Skip to content
This repository has been archived by the owner on Nov 17, 2023. It is now read-only.

Fixing autograd images #17243

Merged
merged 1 commit into from
Jan 14, 2020
Merged

Fixing autograd images #17243

Changes from all commits
Commits
Show all changes
1 commit
Select commit
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
8 changes: 4 additions & 4 deletions docs/python_docs/python/tutorials/packages/autograd/index.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -23,7 +23,7 @@

Gradients are fundamental to the process of training neural networks, and tell us how to change the parameters of the network to improve its performance.

![auto-gradient](/api/python/docs/_static/autograd_images/autograd_gradient.png)
![auto-gradient](http://mxnet.incubator.apache.org/api/python/docs/_static/autograd_images/autograd_gradient.png)

### Long Answer:

Expand All @@ -39,7 +39,7 @@ Assuming we've calculated the gradient of each parameter with respect to the los

We differentiate. [MXNet Gluon](/api/python/docs/tutorials/packages/gluon/index.html) uses Reverse Mode Automatic Differentiation (`autograd`) to backprogate gradients from the loss metric to the network parameters.

![forward-backward](/api/python/docs/_static/autograd_images/autograd_forward_backward.png)
![forward-backward](http://mxnet.incubator.apache.org/api/python/docs/_static/autograd_images/autograd_forward_backward.png)

### Long Answer:

Expand Down Expand Up @@ -217,7 +217,7 @@ def f(x):

We can plot the resultant function for $x$ between 0 and 1, and we should recognise certain functions in segments of $x$. Starting with a quadratic curve from 0 to 1/2, we have a cubic curve from 1/2 to 2/3, a quartic from 2/3 to 3/4 and finally a flatline.

![control-flow](/api/python/docs/_static/autograd_images/autograd_control_flow.png)
![control-flow](https://mxnet.incubator.apache.org/api/python/docs/_static/autograd_images/autograd_control_flow.png)

Using `autograd`, let's now find the gradient of this arbritrary function. We don't have a vectorized function in this case, because of the control flow, so let's also create a function to calculate the gradient using `autograd`.

Expand All @@ -234,7 +234,7 @@ grads = [get_grad(f, x).asscalar() for x in xs]
print(grads)
```

![flow-grad](/api/python/docs/_static/autograd_images/autograd_control_flow_grad.png)
![flow-grad](https://mxnet.incubator.apache.org/api/python/docs/_static/autograd_images/autograd_control_flow_grad.png)

We can calculate the gradients by hand in this situation (since it's a toy example), and for the four segments discussed before we'd expect $2x$, $3x^2$, $4x^3$ and 0. As a spot check, for $x=0.6$ the hand calculated gradient would be $3x^2=1.08$, which equals `1.08` as computed by `autograd`.

Expand Down