Convergent Rating and Billing Engine怎么翻译啊

psg20042022-10-04 11:39:541条回答

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shpaso 共回答了19个问题 | 采纳率73.7%
Convergent Rating and Billing
1年前

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这个英文微积分怎么做?第一问是选择有两个答案一个是divergent 一个是convergent?
这个英文微积分怎么做?


第一问是选择有两个答案
一个是divergent 一个是convergent?
hdhu1年前3
5227542 共回答了21个问题 | 采纳率90.5%
(i) 他是问你积分是否收敛.
convergent = 收敛
divergent = 发散
(ii) 如果收敛,算出积分值.如果发散,输入333.
x^(-5/3) 的积分是收敛的.
设 u = 3x + 5
du = 3dx
dx = du / 3
原式
= ∫ (3x + 5) ^ (-5/3) dx
= (1/3) ∫ u ^ (-5/3) du
= (1/3) (-3/2) u ^ (-2/3)
= (-1/2) u ^ (-2/3)
= (-1/2) (3x + 5) ^ (-2/3)
= 0 - (-1/2) (3 + 5) ^ (-2/3)
= 1/8
准一维流动.Consider a convergent-divergent duct with exit and thr
准一维流动.
Consider a convergent-divergent duct with exit and throat areas of 0.5 m2 and 0.25m2,respectively.The inlet reservoir pressure is 1 atm and the exit static pressure is 0.6 atm.For this pressure ratio,the flow will be supersonic in a portion of the nozzle,terminating with a normal shock inside the nozzle.Calculate the local area ratio (A/A*) at which the shock is located inside the nozzle.
晕ss1了1年前1
cxghb 共回答了22个问题 | 采纳率95.5%
Modern compressible flow by John D Anderson
Eq. (5.28)
Eq. (5.24)
= (0.6/1) × (0.5/0.25) = 1.2
Me2 = -1/(1.4 - 1) + √(1/(1.4-1)^2 +2/(1.4-1)×〖(2/(1.4+1))〗^((1.4+1)/(1.4-1))×〖(1/1.2)〗^2 ) = 0.223
From Table Isentropic flow properties
Mepoe/pe
0.2201.0343
Eq. (5.29)
po2/po1 = 1.0343 × (0.6/1) = 0.621
From Table Normal shock properties
Mpo2/po1
1.22000.6281
From Table Isentropic flow properties
MA/A*
1.22001.0366
The local area ratio at which the shock is located inside the nozzle is 1.0366
英语翻译If the method is convergent,L2 norm of the difference ve
英语翻译
If the method is convergent,L2 norm of the difference vector,Δp,and the residual vector,A(p),converge to zero [see 12].We are reporting convergence of both of
these vectors.For better understanding the error reducing property of these methods,we report variation of ‖A(pk)‖L2/‖A(p0)‖L2 and ‖Δ(pk)‖L2/‖Δ(p0)‖L2
with iterations (k).
We performed three experiments with different initialization in the algorithms (7) and (9).In the first test,let the initial vector be zero for both the algorithms.Figure 2 reports the result.Figure 2(a) presents convergence of the residual vector while the Figure 2(b) presents convergence of the difference vector.In these figures,NIM stands for Newton Iterative Method while AJNIM stands for Alterted Newton Iterative Method.These figures show that both the methods converges at the same rate (quadratically),but still the Altered Jacobian Newton Iterative Method is better in reducing the error.
In the second case,let us select initial vector whose elements 10.Figure 3 presents comparison of the two methods for an initial vector whose elements are 10.It can be seen in the Figures 3(a) and 3(b) that the Altered Jacobian
Newton Iterative Method converges faster than the Newton Iterative Method.Let us finally take an initial vector with elements equal to 100.Figure 4 presents comparison
of the two methods for an initial vector whose elements are 100.The Figures 4(a) and 4(b) show that the Newton Iterative Method is not converging while the Altered
Jacobian Newton Iterative Method still converges.The table 1 presents error after 10 iterations of the two methods.These experiments does show the independence of the convergence of the Altered Jacobian Newton Iterative Method with respect to initialization.We saw that the Newton Iterative Method converges quadratically for the first case (initial guess is zero vector) but its convergence rate decreases as we selected other initial guesses.On the other hand,for all the initial guesses the Altered Jacobian Newton Iterative Method converges quadratically.
We have developed a nonlinear algorithm named Altered Jacobian Newton Iterative Method for solving system nonlinear equations formed from discretization of nonlinear elliptic problems.Presented numerical work shows
that the Altered Jacobian Newton Iterative Method is robust with respect to the initialization.简单翻译下,大概符合就好
无不味1年前1
jsnthajx 共回答了16个问题 | 采纳率81.3%
假如方法收敛,L2范数的差别向量,Δp,而残余向量,一个(p)、收敛到零(参见第12].我们报道了收敛性方面的
这些向量.为更好的理解错误减少财产的这些方法,我们报告一个pk的变化)‖‖L2 /‖(并)‖一ΔL2和‖pk‖L2 /‖Δ‖L2(并)
与迭代(k).
我们进行了三个实验用不同的初始化的演算法(7)、(9).在第一次试验中,让初始向量的算法都为零.图2报告结果.图2(a)提出了收敛的残余向量,而图2(b)提出的收敛性差异向量.在这些数字,尼莫代表牛顿迭代法而AJNIM代表Alterted牛顿迭代法.这些数据表明,两岸方法收敛速度相同(quadratically),但仍然改变后的雅可比牛顿迭代方法更好地减少错误.
在第二种情况下,让我们选择初始向量的元素10.图3这两种方法进行了比较,一开始它的元素是矢量10.可以看出,在这些数字3(a)和(b),3改变雅可比
牛顿迭代法的收敛速度比牛顿迭代法.让我们最后采取一个初始向量与元素等于100.呈现比较图4
这两种方法的,一开始它的元素是矢量100.这些数字4(a)和4个(b)表明,牛顿迭代法是不收敛而改变
雅可比牛顿迭代法仍然收敛.表1礼物的失误后10次迭代的两种方法.这些实验的确展现的独立性的收敛性改变雅可比牛顿迭代法对初始化.我们看见牛顿迭代法的收敛quadratically第一个病例(初始猜测是零矢量),但其收敛速度会随我们选其他最初的猜测.另一方面,对于所有的初始猜改变收敛quadratically雅可比牛顿迭代法.
我们已经发展了一种非线性算法改变雅可比牛顿迭代法命名为解决了系统非线性方程的离散化,形成的非线性椭圆问题.数值作品表现了
这改变了雅可比牛顿迭代法具有良好的鲁棒性和对初始化.
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