Fast transient stability solution using taylor expansion and energy function Three methods of fast stability calculation are considered for on-line monitoring of power system stability. From practical viewpoints, method (1) requires too long a computing time; method (2) also requires too long a prep...
Proposition 3: Taylor Expansion 设f: \mathbb R^n \to \mathbb R,f \in C^2,那么存在\tau_1, \tau_2, \tau_3 \in (0, 1),使得 (1)f(x + p) = f(x) + \nabla f(x+\tau_1 p)^T p (2)f(x+p) = f(x) + \nabla f(x)^T p + \frac 12 p^T \nabla ^2 f(x + ...
aSimilarly, the Taylor expansion of measurement transition function h is given as in (7) and (8). The Jacobian Gt is the value of first derivative of gat the point µt-1. 同样,测量转移函数h泰勒扩展在7和 (8) 被给 (和)。 Jacobian Gt是gat最初倒数的价值点µt-1。[translate]...
如果变化 \Delta y 不是太大,非线性关系可以应用初值周围的泰勒展开式(Taylor expansion)来近似: P_1 = P_0 + f '(y_0)\Delta y + \frac{1}{2}f ''(y_0)(\Delta y)^2 +· ·· 这种展开可以推广到函数依赖于两个或多个变量的情况。对于债券来说,一阶导数与久期有关,二阶导数与凸性有关...
理想气体的问题 已知Taylor Expansion是1/(1-x)=1+x+x2+x3+...(那个数字是x的次方) 把Van der Waals气压转换为理想气体方程,假设V足够大 答案 丫,这题很好玩哪,正好很久没有碰过了,哈哈,玩一下Van der Waals:(p+a/Vm^2)(Vm-b)=RTso p=RT/(Vm-b)-a/Vm^2use the Taylor Expansion (b/Vmb...
We prove that, for every r ϵ n, the best kp-approximations have a Taylor expansion of order r of the form hp=h1∗+∑l=1rαl(p−1)l+γp(r) for some αl ϵ rn, 1 ≤ l ≤ r, and γp(r) ϵ rn with |γp(r)|=o((p−1)r+1)....
taylor’s expansion 泰勒展开式taylor’s series 泰勒级数taylor’s theorem 泰勒定理tension 张力term 项terminal box 终端框terminal point 终点terminal side 终边terminal velocity 终端速度terminating decimal 有尽小数tesselation 密铺;铺嵌;嵌砌test ciiterion 检验标准test of significance 显著性检验tetrahedron ...
1.1 Ito-Taylor Expansion First, let us recall how we can obtain a Taylor expansion from an integral representation for the deterministic case. Consider the autonomous ODE, d dt X(t) = a [X(t)] . Let f be a function of X(t) , then the evolution of the function f is governed by ...
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假如mgf 存在的话,记为 M_X(t) , 我们可以在零点附近作Taylor Expansion, 有 M_X(t)=\sum_{k=0}^\infty \frac{t^k}{k!}E[X^k] . 如果已知所有的 E[X^k],k=0,1,2,... ,而且函数项级数 \sum_{k=0}^\infty \frac{t^k}{k!}E[X^k] 拥有正的收敛半径,那么 mgf 就在零点的一个...