Merge pull request #1493 from d2l-ai/master

Release v0.15.0

Author: astonzhang

README.md

@@ -63,4 +63,4 @@ This open source book is made available under the Creative Commons Attribution-S
 
 The sample and reference code within this open source book is made available under a modified MIT license. See the [LICENSE-SAMPLECODE](LICENSE-SAMPLECODE) file.
 
-[Chinese version](https://github.com/d2l-ai/d2l-zh) | [Discuss and report issues](https://discuss.d2l.ai/) | [Other Information](INFO.md)
+[Chinese version](https://github.com/d2l-ai/d2l-zh) | [Discuss and report issues](https://discuss.d2l.ai/) | [Code of conduct](CODE_OF_CONDUCT.md) | [Other Information](INFO.md)

chapter_appendix-mathematics-for-deep-learning/distributions.md

@@ -334,7 +334,7 @@ def binom(n, k):
         comb = comb * (n - i) // (i + 1)
     return comb
 
-pmf = torch.tensor([p**i * (1-p)**(n - i) * binom(n, i) for i in range(n + 1)])
+pmf = d2l.tensor([p**i * (1-p)**(n - i) * binom(n, i) for i in range(n + 1)])
 
 d2l.plt.stem([i for i in range(n + 1)], pmf, use_line_collection=True)
 d2l.plt.xlabel('x')
@@ -672,7 +672,7 @@ d2l.plot(x, np.array([phi(y) for y in x.tolist()]), 'x', 'c.d.f.')
 ```{.python .input}
 #@tab pytorch
 def phi(x):
-    return (1.0 + erf((x - mu) / (sigma * torch.sqrt(torch.tensor(2.))))) / 2.0
+    return (1.0 + erf((x - mu) / (sigma * torch.sqrt(d2l.tensor(2.))))) / 2.0
 
 d2l.plot(x, torch.tensor([phi(y) for y in x.tolist()]), 'x', 'c.d.f.')
 ```

chapter_appendix-mathematics-for-deep-learning/eigendecomposition.md

@@ -538,7 +538,7 @@ for all practical purposes, our random vector has been transformed
 into the principle eigenvector!
 Indeed this algorithm is the basis 
 for what is known as the *power iteration*
-for finding the largest eigenvalue and eigenvector of a matrix.  For details see, for example, :cite:`Van-Loan.Golub.1983`.
+for finding the largest eigenvalue and eigenvector of a matrix. For details see, for example, :cite:`Van-Loan.Golub.1983`.
 
 ### Fixing the Normalization
 

chapter_appendix-mathematics-for-deep-learning/geometry-linear-algebraic-ops.md

@@ -1,7 +1,7 @@
 # Integral Calculus
 :label:`sec_integral_calculus`
 
-Differentiation only makes up half of the content of a traditional calculus education.  The other pillar, integration, starts out seeming a rather disjoint question, "What is the area underneath this curve?"  While seemingly unrelated, integration is tightly intertwined with the differentiation via what is known as the *fundamental theorem of calculus*.  
+Differentiation only makes up half of the content of a traditional calculus education.  The other pillar, integration, starts out seeming a rather disjoint question, "What is the area underneath this curve?"  While seemingly unrelated, integration is tightly intertwined with the differentiation via what is known as the *fundamental theorem of calculus*.
 
 At the level of machine learning we discuss in this book, we will not need a deep understanding of integration. However, we will provide a brief introduction to lay the groundwork for any further applications we will encounter later on.
 
@@ -187,7 +187,7 @@ We will instead take a different approach.  We will work intuitively with the no
 
 ## The Fundamental Theorem of Calculus
 
-To dive deeper into the theory of integration, let us introduce a function 
+To dive deeper into the theory of integration, let us introduce a function
 
 $$
 F(x) = \int_0^x f(y) dy.
@@ -201,10 +201,10 @@ $$
 
 This is a mathematical encoding of the fact that we can measure the area out to the far end-point and then subtract off the area to the near end point as indicated in :numref:`fig_area-subtract`.
 
-![Visualizing why we may reduce the problem of computing the area under a curve between two points to computing the area to the left of a point.](../img/SubArea.svg)
+![Visualizing why we may reduce the problem of computing the area under a curve between two points to computing the area to the left of a point.](../img/sub-area.svg)
 :label:`fig_area-subtract`
 
-Thus, we can figure out what the integral over any interval is by figuring out what $F(x)$ is.  
+Thus, we can figure out what the integral over any interval is by figuring out what $F(x)$ is.
 
 To do so, let us consider an experiment.  As we often do in calculus, let us imagine what happens when we shift the value by a tiny bit.  From the comment above, we know that
 
@@ -259,7 +259,7 @@ First, suppose that we have a function which is itself an integral:
 
 $$
 F(x) = \int_0^x f(y) \; dy.
-$$ 
+$$
 
 Let us suppose that we want to know how this function looks when we compose it with another to obtain $F(u(x))$.  By the chain rule, we know
 
@@ -286,16 +286,16 @@ $$\int_{u(0)}^{u(x)} f(y) \; dy = \int_0^x f(u(y))\cdot \frac{du}{dy} \;dy.$$
 
 This is the *change of variables* formula.
 
-For a more intuitive derivation, consider what happens when we take an integral of $f(u(x))$ between $x$ and $x+\epsilon$. For a small $\epsilon$, this integral is approximately $\epsilon f(u(x))$, the area of the associated rectangle.  Now, let us compare this with the integral of $f(y)$ from $u(x)$ to $u(x+\epsilon)$.  We know that $u(x+\epsilon) \approx u(x) + \epsilon \frac{du}{dx}(x)$, so the area of this rectangle is approximately $\epsilon \frac{du}{dx}(x)f(u(x))$.  Thus, to make the area of these two rectangles to agree, we need to multiply the first one by $\frac{du}{dx}(x)$ as is illustrated in :numref:`fig_rect-transform`.  
+For a more intuitive derivation, consider what happens when we take an integral of $f(u(x))$ between $x$ and $x+\epsilon$. For a small $\epsilon$, this integral is approximately $\epsilon f(u(x))$, the area of the associated rectangle.  Now, let us compare this with the integral of $f(y)$ from $u(x)$ to $u(x+\epsilon)$.  We know that $u(x+\epsilon) \approx u(x) + \epsilon \frac{du}{dx}(x)$, so the area of this rectangle is approximately $\epsilon \frac{du}{dx}(x)f(u(x))$.  Thus, to make the area of these two rectangles to agree, we need to multiply the first one by $\frac{du}{dx}(x)$ as is illustrated in :numref:`fig_rect-transform`.
 
-![Visualizing the transformation of a single thin rectangle under the change of variables.](../img/RectTrans.svg)
+![Visualizing the transformation of a single thin rectangle under the change of variables.](../img/rect-trans.svg)
 :label:`fig_rect-transform`
 
 This tells us that
 
 $$
 \int_x^{x+\epsilon} f(u(y))\frac{du}{dy}(y)\;dy = \int_{u(x)}^{u(x+\epsilon)} f(y) \; dy.
-$$  
+$$
 
 This is the change of variables formula expressed for a single small rectangle.
 
@@ -404,7 +404,7 @@ ax.set_zlim(0, 1)
 ax.dist = 12
 ```
 
-We write this as 
+We write this as
 
 $$
 \int_{[a, b]\times[c, d]} f(x, y)\;dx\;dy.
@@ -416,7 +416,7 @@ $$
 \int_{[a, b]\times[c, d]} f(x, y)\;dx\;dy = \int_c^{d} \left(\int_a^{b} f(x, y) \;dx\right) \; dy.
 $$
 
-Let us see why this is.  
+Let us see why this is.
 
 Consider the figure above where we have split the function into $\epsilon \times \epsilon$ squares which we will index with integer coordinates $i, j$.  In this case, our integral is approximately
 
@@ -430,16 +430,16 @@ $$
  \sum _ {j} \epsilon \left(\sum_{i} \epsilon f(\epsilon i, \epsilon j)\right).
 $$
 
-![Illustrating how to decompose a sum over many squares as a sum over first the columns (1), then adding the column sums together (2).](../img/SumOrder.svg)
+![Illustrating how to decompose a sum over many squares as a sum over first the columns (1), then adding the column sums together (2).](../img/sum-order.svg)
 :label:`fig_sum-order`
 
-The sum on the inside is precisely the discretization of the integral 
+The sum on the inside is precisely the discretization of the integral
 
 $$
 G(\epsilon j) = \int _a^{b} f(x, \epsilon j) \; dx.
 $$
 
-Finally, notice that if we combine these two expressions we get 
+Finally, notice that if we combine these two expressions we get
 
 $$
 \sum _ {j} \epsilon G(\epsilon j) \approx \int _ {c}^{d} G(y) \; dy = \int _ {[a, b]\times[c, d]} f(x, y)\;dx\;dy.
@@ -466,9 +466,9 @@ $$
 $$
 
 ## Change of Variables in Multiple Integrals
-As with single variables in :eqref:`eq_change_var`, the ability to change variables inside a higher dimensional integral is a key tool.  Let us summarize the result without derivation.  
+As with single variables in :eqref:`eq_change_var`, the ability to change variables inside a higher dimensional integral is a key tool.  Let us summarize the result without derivation.
 
-We need a function that reparameterizes our domain of integration.  We can take this to be $\phi : \mathbb{R}^n \rightarrow \mathbb{R}^n$, that is any function which takes in $n$ real variables and returns another $n$.  To keep the expressions clean, we will assume that $\phi$ is *injective* which is to say it never folds over itself ($\phi(\mathbf{x}) = \phi(\mathbf{y}) \implies \mathbf{x} = \mathbf{y}$).  
+We need a function that reparameterizes our domain of integration.  We can take this to be $\phi : \mathbb{R}^n \rightarrow \mathbb{R}^n$, that is any function which takes in $n$ real variables and returns another $n$.  To keep the expressions clean, we will assume that $\phi$ is *injective* which is to say it never folds over itself ($\phi(\mathbf{x}) = \phi(\mathbf{y}) \implies \mathbf{x} = \mathbf{y}$).
 
 In this case, we can say that
 
@@ -486,7 +486,7 @@ D\boldsymbol{\phi} = \begin{bmatrix}
 \end{bmatrix}.
 $$
 
-Looking closely, we see that this is similar to the single variable chain rule :eqref:`eq_change_var`, except we have replaced the term $\frac{du}{dx}(x)$ with $\left|\det(D\phi(\mathbf{x}))\right|$.  Let us see how we can to interpret this term.  Recall that the $\frac{du}{dx}(x)$ term existed to say how much we stretched our $x$-axis by applying $u$.  The same process in higher dimensions is to determine how much we stretch the area (or volume, or hyper-volume) of a little square (or little *hyper-cube*) by applying $\boldsymbol{\phi}$.  If $\boldsymbol{\phi}$ was the multiplication by a matrix, then we know how the determinant already gives the answer.  
+Looking closely, we see that this is similar to the single variable chain rule :eqref:`eq_change_var`, except we have replaced the term $\frac{du}{dx}(x)$ with $\left|\det(D\phi(\mathbf{x}))\right|$.  Let us see how we can to interpret this term.  Recall that the $\frac{du}{dx}(x)$ term existed to say how much we stretched our $x$-axis by applying $u$.  The same process in higher dimensions is to determine how much we stretch the area (or volume, or hyper-volume) of a little square (or little *hyper-cube*) by applying $\boldsymbol{\phi}$.  If $\boldsymbol{\phi}$ was the multiplication by a matrix, then we know how the determinant already gives the answer.
 
 With some work, one can show that the *Jacobian* provides the best approximation to a multivariable function $\boldsymbol{\phi}$ at a point by a matrix in the same way we could approximate by lines or planes with derivatives and gradients. Thus the determinant of the Jacobian exactly mirrors the scaling factor we identified in one dimension.
 
@@ -502,7 +502,7 @@ $$
 \int _ 0^\infty \int_0 ^ {2\pi} e^{-r^{2}} \left|\det(D\mathbf{\phi}(\mathbf{x}))\right|\;d\theta\;dr,
 $$
 
-where 
+where
 
 $$
 \left|\det(D\mathbf{\phi}(\mathbf{x}))\right| = \left|\det\begin{bmatrix}
@@ -517,7 +517,7 @@ $$
 \int _ 0^\infty \int _ 0 ^ {2\pi} re^{-r^{2}} \;d\theta\;dr = 2\pi\int _ 0^\infty re^{-r^{2}} \;dr = \pi,
 $$
 
-where the final equality follows by the same computation that we used in section :numref:`integral_example`.  
+where the final equality follows by the same computation that we used in section :numref:`integral_example`.
 
 We will meet this integral again when we study continuous random variables in :numref:`sec_random_variables`.