是梯度的用途数量＆＃x27;在Gradienttape Limited中？

Question

我正在尝试解决以下问题：

因此，我尝试使用tf.gradienttape 4次尝试。以下是我的代码：

import tensorflow as tf
import matplotlib.pyplot as plt
import numpy as np
import math
from scipy import io 

tf.random.set_seed(1234)
np.random.seed(1234)

n_u=20
n_f=30

t_u=tf.cast(tf.linspace(0,1,n_u),tf.float32)
t_u=tf.reshape(t_u,shape=(n_u,1))
t_u=tf.Variable(t_u)

x_u=tf.cast(tf.linspace(0,1,n_u),tf.float32)
x_u=tf.reshape(x_u,shape=(n_u,1))
x_u=tf.Variable(x_u)

t_f=tf.cast(tf.linspace(0,1,n_f),tf.float32)
t_f=tf.reshape(t_f,shape=(n_f,1))
t_f=tf.Variable(t_f)

x_f=tf.cast(tf.linspace(0,1,n_f),tf.float32)
x_f=tf.reshape(x_f,shape=(n_f,1))
x_f=tf.Variable(x_f)

def cov(x,x_,t,t_,sigma=0.5, weight_x=-1, weight_t=-3,repeat = False):
    sigma = tf.math.log(1+tf.math.exp(tf.constant(sigma,dtype=tf.float32)))
    weight_x = tf.math.log(1+tf.math.exp(tf.constant(weight_x,dtype=tf.float32)))
    weight_t = tf.math.log(1+tf.math.exp(tf.constant(weight_t,dtype=tf.float32)))

    if repeat:
        x=tf.repeat(x,x_.shape[0],axis=1)
        x_=tf.transpose(tf.repeat(x_,x.shape[0],axis=1))
        t=tf.repeat(t,t_.shape[0],axis=1)
        t_=tf.transpose(tf.repeat(t_,t.shape[0],axis=1))
    return sigma**2*tf.math.exp(-1/2* weight_x*(x-x_)**2 -1/2 * weight_t*(t-t_)**2)
#======================The above are the settings  =================================
def K_12(x,x_,t,t_):
    x=tf.repeat(x,x_.shape[0],axis=1)
    x_=tf.transpose(tf.repeat(x_,x.shape[0],axis=1))
    t=tf.repeat(t,t_.shape[0],axis=1)
    t_=tf.transpose(tf.repeat(t_,t.shape[0],axis=1))

    with tf.GradientTape() as g:
        g.watch(x_)
        with tf.GradientTape(persistent=True) as gg:
            gg.watch(x_)
            gg.watch(t_)
            with tf.GradientTape() as ggg:
                ggg.watch(x)
                with tf.GradientTape(persistent=True) as gggg:
                    gggg.watch(x)
                    gggg.watch(t)
                    K_uu=cov(x,x_,t,t_)
                K_x=gggg.gradient(K_uu,x)
                K_t=gggg.gradient(K_uu,t)
            K_xx=ggg.gradient(K_x,x)
            K_f=K_t-1*K_xx
        K_f_x=gg.gradient(K_f,x_)
        K_f_t=gg.gradient(K_f,t_)
    K_f_xx=g.gradient(K_f_x,x_)
    return K_t - K_f_xx
print(K_12(x_u,x_f,t_u,t_f))

结果很好，除了两个点：k_f_xx [0,0]和k_f_xx [-1，-1] 对于上述两个点，获得NAN值。对于三个梯度时间，没有获得NAN值，但是对于四个渐变时间，“ NAN”值出现了。因此，我想知道使用超过四个渐变tape是否有任何问题？如果没有，那是什么问题？