7. 计算化学的低标度 Laplace-Transform 方法格点导出思路#
创建时间：2022-10-26
在这份文档中，我们会首先介绍 Laplace-Transform 的定义与意义。随后给出关于 Laplace-Transform 的格点优化思路。最后我们呈现了多种情况下的预优化格点，并比较了不同的积分区间下，格点的数量对 Laplace-Transform 作为近似的误差的影响程度。
import matplotlib.pyplot as plt
from matplotlib_inline.backend_inline import set_matplotlib_formats
%matplotlib inline
set_matplotlib_formats('svg')
import numpy as np
import pandas as pd
import pickle
from scipy import integrate, optimize, interpolate
from scipy.special import expi, exp1, roots_laguerre
from numpy import exp
7.1. Laplace-Transform 方法简介#
7.1.1. 对 \(1/x\) 函数的变换#
计算化学中，特别是类似于 MP2 或 MPn 方法、或者 RPA 方法，会对表达式 \(1/x\) 进行下述变换：
\[
\frac{1}{x} = \int_0^{+\infty} \mathrm{e}^{- t x} \, \mathrm{d} t \simeq \mathrm{LT} (\boldsymbol{t}, \boldsymbol{w}, x) \mathrel{:=} \sum_\alpha^\tau w_\alpha \mathrm{e}^{- t_\alpha x}
\]
对于第二个约等号，它就是将解析的积分转换为数值的积分；其数值格点为 \(\boldsymbol{t}\) 或记为 \(t_\alpha\)，格点权重为 \(\boldsymbol{w}\) 或记为 \(w_\alpha\)。
对于 Laplace-Transform 近似函数 \(\mathrm{LT} (\boldsymbol{t}, \boldsymbol{w}, x)\)，其程序化实现可以通过下述函数给出：
def get_LT(t, w):
t = np.asarray(t).reshape(-1, 1)
w = np.asarray(w).reshape(-1, 1)
def inner(x):
res = (w * np.exp(- t * x)).sum(axis=0)
return res if res.size > 1 else res[0]
return inner
在 \([0,+\infty)\) 上的格点积分通常会选用 Gauss-Laguerre。因此我们不妨以这个格点，进行简单的计算。如果现在的 \(x = 2\)，那么我们预期的 \(\mathrm{LT} (\boldsymbol{t}, \boldsymbol{w}, x) \simeq 1/x = 0.5\)。下述程序的 t_GL20 与 w_GL20 分别是 20 阶 Gauss-Laguerre 格点的坐标与权重。
t_GL20 = roots_laguerre(20)[0]
w_GL20 = np.exp(t_GL20) * roots_laguerre(20)[1]
get_LT(t_GL20, w_GL20)(2)
0.4999999999999992
7.1.2. Laplace-Transform 在张量运算的意义#
事实上 Laplace-Transform 是将一个非常简单的倒数运算，转换为一个非常复杂的格点积分。在大多数情况下，这样的计算方式是非常划不来的。
但对于一些张量运算问题，这种化简为繁的变换反而会给出更优的计算复杂度。在化学中较早利用这一点的文献可以参考 Häser JCP 1991 [1]。以下述计算问题为例：
\[
E = \sum_{ijk}^n \frac{A_{ik} A_{jk}}{\varepsilon_i + \varepsilon_j}
\]
出于方便，我们令角标 \(i, j, k\) 所对应的维度均为 \(n = 1000\)。该问题有两种计算上先后的方法：
\[\begin{split}
\begin{align*}
B_{jk} = \sum_i^n \frac{A_{ik}}{\varepsilon_i + \varepsilon_j}, \; E &= \sum_{jk}^n B_{jk} A_{jk}; \text{or} \\
C_{ij} = \sum_k^n A_{ik} A_{jk}, \; E &= \sum_{ij}^n \frac{C_{ij}}{\varepsilon_i + \varepsilon_j}
\end{align*}
\end{split}\]
不论哪种计算方法，总是有单步的运算涉及到 3 个角标，因此我们认为计算复杂度为 \(O(n^3) \sim 10^{9}\)。
但若引入 Laplace-Transform，那么计算问题就化为
\[\begin{split}
\begin{align*}
E &= \sum_{ijk}^n \frac{A_{ik} A_{jk}}{\varepsilon_i + \varepsilon_j} \\
&\sim \sum_{ijk}^n \sum_\alpha^\tau w_\alpha \mathrm{e}^{- t_\alpha (\varepsilon_i + \varepsilon_j)} A_{ik} A_{jk} \\
&= \sum_{k}^n \sum_\alpha^\tau \left( w_\alpha \cdot \sum_i^n A_{ik} \mathrm{e}^{- t_\alpha \varepsilon_i} \cdot \sum_j^n A_{jk} \mathrm{e}^{- t_\alpha \varepsilon_j} \right)
\end{align*}
\end{split}\]
其中 \(\alpha\) 是 Laplace-Transform 所需要的格点角标，不妨认为其格点总数 \(\tau \sim 20\) (事实上这个数量在实际问题中可以更低)。现在令
\[
D_{k\alpha} = \sum_i^n A_{ik} \mathrm{e}^{- t_\alpha \varepsilon_i}
\]
该步的计算过程牵涉到角标 \(i k \alpha\)，计算复杂度是 \(O(n^2 \tau) \sim 2 \times 10^7\)。随后一步是
\[
E = \sum_{k}^n \sum_\alpha^\tau w_\alpha D_{k \alpha}^2
\]
这一步只牵涉到角标 \(k \alpha\)，其计算消耗不关键。因此 Laplace-Transform 下整个算术计算量大概是 \(2 \times 10^7\)，比起未变换的方法要快上 \(n / \tau \sim 50\) 倍。至此，化简为繁的 Laplace-Transform 在一部分张量运算中的意义就凸显出来了。
7.2. Laplace-Transform 作为近似的准确性#
7.2.1. 20 阶 Gauss-Laguerre 格点准确性的绘图展示#
我们先前已经验证了 \(x = 2\) 时，在 20 阶 Gauss-Laguerre 格点下的 Laplace-Transform 的合理性了。
我们进而会再问
Laplace-Transform 作为近似，有多准确，在何种精度要求下可以使用；
格点数 \(\tau\) 与精度之间的关系怎样，多大程度上可以节省格点数量。
这节先解决第一个问题。我们首先以两种格点呈现 Laplace-Transform 准确性：
20 阶 Gauss-Laguerre 格点 (记为 GL-20)，格点坐标变量为
t_GL20、权重坐标变量为w_GL20；这组格点我们已经生成过了；10 点优化后的格点 (记为 L2log-10)，格点坐标变量为
t_L2log10、权重坐标变量为w_L2log10；这组格点定义在下面的代码块中。
t_L2log10 = 10**np.array([
-2.53336424, -1.68532097, -1.0947392 , -0.5937269 , -0.1473786 ,
0.2593688 , 0.63558393, 0.98782998, 1.32281566, 1.65472166])
w_L2log10 = 10**np.array([
-2.09510186, -1.50144309, -1.00290442, -0.55908055, -0.1570772 ,
0.21293416, 0.55798319, 0.88444692, 1.20302408, 1.55159606])
第一张图展示了函数 \(1/x\) 与 \(\mathrm{LT} (\boldsymbol{t}, \boldsymbol{w}, x)\) 的大体趋势。从下图中，我们能看到的情况是：
对于格点 GL-20，并不是所有的 Laplace-Transform 函数 \(\mathrm{LT} (\boldsymbol{t}, \boldsymbol{w}, x)\) 的结果都能接近真值 \(1/x\)，而只有 \((0.1, 20)\) 区间内的结果接近；
对于格点 L2log-10，则在 \((0.1, 500)\) 区间内的结果都接近真值 \(1/x\)。
Show code cell source
fig, ax = plt.subplots(figsize=(5, 4))
x = np.logspace(-4, 4, 10000)
ax.plot(x, 1 / x, color="black", linewidth=1, label="1 / x")
ax.plot(x, get_LT(t_GL20, w_GL20)(x), label=r"GL-20")
ax.plot(x, get_LT(t_L2log10, w_L2log10)(x), label=r"L2log-10")
ax.fill_between([0.1, 500], [1e10, 1e10], color="C2", alpha=0.2)
ax.set_xscale("log")
ax.set_yscale("log")
ax.set_xlim(1e-4, 1e4)
ax.set_ylim(1e-4, 1e4)
ax.set_xlabel("$x$")
ax.set_ylabel("$y$")
ax.set_title("General trend of $1 / x \sim LT(t,w,x)$")
ax.legend()
fig.tight_layout()
若我们绘制绝对误差图，则可以更加明显地看到这个现象。
GL-20 近似的函数在 \((0.2, 9)\) 的区间内，可以非常接近 \(1/x\)；但一旦超过这个区间，则近似的准确性急剧下降。
而对于 L2log-10 格点，在 \((0.1, 500)\) 的区间内，近似函数的误差在 \(10^{-4}\) 以内波动；而超过这一区间，近似准确性也同样急剧下降。
Show code cell source
fig, ax = plt.subplots(figsize=(5, 3))
ax.plot(x, get_LT(t_GL20, w_GL20)(x) - 1 / x, label="GL-20")
ax.plot(x, get_LT(t_L2log10, w_L2log10)(x) - 1 / x, label="L2log-10")
ax.fill_between([0.1, 500], [-1e-4, -1e-4], [1e-4, 1e-4], color="C2", alpha=0.2)
ax.set_xscale("log")
ax.set_xlim(1e-2, 1e3)
ax.set_ylim(-3e-4, 3e-4)
ax.set_xlabel("$x$")
ax.set_ylabel("Deviation")
ax.set_title("Error function $LT(t,w,x) - 1/x$")
ax.legend()
fig.tight_layout()
在化学中，我们对这类近似函数的要求有两点：
近似的绝对误差应当在 \(10^{-4}\) 级别；这对应的能量误差是 \(10^{-4}\) Hartree，即 0.06 kcal/mol。之所以要有这个精度要求，是因为化学精度为 1 kcal/mol。为了避免数值计算误差影响化学的认识，我们希望将误差控制在非常小的范围内。
至少应当保证区间 \(x \in (0.1, 500)\) 内的数值有较好的近似。对于 MP2 方法而言，\(x\) 就是轨道能级的差值 \(\varepsilon_a + \varepsilon_b - \varepsilon_i - \varepsilon_j\)；对于一般的分子而言，该数值常见的区间即是 \((0.1, 500)\)。
因此，如果只是使用 20 阶的 Gauss-Laguerre 格点 GL-20，是无法满足计算化学对求取 \(1/x\) 运算的需求的。相比而言，误差稍大但可接受的 L2log-10 格点不仅数量更少，并且更为适合用于化学的实际问题。
7.2.2. 误差量标#
在进一步考虑优化格点前，我们先规定近似函数的误差值如何确定。
其中一种误差量标是 \(L_2^\mathrm{log}\) 量标。以格点 \(\boldsymbol{t}\) 与相应权重 \(\boldsymbol{w}\) 为变量，令下述积分损失函数 \(I_2^\mathrm{log}\) 为
\[\begin{split}
\begin{align*}
I_2^\mathrm{log} (\boldsymbol{t}, \boldsymbol{w}) &= \int_{x_\min}^{x_\max} \left( \mathrm{LT} (\boldsymbol{t}, \boldsymbol{w}, x) - \frac{1}{x} \right)^2 \mathrm{d} \log x \\
&= \int_{x_\min}^{x_\max} \frac{1}{x} \left( \sum_\alpha^\tau w_\alpha \mathrm{e}^{- t_\alpha x} - \frac{1}{x} \right)^2 \mathrm{d} x \\
&= \frac{1}{2} \left( \frac{1}{x_\min^2} - \frac{1}{x_\max^2} \right) + \sum_{\alpha \beta}^\tau w_\alpha w_\beta \left( E_1 \big( (t_\alpha + t_\beta) x_\min \big) - E_1 \big( (t_\alpha + t_\beta) x_\max \big) \right) \\
&\quad + 2 \sum_\alpha^\tau w_\alpha \left( t_\alpha \big( E_1 (t_\alpha x_\min) - E_1 (t_\alpha x_\max) \big) - \frac{\mathrm{e}^{- t_\alpha x_\min}}{x_\min} + \frac{\mathrm{e}^{- t_\alpha x_\max}}{x_\max} \right)
\end{align*}
\end{split}\]
并令误差函数 \(L_2^\mathrm{log}\) 为
\[
L_2^\mathrm{log} (\boldsymbol{t}, \boldsymbol{w}) = \sqrt{\frac{I_2^\mathrm{log} (\boldsymbol{t}, \boldsymbol{w})}{\log x_\max - \log x_\min}}
\]
损失函数 \(I_2^\mathrm{log}\) 可以用于求取极小值；误差函数 \(L_2^\mathrm{log}\) 可以用于确定与比较误差数值。
上面所出现的函数 \(E_1\) 是指数积分函数 (参考 NIST 数学函数 [2])：
\[
E_1 (z) = \int_z^{+\infty} \frac{\mathrm{e}^{-t}}{t} \mathrm{d} t
\]
下述程序 loss_L2log 与 err_L2log 分别是求取 \(I_2^\mathrm{log}\) 与 \(L_2^\mathrm{log}\) 的函数。
def loss_L2log(t, w, x_min, x_max):
t2 = t[:, None] + t
w2 = w[:, None] * w
l = 0.5 * (1 / x_min**2 - 1 / x_max**2)
l += (w2 * (exp1(x_min * t2) - exp1(x_max * t2))).sum()
l += 2 * w @ (t * (exp1(x_min * t) - exp1(x_max * t)) - exp(- x_min * t) / x_min + exp(- x_max * t) / x_max)
return l
def err_L2log(t, w, x_min, x_max):
return np.sqrt(loss_L2log(t, w, x_min, x_max) / (np.log(x_max) - np.log(x_min)))
我们可以对先前考察过的格点 GL20 (Gauss-Laguerre 20 格点) 与 L2log-10 (优化的 10 格点) 给出 \((0.1, 500)\) 区间内的 \(L_2\) 误差值：
err_L2log(t_GL20, w_GL20, 0.1, 500)
0.005887670938109512
err_L2log(t_L2log10, w_L2log10, 0.1, 500)
4.676641107400367e-05
7.2.3. 权重的确定方式#
若已知格点 \(\boldsymbol{t}\) 使得 \(I_2^\mathrm{log} (\boldsymbol{t}, \boldsymbol{w})\) 可以取到最小值，那么其对应的格点权重 \(\boldsymbol{w}\) 可以通过下述方式给出：
\[
\boldsymbol{w}: \frac{\partial I_2^\mathrm{log} (\boldsymbol{t}, \boldsymbol{w})}{\partial \boldsymbol{w}} = \boldsymbol{0}
\]
该求取过程可以化为下述向量计算问题：
\[
\mathbf{A} \boldsymbol{w} = \boldsymbol{b}
\]
其中，
\[\begin{split}
\begin{align*}
A_{\alpha \beta} &= \int_{x_\min}^{x_\max} \frac{\mathrm{e}^{- (t_\alpha + t_\beta) x}}{x} \mathrm{d} x \\
&= E_1 \big( (t_\alpha + t_\beta) x_\min \big) - E_1 \big( (t_\alpha + t_\beta) x_\max \big) \\
b_\alpha &= \int_{x_\min}^{x_\max} \frac{\mathrm{e}^{- t_\alpha x}}{x^2} \mathrm{d} x \\
&= \frac{\mathrm{e}^{- t_\alpha x_\min}}{x_\min} - \frac{\mathrm{e}^{- t_\alpha x_\max}}{x_\max} - t_\alpha \big( E_1 (t_\alpha x_\min) - E_1 (t_\alpha x_\max) \big)
\end{align*}
\end{split}\]
下述函数可以求得特定格点 \(\boldsymbol{t}\) 下对应的权重 \(\boldsymbol{w}\)：
def get_w_L2log(t, x_min, x_max):
t2 = t[:, None] + t
a = exp1(t2 * x_min) - exp1(t2 * x_max)
b = exp(- t * x_min) / x_min - exp(- t * x_max) / x_max - t * (exp1(t * x_min) - exp1(t * x_max))
return np.linalg.solve(a, b)
7.2.4. 格点优化#
到此为止，我们可以实现格点优化了。但其中仍然有一些实现细节需要关注：
我们的目标误差是非常小的，\(L_2^\mathrm{log}\) 可能在 \(10^{-5}\) 量级，而 \(I_2^\mathrm{log}\) 可能在 \(10^{-8}\) 量级。将 \(I_2^\mathrm{log}\) 交给库函数进行优化时，很可能因为函数值变化太小而不会优化到底。因此，我们可以对 \(I_2^\mathrm{log}\) 乘上较大的数 (\(10^{7}\)) 以避免优化不进行。
由于格点可能非常小，格点更适合以其对数 \(\log \boldsymbol{t}\) 而非直接对 \(\boldsymbol{t}\) 进行优化。
对于 L-BFGS-B 或 BFGS 等直接利用梯度信息的方法，由于数值误差的存在，需要对数值导数的差分值作额外设定，并且有可能无法优化到最佳情况。在这些梯度方法后，最好接上 Simplex 优化方法譬如 Nelder-Mead 等，并且可能需要使用多次，经常可以再将误差缩小一些。
一些特殊情况下，优化过程中，两个格点距离太近；这导致权重数值会存在巨大的一正一负。这容易产生严重的数值误差，从而给出错误的负的损失函数。遇到这种情况需要手动重新优化。
下述函数过程就是对 \((x_\min, x_\max) = (0.1, 500)\) 区间，对于 10 格点情形，以 \([10^{-3}, 10^{2.5}]\) 为 \(\boldsymbol{t}\) 的取值区间的等比数列为初猜，作以 \(I_2^\mathrm{log}\) 的优化。优化后的结果中，res.x 是 \(\log \boldsymbol{t}\)，而 res.fun 是 \(10^7 I_2^\mathrm{log}\) 的数值。
def loss_work_L2log(x_min, x_max):
def inner(t_log10):
t = 10**t_log10
w = get_w_L2log(t, x_min, x_max)
l = loss_L2log(t, w, x_min, x_max)
return l * 1e7
return inner
t_init = np.linspace(-3, 2.5, 10)
res = optimize.minimize(loss_work_L2log(0.1, 500), t_init,
method="L-BFGS-B", options={"maxiter": 100, "eps": 1e-5})
print("I2log Optimize after L-BFGS-B :", res.fun)
res = optimize.minimize(loss_work_L2log(0.1, 500), res.x,
method="Nelder-Mead", options={"maxiter": 5000})
print("I2log Optimize after Nelder-Mead, iter 1:", res.fun)
res = optimize.minimize(loss_work_L2log(0.1, 500), res.x,
method="Nelder-Mead", options={"maxiter": 5000})
print("I2log Optimize after Nelder-Mead, iter 2:", res.fun)
I2log Optimize after L-BFGS-B : 0.47177181272672897
I2log Optimize after Nelder-Mead, iter 1: 0.18628000475473527
I2log Optimize after Nelder-Mead, iter 2: 0.18627957842909382
res.x
array([-2.53341571, -1.68532385, -1.09473744, -0.59372725, -0.14735734,
0.25941341, 0.63564138, 0.98788772, 1.32286416, 1.6547582 ])
np.log10(get_w_L2log(10**res.x, 0.1, 500))
array([-2.0951377 , -1.50142473, -1.00291163, -0.55906999, -0.1570291 ,
0.21299961, 0.55804858, 0.88449758, 1.20305738, 1.55161723])
事实上我们之前使用的 L2log-10 格点 t_L2log10 就是通过这种方式导出的。可以验证，10**res.x 的结果与 t_L2log10 是非常接近的。
7.3. 更普遍的格点#
对于一般情况而言，上述的 \((x_\min, x_\max) = (0.1, 500)\) 区间内 10 点格点，已经足够用于一般的化学问题了。但如果遇到更普遍的问题，比如分子体系或基组过大 (以至于分母数值需要上千)，或者 HOMO/LUMO gap 数值太小 (小于 0.1 Hartree)，那么上述的格点就不适用了。
同时，如果实际的问题要求并不高，只需要在区间 \((x_\min, x_\max) = (0.5, 50)\) 能计算准确即可 (譬如水分子的 B3LYP/6-31G)，那么 10 个积分格点甚至都太多了。
由于当前问题作为最小值优化问题，其存在较大的潜在数值误差问题；因此实际上不适合依据化学问题的需求，自动化地给出当前问题最佳的格点。这更适合预先手动优化一些格点；而后依据特定问题和特定需求，代入合适的预优化的格点。这一节我们就对一些可以预见的情况作提前优化。
在优化中所使用到的一些方便的函数被折叠在下方。
Show code cell source
def print_outputs(res, x_min, x_max, nspace=0, ret=False):
x_list = np.logspace(np.log10(x_min), np.log10(x_max), 100000)
t1 = np.sort(res.x)
t2 = get_w_L2log(10**t1, x_min, x_max)
t3 = np.log10(np.abs(t2))
t4 = np.sign(t2)
t5 = err_L2log(10**t1, t2, x_min, x_max)
t6 = np.abs(get_LT(10**t1, t2)(x_list) - 1 / x_list).max()
print(" "*nspace + "\"grid_log\": " + "[" + ", ".join(["{:18.10}".format(i).strip() for i in t1]).strip() + "],")
print(" "*nspace + "\"weight_log\": " + "[" + ", ".join(["{:18.10}".format(i).strip() for i in t3]).strip() + "],")
print(" "*nspace + "\"weight_sgn\": " + "[" + ", ".join(["{:}".format(int(i)) for i in t4]).strip() + "],")
print(" "*nspace + "\"L2_err\": {:10.4e},".format(t5))
print(" "*nspace + "\"max_err\": {:10.4e},".format(t6))
if ret:
return {
"grid_log": t1,
"weight_log": t3,
"weight_sgn": np.int(t4),
"L2_err": t5,
"max_err": t6,
}
Show code cell source
def scale_points(x, tau, x_min=None, x_max=None):
x = np.sort(x)
x_min = np.min(x) if x_min is None else x_min
x_max = np.max(x) if x_max is None else x_max
spl = interpolate.interp1d(np.linspace(0, 1, res.x.size), np.sort(res.x))
new_x = spl(np.linspace(0, 1, tau))
new_x = x_min + (new_x - new_x.min()) * (x_max - x_min) / (new_x.max() - new_x.min())
return new_x
优化的结果放入变量 optimized_LT 中。由于数据量太大，我们将该结果也进行了折叠。
Show code cell source
optimized_LT = {
(0.01, 100): {
5: {
"grid_log": [-0.4437405929, 0.5680664814, 1.278880164, 1.859062781, 2.371080954],
"weight_log": [0.04714588325, 0.8394718905, 1.441894483, 1.951473521, 2.431730904],
"weight_sgn": [1, 1, 1, 1, 1],
"L2_err": 4.2277e-02,
"max_err": 1.8173e-01,
},
6: {
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"grid_log": [-2.942906101, -2.203284471, -1.768367303, -1.41588326, -1.0945848, -0.7981809782, -0.5249240222, -0.266584381, -0.01578407605, 0.2270887103, 0.4596841524, 0.6835659818, 0.9035116479, 1.123825851, 1.333846088],
"weight_log": [-2.529918185, -2.124445553, -1.828663165, -1.529786058, -1.242525954, -0.9826332567, -0.7410212044, -0.4994982315, -0.260238953, -0.03432615921, 0.1790458313, 0.3902777234, 0.6073510605, 0.8301848354, 0.9577214119],
"weight_sgn": [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
"L2_err": 7.6906e-08,
"max_err": 2.4408e-07,
},
},
(0.5, 1000): {
5: {
"grid_log": [-2.141933872, -1.130462121, -0.4198380307, 0.1602391233, 0.6721994873],
"weight_log": [-1.651169131, -0.8591923227, -0.2569202816, 0.2525890211, 0.7328088128],
"weight_sgn": [1, 1, 1, 1, 1],
"L2_err": 8.9646e-04,
"max_err": 3.6300e-03,
},
6: {
"grid_log": [-2.46656468, -1.477871714, -0.7798944275, -0.2080702904, 0.2892315188, 0.7458235705],
"weight_log": [-1.984119229, -1.21588992, -0.623468033, -0.1225795209, 0.3242280442, 0.7643250616],
"weight_sgn": [1, 1, 1, 1, 1, 1],
"L2_err": 3.4038e-04,
"max_err": 1.4825e-03,
},
7: {
"grid_log": [-2.654583644, -1.724011285, -1.062754423, -0.5131328964, -0.03202693658, 0.4028991942, 0.8141120125],
"weight_log": [-2.192099764, -1.48909935, -0.9260705161, -0.4427128195, -0.01185250248, 0.3863996012, 0.7933123283],
"weight_sgn": [1, 1, 1, 1, 1, 1, 1],
"L2_err": 1.2365e-04,
"max_err": 5.6806e-04,
},
8: {
"grid_log": [-2.776939975, -1.901898542, -1.2849329, -0.7646937743, -0.304305859, 0.1135646032, 0.5010618569, 0.8756300503],
"weight_log": [-2.331338148, -1.699208828, -1.175124592, -0.7153241586, -0.3016838197, 0.07861775115, 0.4392756867, 0.8192359317],
"weight_sgn": [1, 1, 1, 1, 1, 1, 1, 1],
"L2_err": 4.4203e-05,
"max_err": 2.1161e-04,
},
9: {
"grid_log": [-2.868033786, -2.035023372, -1.461059957, -0.9729962473, -0.536205957, -0.1365672091, 0.234652578, 0.5852655417, 0.9301594711],
"weight_log": [-2.43361206, -1.862759815, -1.381499717, -0.948696923, -0.5543830335, -0.1898124676, 0.1525253465, 0.4837786972, 0.841964583],
"weight_sgn": [1, 1, 1, 1, 1, 1, 1, 1, 1],
"L2_err": 1.5645e-05,
"max_err": 7.9089e-05,
},
10: {
"grid_log": [-2.940970083, -2.137355626, -1.600806778, -1.144679433, -0.7327939934, -0.3527904162, 0.002226689152, 0.3373838477, 0.6584408518, 0.9788453339],
"weight_log": [-2.513575583, -1.99059065, -1.551982903, -1.147880487, -0.7744921742, -0.426409442, -0.09864656733, 0.2140348263, 0.5217362344, 0.8620598777],
"weight_sgn": [1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
"L2_err": 5.5012e-06,
"max_err": 2.9775e-05,
},
11: {
"grid_log": [-2.987934083, -2.200191521, -1.687271773, -1.253901902, -0.861328637, -0.4982743103, -0.1593247574, 0.1592706998, 0.4611030471, 0.7514970935, 1.043874483],
"weight_log": [-2.564106623, -2.068951649, -1.660628141, -1.278778981, -0.9231562347, -0.591621035, -0.2809636622, 0.01222850814, 0.2930274731, 0.57266818, 0.8902159769],
"weight_sgn": [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
"L2_err": 2.0370e-06,
"max_err": 8.6320e-06,
},
12: {
"grid_log": [-3.013531245, -2.233658429, -1.73406301, -1.315692839, -0.9373909238, -0.5870955973, -0.2592227723, 0.04977192332, 0.3422630276, 0.6203925166, 0.8882544585, 1.15849579],
"weight_log": [-2.591405189, -2.110727269, -1.721530425, -1.35659529, -1.015154047, -0.695484701, -0.3946386247, -0.1102923529, 0.1591588726, 0.4171337571, 0.6744942437, 0.9720415772],
"weight_sgn": [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
"L2_err": 1.0836e-06,
"max_err": 5.5331e-06,
},
13: {
"grid_log": [-3.043264589, -2.270841864, -1.783577996, -1.379295253, -1.014940462, -0.6772147949, -0.359518928, -0.05822148155, 0.2284712989, 0.5022052928, 0.7665197651, 1.031659845, 1.404852665],
"weight_log": [-2.622791166, -2.155803071, -1.784532695, -1.436115694, -1.109120313, -0.8005954235, -0.5072161476, -0.2279964547, 0.03762443438, 0.2927923043, 0.5468861876, 0.8319667336, 1.941076997],
"weight_sgn": [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
"L2_err": 5.7991e-07,
"max_err": 3.1209e-06,
},
14: {
"grid_log": [-3.106467772, -2.348250909, -1.888244198, -1.518200737, -1.186054479, -0.8752163279, -0.5804036756, -0.2986270759, -0.02805789211, 0.2317558622, 0.4817168192, 0.7245869953, 0.9653549767, 1.217119151],
"weight_log": [-2.688996512, -2.249623669, -1.92293513, -1.616120097, -1.318520298, -1.032596205, -0.759017907, -0.4954756421, -0.2423399802, -0.0001797033243, 0.2345083942, 0.4683128462, 0.7123724965, 1.009719507],
"weight_sgn": [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
"L2_err": 1.2158e-07,
"max_err": 8.0588e-07,
},
15: {
"grid_log": [-3.221174784, -2.461502579, -1.999582162, -1.627525451, -1.296707804, -0.9901378469, -0.6984948391, -0.4188869562, -0.1512141498, 0.1060236148, 0.3551824101, 0.5978160773, 0.8354841681, 1.074224403, 1.35647128],
"weight_log": [-2.80324342, -2.361619359, -2.031748387, -1.724314864, -1.433925417, -1.153556046, -0.8805472628, -0.6196141612, -0.3706018242, -0.1289032498, 0.1078324208, 0.3397817121, 0.5713530044, 0.8252843843, 1.336388334],
"weight_sgn": [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
"L2_err": 7.8697e-08,
"max_err": 2.6638e-07,
},
},
}
该字典为三层分级
积分有效区间 \((x_\min, x_\max)\)
格点数量 \(\tau\)
格点参数
使用该变量的方式是：
grid_log是以 10 为底的格点对数 \(\log_{10} (\boldsymbol{t})\)。示例：为获得积分区间是 \((0.1, 500)\)、格点数为 10 的格点坐标 \(\boldsymbol{t}\)，使用下述程序可以给出：
t = 10**np.array(optimized_LT[(0.1, 500)][10]["grid_log"])
t
array([2.92721126e-03, 2.06253744e-02, 8.03194080e-02, 2.54496403e-01,
7.11122272e-01, 1.81414536e+00, 4.31441950e+00, 9.71066315e+00,
2.10061356e+01, 4.51216504e+01])
weight_log是以 10 为底的权重绝对值对数 \(\log_10 (|\boldsymbol{w}|)\)。weight_sgn是权重符号 \(\mathrm{sgn} (\boldsymbol{w})\)。尽管绝大多数情况下权重的符号都是正，但在我们优化的结果中，下述情形的符号有负的情况：\((x_\min, x_\max) = (0.5, 100)\), \(\tau = 13, 15\)；
\((x_\min, x_\max) = (0.5, 200)\), \(\tau = 14\)；
示例：为获得积分区间是 \((0.1, 500)\)、格点数为 10 的格点权重 \(\boldsymbol{w}\)，使用下述程序可以给出：
w = optimized_LT[(0.1, 500)][10]["weight_sgn"] * 10**np.array(optimized_LT[(0.1, 500)][10]["weight_log"])
w
array([8.02966560e-03, 3.14911495e-02, 9.92012078e-02, 2.75559297e-01,
6.95335779e-01, 1.63022356e+00, 3.60905672e+00, 7.65579007e+00,
1.59484275e+01, 3.55989272e+01])
L2_err是 \(L_2^\mathrm{log}\) 误差值，如上文定义。max_err是积分有效区间内的最大绝对值误差 \(\max_{x \in [x_\min, x_\max]} | \mathrm{LT} (\boldsymbol{t}, \boldsymbol{w}, x) - 1 / x |\)。
我们可以看一些优化数据的表现。下面的表格中，行头的第一行是积分下限 \(x_\min\)、第二行是积分上限 \(x_\max\)；列头是格点数量 \(\tau\)。表格数据是误差数值 \(L_2^\mathrm{log}\)。
对于积分上限确定为 \(x_\max = 500\) 的情况，我们会发现，随着 \(x_\min\) 的上升，为达成相同的精度，可以使用更少的格点数量。当 \(x_\min = 0.01\) 时需要 13 个格点才能达到 \(L_2^\mathrm{log} < 10^{-4}\)；而 \(x_\min = 0.5\) 时，只需要 7 个格点即可达成，相当于可以节省一半的计算量。
Show code cell source
l2_err_optimized_LT = {}
for (x_min, x_max) in optimized_LT:
l2_err_optimized_LT[(x_min, x_max)] = {}
for tau in optimized_LT[(x_min, x_max)]:
l2_err_optimized_LT[(x_min, x_max)][tau] = optimized_LT[(x_min, x_max)][tau]["L2_err"]
df = pd.DataFrame(l2_err_optimized_LT)
df1 = df[[i for i in df.columns if i[1] == 500]]
df1.style.applymap(lambda v: "background-color: #E2F0D9;" if v < 1e-6 else
"background-color: #DAE3F3;" if 1e-6 <= v and v < 1e-5 else
"background-color: #FFFFFF;" if 1e-5 <= v and v < 1e-4 else
"background-color: #FFF2CC;" if 1e-4 <= v and v < 1e-3 else
"background-color: #FBE5D6;" if 1e-3 <= v else None) \
.format("{:.2e}")
| 0.010000 | 0.020000 | 0.050000 | 0.100000 | 0.200000 | 0.500000 | |
|---|---|---|---|---|---|---|
| 500 | 500 | 500 | 500 | 500 | 500 | |
| 5 | 3.91e-02 | 2.02e-02 | 8.46e-03 | 4.38e-03 | 2.24e-03 | 8.48e-04 |
| 6 | 1.69e-02 | 8.74e-03 | 3.64e-03 | 1.85e-03 | 8.81e-04 | 2.86e-04 |
| 7 | 7.77e-03 | 3.99e-03 | 1.62e-03 | 7.65e-04 | 3.30e-04 | 9.41e-05 |
| 8 | 3.72e-03 | 1.88e-03 | 7.05e-04 | 3.05e-04 | 1.21e-04 | 3.06e-05 |
| 9 | 1.82e-03 | 8.78e-04 | 2.98e-04 | 1.20e-04 | 4.42e-05 | 9.87e-06 |
| 10 | 8.85e-04 | 3.98e-04 | 1.24e-04 | 4.68e-05 | 1.59e-05 | 3.16e-06 |
| 11 | 4.19e-04 | 1.78e-04 | 5.15e-05 | 1.82e-05 | 5.76e-06 | 1.48e-06 |
| 12 | 1.96e-04 | 7.98e-05 | 2.13e-05 | 6.99e-06 | 2.71e-06 | 7.58e-07 |
| 13 | 9.57e-05 | 3.59e-05 | 9.09e-06 | 2.71e-06 | 1.16e-06 | 3.28e-07 |
| 14 | 5.69e-05 | 1.93e-05 | 4.32e-06 | 1.17e-06 | 4.33e-07 | 9.85e-08 |
| 15 | 4.50e-05 | 1.33e-05 | 2.10e-06 | 5.89e-07 | 2.11e-07 | 7.69e-08 |
随后我们看到积分下限确定为 \(x_\min = 0.1\) 时，随着 \(x_\max\) 的增大，尽管达到相同精度所需要的格点数确实增加，但增加的数量一般不大。处理 \(x_\min = 0.1\) 的情形一般需要 10 个格点。
Show code cell source
df1 = df[[i for i in df.columns if i[0] == 0.1]]
df1.style.applymap(lambda v: "background-color: #E2F0D9;" if v < 1e-6 else
"background-color: #DAE3F3;" if 1e-6 <= v and v < 1e-5 else
"background-color: #FFFFFF;" if 1e-5 <= v and v < 1e-4 else
"background-color: #FFF2CC;" if 1e-4 <= v and v < 1e-3 else
"background-color: #FBE5D6;" if 1e-3 <= v else None) \
.format("{:.2e}")
| 0.100000 | ||||
|---|---|---|---|---|
| 100 | 200 | 500 | 1000 | |
| 5 | 4.24e-03 | 4.48e-03 | 4.38e-03 | 4.23e-03 |
| 6 | 1.43e-03 | 1.70e-03 | 1.85e-03 | 1.82e-03 |
| 7 | 4.71e-04 | 6.18e-04 | 7.65e-04 | 8.11e-04 |
| 8 | 1.53e-04 | 2.21e-04 | 3.05e-04 | 3.52e-04 |
| 9 | 4.93e-05 | 7.82e-05 | 1.20e-04 | 1.49e-04 |
| 10 | 1.58e-05 | 2.75e-05 | 4.68e-05 | 6.21e-05 |
| 11 | 5.17e-06 | 9.81e-06 | 1.82e-05 | 2.57e-05 |
| 12 | 2.01e-06 | 3.91e-06 | 6.99e-06 | 1.07e-05 |
| 13 | 1.03e-06 | 1.70e-06 | 2.71e-06 | 4.35e-06 |
| 14 | 5.52e-07 | 1.20e-06 | 1.17e-06 | 1.85e-06 |
| 15 | 5.21e-07 | 7.66e-07 | 5.89e-07 | 9.87e-07 |