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autograd video and image link fixes and removing symbol tutorials #17227

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Jan 7, 2020
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autograd video and image link fixes and removing symbol tutorials #17227

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10 changes: 0 additions & 10 deletions docs/python_docs/python/api/symbol/index.rst
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Expand Up @@ -37,16 +37,6 @@ The following example shows how you might build a simple expression with the Sym
(a, b, c)


Tutorials
---------
.. container:: cards

.. card::
:title: Symbol Guide
:link: ../../tutorials/packages/symbol/

The Symbol guide. Start here!

Symbol Package
--------------

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14 changes: 7 additions & 7 deletions docs/python_docs/python/tutorials/packages/autograd/index.md
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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](//_static/autograd_images/autograd_gradient.png)
![auto-gradient](/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](//_static/autograd_images/autograd_forward_backward.png)
![forward-backward](/api/python/docs/_static/autograd_images/autograd_forward_backward.png)

### Long Answer:

Expand All @@ -56,7 +56,7 @@ Stage 2. Work backwards through this record and evaluate the partial derivatives

<p style="text-align:center">
<video width="600" controls playsinline autoplay muted loop>
<source src="/_static/autograd_images/autograd_graph.mp4" type="video/mp4">
<source src="/api/python/docs/_static/autograd_images/autograd_graph.mp4" type="video/mp4">
</video>
</p>

Expand Down Expand Up @@ -173,7 +173,7 @@ net[0].weight.grad_req = 'null'

<p style="text-align:center">
<video width="600" controls playsinline autoplay muted loop>
<source src="../../../../_static/autograd_images/autograd_grad_req.mp4" type="video/mp4">
<source src="/api/python/docs/_static/autograd_images/autograd_grad_req.mp4" type="video/mp4">
</video>
</p>

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](//_static/autograd_images/autograd_control_flow.png)
![control-flow](/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](//_static/autograd_images/autograd_control_flow_grad.png)
![flow-grad](/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 All @@ -247,7 +247,7 @@ Most of the time `autograd` will be aware of the complete computational graph, a

<p style="text-align:center">
<video width="600" controls playsinline autoplay muted loop>
<source src="../../../../_static/autograd_images/autograd_head_grad.mp4" type="video/mp4">
<source src="/api/python/docs/_static/autograd_images/autograd_head_grad.mp4" type="video/mp4">
</video>
</p>

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