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# Runge-Kutta method, Matrix exponential

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# Runge-Kutta method, Matrix exponential
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## PH
• 2023-05-22 : First posting.
## TOC ## Differential equation with matrices (RLC circuit) \begin{align*} \frac{d x_k}{d t} &= A_{ki} x_i + B_{ki} y_i \\ y_k &= C_{ki} x_i + D_{ki} y_i \end{align*} where dummy (repeated) indices means summation over all (e.g. A_{ki} x_i \equiv \sum_{i} A_{ki} x_i). ## \exp(A_{ij}) Taylor series definition of matrix exponential . \exp(A_{ij}) = \sum_{k=0}^{\infty} \frac{1}{k!} A^k If A_{ij} can be diagonalized as A_{ij} = U_{ik} \delta_{kl} D_{kl} U^{-1}_{lj} , then \exp(A_{ij}) = \exp(U_{ik} \delta_{kl} D_{kl} U^{-1}_{lj}) = U_{ik} \exp(\delta_{kl} D_{kl}) U^{-1}_{lj} where \exp(\delta_{kl} D_{kl}) = \delta_{kl} \exp(D_{kl}) . Explicitely speaking, A = \begin{bmatrix} a_1 & 0 & \cdots & 0 \\ 0 & a_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_n \end{bmatrix} , then its exponential can be obtained by exponentiating each entry on the main diagonal: e^A = \begin{bmatrix} e^{a_1} & 0 & \cdots & 0 \\ 0 & e^{a_2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & e^{a_n} \end{bmatrix} . ## Evolution of variables Let an evolution of $\Psi$ from the initial value be specified as follows: \Psi \big( \{ x^{\mu} \} \big) = \Psi \big( \{ x^{\mu} (\lambda) \} \big) . The evolution equation is \frac{d \Psi \big( \{ x^{\mu} (\lambda) \} \big)}{d \lambda} = f \Big( \{ x^{\mu} (\lambda) \} ; \lambda \Big) . Then we want to find the final value $\Psi_f$ from the initial value $\Psi_i$ where \Psi_f \equiv \Psi \big( \{ x^{\mu} (\lambda_f) \} \big) ~~~ \text{and} ~~~ \Psi_i \equiv \Psi \big( \{ x^{\mu} (\lambda_i) \} \big) . And let's think the case that the path $x^{\mu} (\lambda)$ is set to be fixed. Then \begin{align*} \Psi_f - \Psi_i & = \int_{\lambda_i}^{\lambda_f} f \Big( \{ x^{\mu} (\lambda) \} ; \lambda \Big) ~ d \lambda \\ &= \int_{\lambda_i}^{\lambda_f} f \Big( \{ x^{\mu} \} ; \lambda \Big) ~ d \lambda \end{align*} where $f()$ is explicitely expressed by $\{ x^{\mu} \}$, $\lambda$. ## Numerical implementation ### Euler's method (Up to 1st-order) \Psi \Big( \big\{ x^{\mu} \bigg( \lambda_i + \frac{\lambda_f - \lambda_i}{N} (n+1) \bigg) \big\} \Big) - \Psi \Big( \big\{ x^{\mu} \bigg( \lambda_i + \frac{\lambda_f - \lambda_i}{N} n \bigg) \big\} \Big) becomes, with defining $h \equiv \frac{\lambda_f - \lambda_i}{N}$, \begin{align*} \Psi \Big( \big\{ x^{\mu} \bigg( \lambda_i + h (n+1) \bigg) \big\} \Big) - \Psi \Big( \big\{ x^{\mu} \bigg( \lambda_i + h n \bigg) \big\} \Big) &= \sum_{j=0}^{n-1} f \Big( \{ x^{\mu} (\lambda_i + h j) \} ; \lambda_i + h j \Big) \cdot h \\ &= \sum_{j=0}^{n-1} f \Big( \lambda_i + h j \Big) \cdot h . \end{align*} ## Let's try to satisfy/content higher orders. We can candidate that the below equation is correct upto the fourth order $\mathrm{O}(h^4)$. Difining $\lambda_n \equiv \lambda_i + h n$, \Psi \big( \{ x^{\mu} (\lambda_{n+1}) \} \big) := \Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big) + h \cdot \sum_k b_k f_k , where \begin{align*} f_0 &= f \Big( \{ x^{\mu} (\lambda_n) \} ; \lambda_n \Big) \\ f_1 &= f \Big( \{ x^{\mu} (\lambda_n + c_1 \cdot h) \} ; \lambda_n + c_1 \cdot h \Big) \\ f_2 &= f \Big( \{ x^{\mu} (\lambda_n + c_2 \cdot h) \} ; \lambda_n + c_2 \cdot h \Big) \\ f_3 &= f \Big( \{ x^{\mu} (\lambda_n + c_3 \cdot h) \} ; \lambda_n + c_3 \cdot h \Big) . \end{align*} Generally $f$ can be written by f = f \Big( \{ x^{\mu} (\lambda_n + c \cdot h) \} ; \lambda_n + c \cdot h \Big) . With index $p$, f_p = f \Big( \{ x^{\mu} (\lambda_n + c_p \cdot h) \} ; \lambda_n + c_p \cdot h \Big) . ### Taylor expansion of $\Psi$ \begin{align*} \Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big) &= \Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big) + \frac{d \Psi (\lambda_{n})}{d \lambda} \cdot h + \frac{1}{2!} \frac{d^2 \Psi (\lambda_{n})}{d \lambda^2} \cdot h^2 + \frac{1}{3!} \frac{d^3 \Psi (\lambda_{n})}{d \lambda^3} \cdot h^3 + \cdots \\ &= \sum_{k=0}^{\infty} \frac{1}{k!} \frac{d^k \Psi (\lambda)}{d \lambda^k} \bigg|_{\lambda = \lambda_n} \cdot h^k ~~ . \end{align*} Since $f_0 = \frac{d \Psi}{d \lambda}$, \begin{align*} \Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big) &= \Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big) + f_0 (\lambda_n) \cdot h + \frac{1}{2!} \frac{d f_0 (\lambda_n)}{d \lambda} \cdot h^2 + \frac{1}{3!} \frac{d^2 f_0 (\lambda_n)}{d \lambda^2} \cdot h^3 + \cdots \\ &= \Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big) + \sum_{k=0}^{\infty} \frac{1}{(k+1)!} \frac{d^k f_0 (\lambda_n)}{d \lambda^k} \cdot h^{k+1} ~~ . \end{align*} The taylor expansion of the Runge-Kutta method becomes \begin{align*} \Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big) &= \Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big) + h \cdot \sum_p b_p f_p (\lambda_n , h) \\ &= \Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big) + h \cdot \sum_p \sum_{k=0}^{\infty} \frac{b_p}{k!} \frac{d^{k} f_p (\lambda ; h)}{d h^{k}} \bigg|_{\lambda=\lambda_n, ~ h=0} \cdot h^{k} , \end{align*} where \begin{align*} \frac{d f_p \big( \lambda ; h \big)}{d h} \bigg|_{\lambda=\lambda_n , ~ h=0} &= \bigg[ \frac{d x^{\mu}}{d h} \frac{\partial}{\partial x^{\mu}} + \frac{d \lambda}{d h} \frac{\partial}{\partial \lambda} \bigg] f_p \bigg|_{\lambda=\lambda_n , ~ h=0} \\ &= c_p \cdot \bigg[ \frac{d x^{\mu}}{d \lambda} \frac{\partial}{\partial x^{\mu}} + \frac{\partial}{\partial \lambda} \bigg] f_p \bigg|_{\lambda=\lambda_n, ~ h=0} \\ &= c_p \cdot \frac{d f_p}{d \lambda} \bigg|_{\lambda=\lambda_n, ~ h=0} \end{align*} and \begin{align*} \frac{d^k f_p \big( \lambda ; h \big)}{d h^k} \bigg|_{\lambda=\lambda_n, ~ h=0} &= \bigg[ c_p \cdot \frac{d}{d \lambda} \bigg]^k f_p \big( \lambda ; h \big) \Bigg|_{\lambda=\lambda_n, ~ h=0} \\ &= c_p^k \cdot \frac{d^k f_p}{d \lambda^k} \big( \lambda ; h \big) \Bigg|_{\lambda=\lambda_n, ~ h=0} . \end{align*} Since \frac{d^k f_p}{d \lambda^k} \bigg|_{\lambda=\lambda_n, ~ h=0} = \frac{d^k f_0}{d \lambda^k} \bigg|_{\lambda=\lambda_n, ~ h=0}, therefore Eq. becomes \begin{align*} \Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big) &= \Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big) + \sum_p \sum_{k=0}^{\infty} \frac{b_p}{k!} \frac{d^{k} f_0 (\lambda ; h)}{d \lambda^{k}} \bigg|_{\lambda=\lambda_n, ~ h=0} \cdot c_p^k ~ h^{k+1} , \end{align*} ## Let's match the coefficients up to $\textrm{O}(h^4)$. Matching the coefficients of Eq. and Eq. , \frac{1}{(k+1)!} = \sum_p \frac{1}{k!} b_p c_p^k Therefore \sum_p b_p c_p^k = \frac{1}{k+1} ~~~~~ \text{for} ~~ k=0, 1, 2, \cdots ### Case $k=0$ \sum_p b_p = 1 . ### Case $k=1$ \sum_p b_p c_p = \frac{1}{2} . ### Case $k=2$ \sum_p b_p c_p^2 = \frac{1}{3} . ### Case $k=3$ \sum_p b_p c_p^3 = \frac{1}{4} . ## Evolution of variables when it explicitely depends on itself. Let's say that the evolution equation is given by \frac{d \Psi \big( \{ x^{\mu} (\lambda) \} \big)}{d \lambda} = f \Big( \{ x^{\mu} (\lambda) \} ; \lambda ; \Psi (\lambda) \Big) . Then we want to find the final value $\Psi_f$ from the initial value $\Psi_i$ where \Psi_f \equiv \Psi \big( \{ x^{\mu} (\lambda_f) \} \big) ~~~ \text{and} ~~~ \Psi_i \equiv \Psi \big( \{ x^{\mu} (\lambda_i) \} \big) . And let's think the case that the path $x^{\mu} (\lambda)$ is set to be fixed. Then \begin{align*} \Psi_f - \Psi_i & = \int_{\lambda_i}^{\lambda_f} f \Big( \{ x^{\mu} (\lambda) \} ; \lambda, \Psi (\lambda) \Big) ~ d \lambda \\ &= \int_{\lambda_i}^{\lambda_f} f \Big( \{ x^{\mu} \} ; \lambda, \Psi \Big) ~ d \lambda \end{align*} where $f()$ is explicitely expressed by $\{ x^{\mu} \}$, $\lambda$, and $\Psi(\lambda)$. ## Numerical implementation to satisfy/content higher orders. We can candidate that the below equation is correct upto the $m$-th order $\mathrm{O}(h^{m})$. Difining $\lambda_n \equiv \lambda_i + h n$, \Psi \big( \{ x^{\mu} (\lambda_{n+1}) \} \big) := \Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big) + h \cdot \sum_k b_k f_k , where \begin{align*} f_0 &= f \Big( \{ x^{\mu} (\lambda_n) \} ; \lambda_n ; \Psi (\lambda_n) \Big) \\ f_1 &= f \Big( \{ x^{\mu} (\lambda_n + c_1 \cdot h) \} ; \lambda_n + c_1 \cdot h ; \Psi (\lambda_n) + h \cdot a_{10} f_0 \Big) \\ f_2 &= f \Big( \{ x^{\mu} (\lambda_n + c_2 \cdot h) \} ; \lambda_n + c_2 \cdot h ; \Psi (\lambda_n) + h \cdot \big( a_{20} f_0 + a_{21} f_1 \big) \Big) \\ f_3 &= f \Big( \{ x^{\mu} (\lambda_n + c_3 \cdot h) \} ; \lambda_n + c_3 \cdot h ; \Psi (\lambda_n) + h \cdot \big( a_{30} f_0 + a_{31} f_1 + a_{32} f_2 \big) \Big) . \end{align*} Generally $f$ can be written by f = f \Big( \{ x^{\mu} (\lambda_n + c \cdot h) \} ; \lambda_n + c \cdot h ; \Psi (\lambda_n) + h \cdot \sum_{m} \big( a_{m} f_{m} \big) \Big) . With index $p$, f_p = f \Big( \{ x^{\mu} (\lambda_n + c_p \cdot h) \} ; \lambda_n + c_p \cdot h ; \Psi (\lambda_n) + h \cdot \sum_{m} \big( a_{pm} f_{m} \big) \Big) . ### Taylor expansion of $\Psi$ \begin{align*} \Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big) &= \Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big) + \frac{d \Psi (\lambda_{n})}{d \lambda} \cdot h + \frac{1}{2!} \frac{d^2 \Psi (\lambda_{n})}{d \lambda^2} \cdot h^2 + \frac{1}{3!} \frac{d^3 \Psi (\lambda_{n})}{d \lambda^3} \cdot h^3 + \cdots \\ &= \sum_{k=0}^{\infty} \frac{1}{k!} \frac{d^k \Psi (\lambda)}{d \lambda^k} \bigg|_{\lambda = \lambda_n} \cdot h^k ~~ . \end{align*} Since $f_0 = \frac{d \Psi}{d \lambda}$, \begin{align*} \Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big) &= \Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big) + f_0 (\lambda_n) \cdot h + \frac{1}{2!} \frac{d f_0 (\lambda_n)}{d \lambda} \cdot h^2 + \frac{1}{3!} \frac{d^2 f_0 (\lambda_n)}{d \lambda^2} \cdot h^3 + \cdots \\ &= \Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big) + \sum_{k=0}^{\infty} \frac{1}{(k+1)!} \frac{d^k f_0 (\lambda_n)}{d \lambda^k} \cdot h^{k+1} ~~ . \end{align*} The taylor expansion of the Runge-Kutta method becomes \begin{align*} \Psi \big( \{ x^{\mu} (\lambda_{n+1} \equiv \lambda_{n} + h) \} \big) &= \Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big) + h \cdot \sum_p b_p f_p (\lambda_n ; h ; \Psi) \\ &= \Psi \big( \{ x^{\mu} (\lambda_{n}) \} \big) + h \cdot \sum_p \sum_{k=0}^{\infty} \frac{b_p}{k!} \frac{d^{k} f_p (\lambda ; h ; \Psi)}{d h^{k}} \bigg|_{\lambda=\lambda_n, ~ h=0} \cdot h^{k} , \end{align*} where \begin{align*} \frac{d f_p \big( \lambda ; h ; \Psi \big)}{d h} \bigg|_{\lambda=\lambda_n , ~ h=0} &= \bigg[ \frac{d x^{\mu}}{d h} \frac{\partial}{\partial x^{\mu}} + \frac{d \lambda}{d h} \frac{\partial}{\partial \lambda} + \frac{d \Psi}{d h} \frac{\partial}{\partial \Psi} \bigg] f_p \bigg|_{\lambda=\lambda_n , ~ h=0} \\ &= \bigg[ c_p \cdot \frac{d x^{\mu}}{d \lambda} \frac{\partial}{\partial x^{\mu}} + c_p \cdot \frac{\partial}{\partial \lambda} + \sum_m \big( a_{pm} f_m + h \cdot a_{pm} \frac{d f_m}{d h} \big) \frac{\partial}{\partial \Psi} \bigg] f_p \bigg|_{\lambda=\lambda_n, ~ h=0} . \end{align*} Since \begin{align*} \frac{d f_0}{d \lambda} \bigg|_{\lambda=\lambda_n} &= \bigg[ \frac{d x^{\mu}}{d \lambda} \frac{\partial}{\partial x^{\mu}} + \frac{\partial}{\partial \lambda} + \frac{d \Psi}{d \lambda} \frac{\partial}{\partial \Psi} \bigg] f_0 \bigg|_{\lambda=\lambda_n} \\ &= \bigg[ \frac{d x^{\mu}}{d \lambda} \frac{\partial}{\partial x^{\mu}} + \frac{\partial}{\partial \lambda} + f_0 \frac{\partial}{\partial \Psi} \bigg] f_0 \bigg|_{\lambda=\lambda_n} ~~ , \end{align*} Eq. becomes \begin{align*} \frac{d f_p \big( \lambda ; h ; \Psi \big)}{d h} \bigg|_{\lambda=\lambda_n , ~ h=0} &= c_p \cdot \bigg( \frac{f_0}{d \lambda} - f_0 \frac{\partial f_0}{\partial \Psi} \bigg) + \sum_m \Big( a_{pm} f_m + h \cdot a_{pm} \frac{f_m}{d h} \Big) \frac{\partial f_p}{\partial \Psi} \\ &= c_p \cdot \frac{f_0}{d \lambda} + \bigg[ \sum_m \Big( a_{pm} f_m + h \cdot a_{pm} \frac{d f_m}{d h} \Big) - c_p f_0 \bigg] \frac{\partial f_p}{\partial \Psi} \bigg|_{\lambda=\lambda_n , ~ h=0} \end{align*} ## Let's match the coefficients up to $\textrm{O}(h^k)$. Matching the coefficients of Eq. and Eq. where $\frac{d f_p \big( \lambda ; h ; \Psi \big)}{d h} \bigg|_{\lambda=\lambda_n , ~ h=0}$ is calculated by Eq. , \sum_p b_p \frac{1}{k!} \frac{d^k f_p}{d h^k} \bigg|_{\lambda=\lambda_n , ~ h=0} = \frac{1}{(k+1)!} \frac{d^k f_0}{d \lambda^k} \bigg|_{\lambda=\lambda_n} \sum_p b_p \frac{d^k f_p}{d h^k} \bigg|_{\lambda=\lambda_n , ~ h=0} = \frac{1}{k+1} \frac{d^k f_0}{d \lambda^k} \bigg|_{\lambda=\lambda_n} ### Order of $h$. $k=0$. \sum_p b_p f_p \bigg|_{\lambda = \lambda_n, ~ h = 0} = f_0 \bigg|_{\lambda = \lambda_n} . Since f_p \bigg|_{\lambda = \lambda_n, ~ h = 0} = f_0 \bigg|_{\lambda = \lambda_n} , therefore \therefore \sum_p b_p = 1 . ### Order of $h^2$. $k=1$. \sum_p b_p \frac{d f_p}{d h} \bigg|_{\lambda=\lambda_n , ~ h=0} = \frac{1}{2} \frac{d f_0}{d \lambda}\bigg|_{\lambda=\lambda_n} . Since \frac{d f_p}{d h} = c_p \frac{\partial f_p}{\partial \lambda} + \sum_m \Big( a_{pm} f_m + h \cdot a_{pm} \frac{d f_m}{d h} \Big) \frac{\partial f_p}{\partial \Psi} , \sum_p b_p \bigg( c_p \frac{\partial f_0}{\partial \lambda}\bigg|_{\lambda=\lambda_n} + \sum_m \Big( a_{pm} f_0 \Big) \frac{\partial f_0}{\partial \Psi} \bigg|_{\lambda=\lambda_n} = \frac{1}{2} \bigg( \frac{\partial f_0}{\partial \lambda}\bigg|_{\lambda=\lambda_n} + \frac{d \Psi}{d \lambda} \frac{\partial f_0}{\partial \Psi}\bigg|_{\lambda=\lambda_n} \bigg) \sum_p b_p \bigg( c_p \frac{\partial f_0}{\partial \lambda}\bigg|_{\lambda=\lambda_n} + \sum_m \Big( a_{pm} f_0 \Big) \frac{\partial f_0}{\partial \Psi} \bigg|_{\lambda=\lambda_n} = \frac{1}{2} \bigg( \frac{\partial f_0}{\partial \lambda}\bigg|_{\lambda=\lambda_n} + f_0 \frac{\partial f_0}{\partial \Psi}\bigg|_{\lambda=\lambda_n} \bigg) Therefore \therefore \sum_p b_p c_p = \frac{1}{2} and \therefore \sum_{p, m} b_p a_{pm} = \frac{1}{2} This could mean that \therefore \sum_m a_{pm} = c_p . ### Order of $h^3$. $k=2$. \sum_{p} b_p \frac{d^2 f_p}{d h^2} \bigg|_{\lambda=\lambda_n , ~ h=0} = \frac{1}{3} \frac{d^2 f_0}{d \lambda^2} \bigg|_{\lambda=\lambda_n} From Eq. , \begin{align*} \frac{d^2 f_p}{d h^2} &= \frac{d}{d h} \bigg[ c_p \frac{\partial f_p}{\partial \lambda} + \sum_m \Big( a_{pm} f_m + h \cdot a_{pm} \frac{d f_m}{d h} \Big) \frac{\partial f_p}{\partial \Psi} \bigg] \\ &= c_p \frac{d}{d h} \bigg( \frac{\partial f_p}{\partial \lambda} \bigg) + \sum_m \Big( 2 a_{pm} \frac{d f_m}{d h} + h \cdot a_{pm} \frac{d^2 f_m}{d h^2} \Big) \frac{\partial f_p}{\partial \Psi} + \sum_m \Big( a_{pm} f_m + h \cdot a_{pm} \frac{d f_m}{d h} \Big) \frac{d}{d h} \bigg( \frac{\partial f_p}{\partial \Psi} \bigg) . \end{align*} And condition $\lambda=\lambda_n$, $h=0$ applies, \begin{align*} \sum_p b_p \Bigg[ c_p \frac{d}{d h} \bigg( \frac{\partial f_p}{\partial \lambda} \bigg) + \sum_m 2 a_{pm} \frac{d f_m}{d h} \frac{\partial f_p}{\partial \Psi} + \sum_m a_{pm} f_m \frac{d}{d h} \bigg( \frac{\partial f_p}{\partial \Psi} \bigg) \Bigg] &= \frac{1}{3} \frac{d}{d \lambda} \Bigg[ \frac{\partial f_0}{\partial \lambda} + f_0 \frac{\partial f_0}{\partial \Psi} \Bigg] \\ &= \frac{1}{3} \Bigg[ \frac{d}{d \lambda} \bigg( \frac{\partial f_0}{\partial \lambda} \bigg) + \frac{d f_0}{d \lambda} \frac{\partial f_0}{\partial \Psi} + f_0 \frac{d}{d \lambda} \bigg( \frac{\partial f_0}{\partial \Psi} \bigg) \Bigg] \end{align*} From the result Eq. , Eq. becomes \begin{align*} \frac{d f_p}{d h} \bigg|_{\lambda=\lambda_n, ~ h=0} &= c_p \frac{\partial f_0}{\partial \lambda} + \sum_m a_{pm} f_0 \frac{\partial f_0}{\partial \Psi} \bigg|_{\lambda=\lambda_n} \\ &= c_p \bigg( \frac{\partial f_0}{\partial \lambda} + f_0 \frac{\partial f_0}{\partial \Psi} \bigg) \bigg|_{\lambda=\lambda_n} \\ &= c_p \frac{d f_0}{d \lambda} \bigg|_{\lambda=\lambda_n} ~ . \end{align*} Then \begin{align*} \sum_p b_p \Bigg[ c_p^2 \frac{d}{d \lambda} \bigg( \frac{\partial f_0}{\partial \lambda} \bigg) + \sum_m 2 a_{pm} c_m \frac{d f_0}{d \lambda} \frac{\partial f_0}{\partial \Psi} + \sum_m a_{pm} f_0 c_p \frac{d}{d \lambda} \bigg( \frac{\partial f_0}{\partial \Psi} \bigg) \Bigg] &= \frac{1}{3} \Bigg[ \frac{d}{d \lambda} \bigg( \frac{\partial f_0}{\partial \lambda} \bigg) + \frac{d f_0}{d \lambda} \frac{\partial f_0}{\partial \Psi} + f_0 \frac{d}{d \lambda} \bigg( \frac{\partial f_0}{\partial \Psi} \bigg) \Bigg] \end{align*} Therefore \sum_p b_p c_p^2 = \frac{1}{3} 2 \sum_{p, m} b_p a_{pm} c_m = \frac{1}{3} \begin{align*} &\sum_{p, m} b_p a_{pm} f_0 c_p = \frac{1}{3} f_0 \\ &\sum_p b_p c_p^2 = \frac{1}{3} \end{align*} ### Order of $h^4$. $k=3$. \sum_p b_p \frac{d^3 f_p}{d h^3} \bigg|_{\lambda=\lambda_n , ~ h=0} = \frac{1}{4} \frac{d^3 f_0}{d \lambda^3} \bigg|_{\lambda=\lambda_n} Then the results are \sum_p b_p c_p^3 = \frac{1}{4} 3 \sum_{p, q} b_p a_{pq} c_q^2 = \frac{1}{4} 6 \sum_{p, q} b_p c_p a_{pq} c_q = \frac{3}{4} 3 \sum_{p, q} b_p c_p^2 a_{pq} = \frac{1}{4} 6 \sum_{p, q, m} b_p a_{pq} a_{qm} c_m = \frac{1}{4} ### Order of $h^5$. $k=4$. \sum_p b_p \frac{d^4 f_p}{d h^4} \bigg|_{\lambda=\lambda_n , ~ h=0} = \frac{1}{5} \frac{d^4 f_0}{d \lambda^4} \bigg|_{\lambda=\lambda_n} The results are \sum_p b_p c_p^4 = \frac{1}{5} 4 \sum_{p, m} b_p a_{pm} c_m^3 = \frac{1}{5} 6 \sum_{p, m} b_p c_p a_{pm} c_m^2 = \frac{3}{5} 2 \sum_{p, m} b_p c_p^2 a_{pm} c_m = \frac{3}{5} ## Written in papers by color ballpoint pen
## Written in papers 2.
## Candidate set $\{ b_p \}$, $\{ c_p \}$, $\{ a_{pm} \}$ As it is not easy to solve the above simultaneous equations for $\{ b_p \}$, $\{ c_p \}$, $\{ a_{pm} \}$, let's just check the well-known coefficients set $\{ b_p \}$, $\{ c_p \}$, $\{ a_{pm} \}$ to satisfy the above simultaneous equations Eq. , , , , , , , , , , , , , $\cdots$ ### Let's consider $b_0 = \frac{1}{6}$, $b_1 = \frac{1}{3}$, $b_2 = \frac{1}{3}$, $b_3 = \frac{1}{6}$ and $c_0 = 0$, $c_1 = \frac{1}{2}$, $c_2 = \frac{1}{2}$, $c_3 = 1$. \begin{matrix} 0 \\ 1/2 & 1/2 \\ 1/2 & 0 & 1/2 \\ 1 & 0 & 0 & 1 \\ & 1/6 & 1/3 & 1/3 & 1/6 \end{matrix}
### Let's consider $b_0 = \frac{1}{8}$, $b_1 = \frac{3}{8}$, $b_2 = \frac{3}{8}$, $b_3 = \frac{1}{8}$ and $c_0 = 0$, $c_1 = \frac{1}{3}$, $c_2 = \frac{2}{3}$, $c_3 = 1$. \begin{matrix} 0 \\ 1/3 & 1/3 \\ 2/3 & -1/3 & 1 \\ 1 & 1 & -1 & 1 \\ & 1/8 & 3/8 & 3/8 & 1/8 \end{matrix}
### Let's consider $b_0 = \frac{16}{135}$, $b_1 = 0$, $b_2 = \frac{6656}{12825}$, $b_3 = \frac{28561}{56430}$, $b_4 = -\frac{9}{50}$, $b_5 = \frac{2}{55}$ and $c_0 = 0$, $c_1 = \frac{1}{4}$, $c_2 = \frac{3}{8}$, $c_3 = \frac{12}{13}$, $c_4 = 1$, $c_5 = \frac{1}{2}$. \begin{matrix} 0 \\ 1/4 & 1/4 \\ 3/8 & 3/32 & 9/32 \\ 12/13 & 1932/2197 & -7200/2197 & 7296/2197 \\ 1 & 439/216 & -8 & 3680/513 & -845/4104 \\ 1/2 & -8/27 & 2 & -3544/2565 & 1859/4104 & -11/40 \\ & 16/135 & 0 & 6656/12825 & 28561/56430 & -9/50 & 2/55 \end{matrix}
### Let's consider $b_0 = \frac{25}{216}$, $b_1 = 0$, $b_2 = \frac{1408}{2565}$, $b_3 = \frac{2197}{4104}$, $b_4 = -\frac{1}{5}$, $b_5 = 0$ and $c_0 = 0$, $c_1 = \frac{1}{4}$, $c_2 = \frac{3}{8}$, $c_3 = \frac{12}{13}$, $c_4 = 1$, $c_5 = \frac{1}{2}$. \begin{matrix} 0 \\ 1/4 & 1/4 \\ 3/8 & 3/32 & 9/32 \\ 12/13 & 1932/2197 & -7200/2197 & 7296/2197 \\ 1 & 439/216 & -8 & 3680/513 & -845/4104 \\ 1/2 & -8/27 & 2 & -3544/2565 & 1859/4104 & -11/40 \\ & 25/216 & 0 & 1408/2565 & 2197/4104 & -1/5 & 0 \end{matrix}