🐛 FIX: changes in source file for pdf build (#136)

* changes for pdf build

* requirements file update

* Update jupyter-book to 0.13.1

* harmonise with lecture-python.myst

* restrict jupyter-server

* FIX: align environment for pdf

* FIX: add latexify to adjust output to png

Co-authored-by: mmcky <[email protected]>

Author: AakashGfude

.github/workflows/ci.yml

@@ -42,12 +42,12 @@ jobs:
           name: build-cache
           path: _build
       # Build Assets (Download Notebooks and PDF via LaTeX)
-      # - name: Build PDF from LaTeX
-      #   shell: bash -l {0}
-      #   run: |
-      #     jb build lectures --builder pdflatex --path-output ./ -n --keep-going
-      #     mkdir -p _build/html/_pdf
-      #     cp -u _build/latex/*.pdf _build/html/_pdf
+      - name: Build PDF from LaTeX
+        shell: bash -l {0}
+        run: |
+          jb build lectures --builder pdflatex --path-output ./ -n --keep-going
+          mkdir -p _build/html/_pdf
+          cp -u _build/latex/*.pdf _build/html/_pdf
       - name: Build Download Notebooks (sphinx-tojupyter)
         shell: bash -l {0}
         run: |

lectures/Manifest.toml

@@ -1,8 +1,8 @@
 # This file is machine-generated - editing it directly is not advised
 
-julia_version = "1.8.3"
+julia_version = "1.8.5"
 manifest_format = "2.0"
-project_hash = "f509bdfe41f89452dd3da3343ba86802b3f9cdba"
+project_hash = "426ee44181a1827c854acad97427614830a3ef74"
 
 [[deps.AMD]]
 deps = ["Libdl", "LinearAlgebra", "SparseArrays", "Test"]
@@ -330,7 +330,7 @@ version = "3.46.0"
 [[deps.CompilerSupportLibraries_jll]]
 deps = ["Artifacts", "Libdl"]
 uuid = "e66e0078-7015-5450-92f7-15fbd957f2ae"
-version = "0.5.2+0"
+version = "1.0.1+0"
 
 [[deps.CompositeTypes]]
 git-tree-sha1 = "02d2316b7ffceff992f3096ae48c7829a8aa0638"
@@ -1005,9 +1005,9 @@ version = "0.4.5"
 
 [[deps.Latexify]]
 deps = ["Formatting", "InteractiveUtils", "LaTeXStrings", "MacroTools", "Markdown", "OrderedCollections", "Printf", "Requires"]
-git-tree-sha1 = "ab9aa169d2160129beb241cb2750ca499b4e90e9"
+git-tree-sha1 = "2422f47b34d4b127720a18f86fa7b1aa2e141f29"
 uuid = "23fbe1c1-3f47-55db-b15f-69d7ec21a316"
-version = "0.15.17"
+version = "0.15.18"
 
 [[deps.LayoutPointers]]
 deps = ["ArrayInterface", "ArrayInterfaceOffsetArrays", "ArrayInterfaceStaticArrays", "LinearAlgebra", "ManualMemory", "SIMDTypes", "Static"]

lectures/Project.toml

@@ -26,6 +26,7 @@ IterativeSolvers = "42fd0dbc-a981-5370-80f2-aaf504508153"
 JuMP = "4076af6c-e467-56ae-b986-b466b2749572"
 KernelDensity = "5ab0869b-81aa-558d-bb23-cbf5423bbe9b"
 LaTeXStrings = "b964fa9f-0449-5b57-a5c2-d3ea65f4040f"
+Latexify = "23fbe1c1-3f47-55db-b15f-69d7ec21a316"
 LeastSquaresOptim = "0fc2ff8b-aaa3-5acd-a817-1944a5e08891"
 LinearAlgebra = "37e2e46d-f89d-539d-b4ee-838fcccc9c8e"
 LinearMaps = "7a12625a-238d-50fd-b39a-03d52299707e"

lectures/dynamic_programming_squared/opt_tax_recur.md

@@ -697,7 +697,6 @@ and $\tau_0$ is the time $t=0$ tax rate.
 In equation {eq}`opt_tax_eqn_10`, it is understood that
 
 ```{math}
-:nowrap:
 
 \begin{aligned}
 \tau_0 = 1 - \frac{u_{l,0}}{u_{c,0}} \\

lectures/tools_and_techniques/geom_series.md

@@ -47,7 +47,7 @@ Below we'll use the following packages:
 
 ```{code-cell} julia
 using LinearAlgebra, Statistics
-using Distributions, LaTeXStrings, Plots, Random, Symbolics
+using Distributions, LaTeXStrings, Plots, Random, Symbolics, Latexify
 ```
 
 ## Key Formulas
@@ -550,9 +550,11 @@ For a reason soon to be revealed, we assume that $G < R$.
 The **present value** of the lease is
 
 $$
-\begin{aligned} p_0  & = x_0 + x_1/R + x_2/(R^2) + \ddots \\
-                 & = x_0 (1 + G R^{-1} + G^2 R^{-2} + \cdots ) \\
-                 & = x_0 \frac{1}{1 - G R^{-1}} \end{aligned}
+\begin{aligned} 
+p_0  & = x_0 + x_1/R + x_2/(R^2) + \ddots \\
+& = x_0 (1 + G R^{-1} + G^2 R^{-2} + \cdots ) \\
+& = x_0 \frac{1}{1 - G R^{-1}} 
+\end{aligned}
 $$
 
 where the last line uses the formula for an infinite geometric series.
@@ -605,7 +607,11 @@ $$
 The present value of this lease is:
 
 $$
-\begin{aligned} \begin{split}p_0&=x_0 + x_1/R  + \dots +x_T/R^T \\ &= x_0(1+GR^{-1}+\dots +G^{T}R^{-T}) \\ &= \frac{x_0(1-G^{T+1}R^{-(T+1)})}{1-GR^{-1}}  \end{split}\end{aligned}
+\begin{aligned}
+\begin{split}
+p_0&=x_0 + x_1/R  + \dots +x_T/R^T \\ &= x_0(1+GR^{-1}+\dots +G^{T}R^{-T}) \\ &= \frac{x_0(1-G^{T+1}R^{-(T+1)})}{1-GR^{-1}}
+\end{split}
+\end{aligned}
 $$
 
 Applying the Taylor series to $R^{-(T+1)}$ about $r=0$ we get:
@@ -629,7 +635,9 @@ $$
 Expanding:
 
 $$
-\begin{aligned} p_0 &=\frac{x_0(1-1+(T+1)^2 rg -r(T+1)+g(T+1))}{1-1+r-g+rg}  \\&=\frac{x_0(T+1)((T+1)rg+r-g)}{r-g+rg} \\ &\approx \frac{x_0(T+1)(r-g)}{r-g}+\frac{x_0rg(T+1)}{r-g}\\ &= x_0(T+1) + \frac{x_0rg(T+1)}{r-g}  \end{aligned}
+\begin{aligned} 
+p_0 &=\frac{x_0(1-1+(T+1)^2 rg -r(T+1)+g(T+1))}{1-1+r-g+rg}  \\&=\frac{x_0(T+1)((T+1)rg+r-g)}{r-g+rg} \\ &\approx \frac{x_0(T+1)(r-g)}{r-g}+\frac{x_0rg(T+1)}{r-g}\\ &= x_0(T+1) + \frac{x_0rg(T+1)}{r-g}  
+\end{aligned}
 $$
 
 We could have also approximated by removing the second term
@@ -756,23 +764,23 @@ G = (1 + g)
 R = (1 + r)
 p0 = x0 / (1 - G * R ^ (-1))
 print("Our formula is")
-p0
+latexify(p0) |> s -> render(s, MIME("image/png"))
 ```
 
 ```{code-cell} julia
 # Partial derivative with respect to g
 dg = Differential(g)
 dp_dg = expand_derivatives(dg(p0))
 print("dp0 / dg is")
-dp_dg
+latexify(dp_dg) |> s -> render(s, MIME("image/png"))
 ```
 
 ```{code-cell} julia
 # Partial derivative with respect to r
 dr = Differential(r)
 dp_dr = expand_derivatives(dr(p0))
 print("dp0 / dr is")
-dp_dr
+latexify(dp_dr) |> s -> render(s, MIME("image/png"))
 ```
 
 We can see that for $\frac{\partial p_0}{\partial r}<0$ as long as

lectures/tools_and_techniques/iterative_methods_sparsity.md

@@ -218,7 +218,7 @@ To solve for the coefficients, we notice that this is a simple system of equatio
 
 $$
 \begin{array}
-    \,y_0 = c_0 + c_1 x_0 + \ldots c_N x_0^N\\
+    cy_0 = c_0 + c_1 x_0 + \ldots c_N x_0^N\\
     \,\ldots\\
     \,y_N = c_0 + c_1 x_N + \ldots c_N x_N^N
 \end{array}
@@ -502,10 +502,10 @@ $$
 Rearrange the $(D + R)x = b$ as
 
 $$
-\begin{align}
+\begin{aligned}
 D x &= b - R x\\
 x &= D^{-1} (b - R x)
-\end{align}
+\end{aligned}
 $$
 
 where, since $D$ is diagonal, its inverse is trivial to calculate with $O(N)$ complexity.
@@ -618,12 +618,12 @@ operations.
 To see an example of a right-preconditioner, consider a matrix $P$ which has a convenient and numerically stable inverse.  Then
 
 $$
-\begin{align}
+\begin{aligned}
 A x &= b\\
 A P^{-1} P x &= b\\
 A P^{-1} y &= b\\
 P x &= y
-\end{align}
+\end{aligned}
 $$
 
 That is, solve $(A P^{-1})y = b$ for $y$, and then solve $P x = y$ for $x$.
@@ -1000,12 +1000,12 @@ before we enumerate them linearly, take a $v\in R^{\mathbf{N}}$ interpreted as a
 For example, if we were implementing the product at the row of $Q$ corresponding to the $(n_1, \ldots, n_M)$ state, then
 
 $$
-\begin{align}
+\begin{aligned}
     Q_{(n_1, \ldots n_M)} \cdot v &=
 \theta \sum_{m=1}^M (n_m < N)  v(n_1, \ldots, n_m + 1, \ldots, n_M)\\
                                         &+ \zeta \sum_{m=1}^M (1 < n_m)  v(n_1, \ldots, n_m - 1, \ldots, n_M)\\
                                         &-\left(\theta\, \text{Count}(n_m < N) + \zeta\, \text{Count}( n_m > 1)\right)v(n_1, \ldots, n_M)
-\end{align}
+\end{aligned}
 $$
 
 Here:
@@ -1182,12 +1182,12 @@ The logic for the adjoint is that for a given $n = (n_1,\ldots, n_m, \ldots n_M)
 Implementing this logic, first in math and then in code,
 
 $$
-\begin{align}
+\begin{aligned}
     Q^T_{(n_1, \ldots, n_M)} \cdot \psi &=
 \theta \sum_{m=1}^M (n_m > 1)  \psi(n_1, \ldots, n_m - 1, \ldots, n_M)\\
                                         &+ \zeta \sum_{m=1}^M (n_m < N)  \psi(n_1, \ldots, n_m + 1, \ldots, n_M)\\
                                         &-\left(\theta\, \text{Count}(n_m < N) + \zeta\, \text{Count}( n_m > 1)\right)\psi(n_1, \ldots, n_M)
-\end{align}
+\end{aligned}
 $$
 
 ```{code-cell} julia

lectures/tools_and_techniques/linear_algebra.md

@@ -1286,7 +1286,6 @@ $$
 To evaluate the function
 
 ```{math}
-:nowrap:
 
 \begin{aligned}
 v(x) &=  -(Ax+ B u)'P(Ax+Bu) - u'Q u \\
@@ -1301,7 +1300,6 @@ For simplicity, denote by $S := (Q + B'PB)^{-1} B'PA$, then $u = -Sx$.
 Regarding the second term $- 2u'B'PAx$,
 
 ```{math}
-:nowrap:
 
 \begin{aligned}
 - 2u'B'PAx &= -2 x'S'B'PAx  \\
@@ -1315,7 +1313,6 @@ are symmetric.
 Regarding the third term $- u'(Q + B'PB) u$,
 
 ```{math}
-:nowrap:
 
 \begin{aligned}
 - u'(Q + B'PB) u &= - x'S' (Q + B'PB)Sx \\
@@ -1329,7 +1326,6 @@ $x'A'PB(Q + B'PB)^{-1}B'PAx$.
 This implies that
 
 ```{math}
-:nowrap:
 
 \begin{aligned}
  v(x) &= - x'A'PAx - 2u'B'PAx - u'(Q + B'PB) u\\

requirements.txt

@@ -1,5 +1,5 @@
 jupyter-book==0.12.3
-sphinxext.rediraffe==0.2.7
-sphinx-tojupyter==0.2.1
 quantecon-book-theme==0.3.1
-markupsafe==2.1.1
+sphinx-tojupyter==0.2.1
+sphinxext.rediraffe==0.2.7
+jupyter-server<2