machine learning - Second derivative in Tensorflow 2.0

Question

I am trying to calculate the second derivative of a simple vector function of a scalar variable f(x) = (x,x^2,x^3) using TF 2.3 with tf.GradientTape.

def f_ab(x):
    return x, x** 2, x** 3

import tensorflow as tf
in1 = tf.cast(tf.convert_to_tensor(tf.Variable([-1,3,0,6]))[:,None],tf.float64)
with tf.GradientTape(persistent=True) as tape2:
    tape2.watch(in1)
    with tf.GradientTape(persistent=True) as tape:
        tape.watch(in1)
        f1,f2,f3 = f_ab(in1)
    df1 = tape.gradient(f1, in1)
    df2 = tape.gradient(f2, in1)
    df3 = tape.gradient(f3, in1)

d2f1_dx2 = tape2.gradient(df1, in1)
d2f2_dx2 = tape2.gradient(df2, in1)
d2f3_dx2 = tape2.gradient(df3, in1)

for some reason, only the last two derivative are correct while the first, d2f1_dx2, turned out to be None.

When I changed f_ab to

def f_ab(x):
    return x** 1, x** 2, x**3   

I got d2f1_dx2 = <tf.Tensor: shape=(1, 4), dtype=float64, numpy=array([[-0., 0., nan, 0.]])> which is "almost" the correct result.

only when I changed f_ab to

def f_ab(inputs_train):
    return tf.math.log(tf.math.exp(x) ), x** 2, x**3

I got the correct result: d2f1_dx2 = <tf.Tensor: shape=(1, 4), dtype=float64, numpy=array([[0., 0., 0., 0.]])>

Has anyone encountered this problem before? why is the straight forward way gives None?

Answer 1

I think it's because the first derivative of x is a constant. So by the time you compute the second derivative Y and f1 are unrelated to each other since f1 is a constant.

In Tensorflow .gradient() method default to None if there is no deffirentiable path in the graph between the 2 variables.

gradient(
    target, sources, output_gradients=None,
    unconnected_gradients=tf.UnconnectedGradients.NONE
)

see Tensorflow doc

You can change this argument by 0 instead of None and you should get the expected result for the derivative of constant.