阵求逆的特性和我们提出的空间域图像移位消去法，可以用较少的计算量求解出卷积方程组，从而快速重建断层像。

We propose a new method-image displacement elimination method, it overcome shortcomings of FDDI and can solve convolution equation set in space domain quickly.

我们利用Vandermonde矩阵求逆的特性和我们提出的空间域图像移位消去法，可以用较少的计算量求解出卷积方程组，从而快速重建断层像。

The properties of the series are discussed and the formula of its auto correlation function is given.

文中研究了多维逆M序列的性质，给出了自相关函数的计算公式。

However, perfect agreement of ages between structures (i.e., scales and otoliths) did not occur. Crappie length at age should be back-calculated using otoliths when more precise age estimates are needed and with scales when fish are too few to be killed.

因此，对于逆推计算的研究，没有确定耳石和鳞片哪个是完美的，但是在较少的样本的前提下，用耳石来对灰刺盖鱼的体长与年龄比鳞片精确一些。

The time domain solutions are obtained by applying the numerical scheme proposed by Crump, for Laplace inversion.

采用Crump方法直接实现Laplce逆变换。文中算例表明Crump方法具有较好的计算精度。

In this part, the characteristics of transient responses of flexible rotor systems with arbitrarily distributed mass on isotropic or anisotropic supports passing through several critical speeds seguencially are examined. Here, the current knowledge of rotor dynamics on the eigenproblem of damped free whirling vibration, the generaliged damped mode theary, the transient forced vibration and so on combined with KBM asymptotic method in the nonlinear vibration theory is employed and extended.

作者综合运用并引伸推广转子动力学关于阻尼自由回旋振动的特征问题、广义阻尼模态理论、瞬态强廹振动等現代理论和方法，结合非线性振动理论中的KBM渐近法，提出自己独創見解和覌点，成功地建立起正回旋和逆回旋两种情况下转子通过临界点的瞬态振动一阶微分方程组，可以在临界点的邻域内从事方程的数值积分计算。

The algorithm is simple and it is not necessary to solve the Diophantine equation and the inverse matri x so that the on-line amounts of computation are reduced greatly.

该算法简单，不必求解Diophantine方程和逆矩阵，大大减小了在线计算量。

Singer Leona Lewis and former Led Zeppelin guitarist Jimmy Page emerged as the bus transformed into a grass-covered carnival float, and the pair combined for a rendition of "Whole Lotta Love".

歌手leona刘易斯和前率领的飞艇的吉他手吉米页出现巴士转化为基层所涵盖的嘉年华花车，和一双合并为一移交&整个lotta爱&。

This is Kate, and that's Erin.

这是凯特，那个是爱朗。