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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Midpoint method ： ウィキペディア英語版 Midpoint method In numerical analysis, a branch of applied mathematics, the midpoint method is a one-step method for numerically solving the differential equation, :$y\text{'}(t)\; =\; f(t,\; y(t)),\; \backslash quad\; y(t\_0)\; =\; y\_0$. The explicit midpoint method is given by the formula :$y\_\; =\; y\_n\; +\; hf\backslash left(t\_n+\backslash frac,y\_n+\backslash fracf(t\_n,\; y\_n)\backslash right),\; \backslash qquad\backslash qquad\; (1e)$ the implicit midpoint method by :$y\_\; =\; y\_n\; +\; hf\backslash left(t\_n+\backslash frac,\backslash frac12\; (y\_n+y\_)\backslash right),\; \backslash qquad\backslash qquad\; (1i)$ for $n=0,\; 1,\; 2,\; \backslash dots$ Here, $h$ is the ''step size'' — a small positive number, $t\_n=t\_0\; +\; n\; h,$ and $y\_n$ is the computed approximate value of $y(t\_n).$ The explicit midpoint method is also known as the modified Euler method, the implicit method is the most simple collocation method, and, applied to Hamiltionian dynamics, a symplectic integrator. The name of the method comes from the fact that in the formula above the function $f$ giving the slope of the solution is evaluated at $t=t\_n+h/2,$ which is the midpoint between $t\_n$ at which the value of ''y''(''t'') is known and $t\_$ at which the value of ''y''(''t'') needs to be found. The local error at each step of the midpoint method is of order $O\backslash left(h^3\backslash right)$, giving a global error of order $O\backslash left(h^2\backslash right)$. Thus, while more computationally intensive than Euler's method, the midpoint method's error generally decreases faster as $h\; \backslash to\; 0$. The methods are examples of a class of higher-order methods known as Runge-Kutta methods. ==Derivation of the midpoint method== The midpoint method is a refinement of the Euler's method :$y\_\; =\; y\_n\; +\; hf(t\_n,y\_n),\backslash ,$ and is derived in a similar manner. The key to deriving Euler's method is the approximate equality :$y(t+h)\; \backslash approx\; y(t)\; +\; hf(t,y(t))\; \backslash qquad\backslash qquad\; (2)$ which is obtained from the slope formula :$y\text{'}(t)\; \backslash approx\; \backslash frac\; \backslash qquad\backslash qquad\; (3)$ and keeping in mind that $y\text{'}\; =\; f(t,\; y).$ For the midpoint methods, one replaces (3) with the more accurate :$y\text{'}\backslash left(t+\backslash frac\backslash right)\; \backslash approx\; \backslash frac$ when instead of (2) we find :$y(t+h)\; \backslash approx\; y(t)\; +\; hf\backslash left(t+\backslash frac,y\backslash left(t+\backslash frac\backslash right)\backslash right).\; \backslash qquad\backslash qquad\; (4)$ One cannot use this equation to find $y(t+h)$ as one does not know $y$ at $t+h/2$. The solution is then to use a Taylor series expansion exactly as if using the Euler method to solve for $y(t+h/2)$: :$y\backslash left(t\; +\; \backslash frac\backslash right)\; \backslash approx\; y(t)\; +\; \backslash fracy\text{'}(t)=y(t)\; +\; \backslash fracf(t,\; y(t)),$ which, when plugged in (4), gives us :$y(t\; +\; h)\; \backslash approx\; y(t)\; +\; hf\backslash left(t\; +\; \backslash frac,\; y(t)\; +\; \backslash fracf(t,\; y(t))\backslash right)$ and the explicit midpoint method (1e). The implicit method (1i) is obtained by approximating the value at the half step $t+h/2$ by the midpoint of the line segment from $y(t)$ to $y(t+h)$ :$y\backslash left(t+\backslash frac\; h2\backslash right)\backslash approx\; \backslash frac12\backslash bigl(y(t)+y(t+h)\backslash bigr)$ and thus :$\backslash frac\backslash approx\; y\text{'}\backslash left(t+\backslash frac\; h2\backslash right)\backslash approx\; k=f\backslash left(t+\backslash frac\; h2,\backslash frac12\backslash bigl(y(t)+y(t+h)\backslash bigr)\backslash right)$ Inserting the approximation $y\_n+h\backslash ,k$ for $y(t\_n+h)$ results in the implicit Runge-Kutta method :$\backslash begin$ k&=f\left(t_n+\frac h2,y_n+\frac h2 k\right)\\ y_&=y_n+h\,k \end which contains the implicit Euler method with step size $h/2$ as its first part. Because of the time symmetry of the implicit method, all terms of even degree in $h$ of the local error cancel, so that the local error is automatically of order $\backslash mathcal\; O(h^3)$. Replacing the implicit with the explicit Euler method in the determination of $k$ results again in the explicit midpoint method. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Midpoint method」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.