为什么 Pytorch autograd 需要标量？

Question

I am working through " Deep Learning for Coders with fastai & Pytorch ". Chapter 4 introduces the autograd function from the PyTorch library on a trivial example.

x = tensor([3.,4.,10.]).requires_grad_()
def f(q): return sum(q**2)
y = f(x)
y.backward()

My question boils down to this: the result of y = f(x) is tensor(125., grad_fn=AddBackward0) , but what does that even mean? Why would I sum the values of three completely different inputs?

I get that using .backward() in this case is shorthand for .backward(tensor[1.,1.,1.]) in this scenario, but I don't see how summing 3 unrelated numbers in a list helps get the gradient for anything. What am I not understanding?

_{I'm not looking for a grad-level explanation here. The subtitle for the book I'm using is AI Applications Without a Ph.D. My experience with gradients is from school is that I should be getting a FUNCTION back, but I understand that isn't the case with Autograd. A graph of this short example would be helpful, but the ones I see online usually include too many parameters or weights and biases to be useful, my mind gets lost in the paths.}

Answer 1

TLDR; the derivative of a sum of functions is the sum of their derivatives

Let x be your input vector made of x_i (where i in [0,n] ), y = x**2 and L = sum(y_i) . You are looking to compute dL/dx , a vector of the same size as x whose components are the dL/dx_j (where j in [0,n] ).

For j in [0,n] , dL/dx_j is simply dy_j/dx_j (derivative of the sum is the sum of derivates and only one of them is different to zero), which is d(x_j**2)/dx_j , ie 2*x_j . Therefore, dL/dx = [2*x_j where j in [0,n]] .

This is the result you get in x.grad when either computing the gradient of x as:

y = f(x)
y.backward()

or the gradient of each components of x separately :

y = x**2
y.backward(torch.ones_like(x))