Auto Differentiation

Hi, I have 2 questions:

  • In question 3: as I understood, derivative of a vector over a vector results in a matrix, since f is a vector of a, I got a vector instead, which I dont understand.
  • In question 5: I have a small problem: my intention is to plot the x_grad after GradientTape, I tried but still error. Hope sb can help me out.
    import matplotlib.pyplot as plt
    import numpy as np

x = tf.range(-10,10,1,dtype=tf.float32)
x = tf.Variable(x)

with tf.GradientTape() as t:
y = np.sin(x)

x_grad = t.gradient(y, x)

x = np.arange(-10,10,0.1)
plt.plot(x, np.sin(x), ‘r’)

#Even when I try to assign value to y, I got None result
y = tf.Variable(tf.zeros_like(x))
with tf.GradientTape() as t:
for i in range(tf.size(y)):

Thank you :smiley:

For question 5, you can plot ‘x_grad’ directly through pl.plt without np.

Hi @randomonlinedude, to calculate the second gradients, you can use tf.hessians():wink:

For question 5, I also ran into some similar problems, like:

TypeError: ResourceVariable doesn't have attribute ....
Y is None object.

I think the reason is that numpy ndarrays are different from tensors and they have different attributes, thus we can’t mix them up. For example, when you serve y as a numpy array in tf.gradient(y,x) method, it will return a None Object.
Also, you can refer to my code below:

import tensorflow as tf
import matplotlib.pyplot as plt

x = tf.range(-10, 10, 0.1)
x = tf.Variable(x)

with tf.GradientTape() as t:
    y = tf.math.sin(x)
x_grad = t.gradient(y, x)

# plotting
x = np.arange(-10, 10, 0.1)
plt.plot(x, np.sin(x), color='r')
plt.plot(x, x_grad.numpy(), color='g')

It works on my local machine.


How do we calculate that gradient of y with respect to x is 4x?

x.t*x is equivalent to x^2 in vector notation. If you take a vector x, then x.T (the transpose of x), and multiply them, you end up with a scalar equivalent to x^2. So really this is taking the derivative of 2x^2, which is 4x.

Heres a longer-winded answer on this as well -
They only discuss the x.T*x case, but multiplying by a scalar 2 just ends up again working out to 4x instead of the 2x they show there.

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In Linear Algebra section, there is a formula: (X.T*X)' = 2X, then multiply both side by 2.

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About Question3: f(x) is a scalar function, not a vector function. You can get the detail in the link:

In question 3 like someone asked previously, the derivative of a vector over a vector is a matrix,
but after calling the gradient method we have a vector,

Is the reason why related to the Backward for Non-Scalar Variables paragraph.
They say:
" Here, (our intent is ) not to calculate the differentiation matrix but rather (the sum of the partial derivatives computed individually for each example ) in the batch."

I think the vector returned by the gradient method is the sum of each column of the matrix generated by the derivative, but I am not totally sure about it…

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typo: “an dot product of x and x is performed”? Should be “a dot product” ?

I believe you are mainly right. The way I understand it is that we have a vector y = [y1,y2, … , yn] and a vector x. If we calculate the derivative of y with respect to x, we will end up with a matrix called the Jacobian. However, since we want the result to be a vector not a matrix, we sum all the elements in vector y, then we calculate the derivative of y with respect to x. Something like this:
y = [y1,y2, … , yn]
y = y1 + y2 + … + yn # Same as y = tf.reduce_sum(y)
dy / dx

There is typo in the doc which causes a lot of confusion. The last statement in 2.5.4, d_grad == d / a should be evaluated to True.