obsidian-publish-mkdocs
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jobindjohn
math blocks in standard obsidian format doesn't render
For some reason my maths blocks doesn't render, however it does in obsidian. The standard way to write math blocks in obsidian is something like
paragraph
$$
maths
$$
paragraph
However I see that with this format the math blocks doesn't render in github either, so I change to the format
paragraph
$$maths$$
paragraphs
Now it renders correctly. However all my notes are written in the first style, there is a way to make that your template supports the use of the first format for math blocks instead of the second?
I not sure why it doesn't work for you. It has been working for me. See the example here: https://jobindjohn.github.io/obsidian-publish-mkdocs/Features/LaTeX%20Math%20Support/
Can you verify if it's not rendering for you after it has been built
I not sure why it doesn't work for you. It has been working for me. See the example here: https://jobindjohn.github.io/obsidian-publish-mkdocs/Features/LaTeX%20Math%20Support/
Can you verify if it's not rendering for you after it has been built
Im not sure what you mean by "after build". The building is automatic in github after I push the files, I could see the files but they doesn't render block maths, just inline ones. I can give you a piece of text so you can debug by yourself what is going on (if you want/have time, of course):
## 2B.9
Because $\lambda ([0,1])=1$ then there is some non-measurable $A\subset [0,1]$. Then define $f:=\chi _A-\chi _{[0,1]\setminus A}$, then $f^{-1}(1)=A$, so $f$ is not measurable, however $|f|=\chi _{[0,1]}$.
---
## 2B.10
Let $A_{n,k}$ the set of numbers in $(0,1)$ with exactly $n$ fives in it firsts $k$ digits in it decimal expansion. It is easy to check that each $A_{n,k}$ is Borel and $A_n:=\bigcap_{k\geqslant 0}A_{n,k}$ is the set of numbers in $(0,1)$ with exactly $n$ fives, so is also Borel.
By last $A:=\bigcup_{n\geqslant 0}A_n$ is the set of numbers in $(0,1)$ with finitely many fives in it decimal expansion, and so it is also Borel, and clearly $A^\complement$ is also Borel.∎
---
## 2B.18
If $f$ is differentiable then it is continuous, so it is Borel measurable. Now note that
$$
f'=\lim_{n \to \infty }n(f(\cdot +1/n)-f)
$$
and every $n(f(\cdot +1/n)-f)$ is also Borel measurable.
---
## 2B.20
Note that $f^g=\exp\left(g\cdot \ln f\right)$, and so everything reduces to see that $\ln f$ and $e^h$ are measurable when $f$ and $h$ are.
---
## 2C.1
Because $\mu (X)\in[0,1)$, but this would imply that there are uncountable measurable subsets $E_t \subsetneq X$ such that $\mu (X)<\mu (E_t)$, what cannot be possible.
---
## 2C.3
Let $\mu (\{k\}):=2^{-k-1}$, now note that $\sum_{k\geqslant 1}2^{-k}=1$ and that each number $E\subset \Bbb N_{\ge 0}$ defines the binary expansion of a number in $(0,1]$. By last set $\mu (\emptyset ):=0$ to complete the measure.
---
## 2C.8
Let $X:=(0,1)$ and $\mathcal{A}:=\{(0,a/2),(a/4,a)\}$ for some $a\in(0,1)$. Then note that
$$
\sigma (\mathcal{A})=\mathcal{A}\cup \{(0,a/4),(a/4,a/2),(a/2,a),(a,1),\ldots ,(0,1),\emptyset \}
$$
Then fixing the values of $\mu$ and $\nu$ in $\mathcal{A}$ and $X$ we have a degree of freedom to choose the values of $(a/4,a/2)$ and $(a,1)$. ∎
---
## 2C.12
Because every singleton is measurable then there is some $f:X\to [0,\infty ]$ such that $\mu (\{x\})=f(x)$, and so it is easy to check that each measure in $(X,\mathcal{S})$ have the form
$$
\mu (E)=\alpha I(E)+\sup\left\{\sum_{x\in D}f(x):D\subset E\,\land\, \#D\leqslant \aleph _0\right\}
$$
for arbitrary $\alpha \in \Bbb R$, and where
$$
I(E):=\begin{cases}
0, & \#E\leqslant \aleph _0\\
1, & \text{ otherwise }
\end{cases}
$$
for each $E\in \mathcal{S}$.
---
## 2D.1
In the next $a_k$ count the number of lists of length $k$ of decimal digits that doesn’t contain a sub-list of consecutive 100 fours:
$$
\begin{align*}
&a_{k+1}=9a_k-8a_{k-100}[k\geqslant 100]-[k=99],\quad a_0=1,\quad A(x):=\sum_{k\geqslant 0}a_kx^k\\
&\therefore\, \sum_{k\geqslant 0}a_{k+1}x^k=9A(x)-8\sum_{k\geqslant 100}a_{k-100}x^k-x^{99}\\
&\iff A(x)-1=9xA(x)-8x^{101}A(x)-x^{100}\\
&\iff A(x)=\frac{1-x^{100}}{1-9x+8x^{101}}
\end{align*}
$$
Now let $B_k$ for the set of numbers in $(0,1)$ such that the first sub-list of 100 consecutive 4s start at the $k+1$ digit of it decimal expansion, and so the decimal expansion of a number in $B_k$ have the form
$$
\underbrace{0,d_1d_2\ldots d_{k-1}}_{\text{ no sublists of 100 consecutive 4s }}\overbrace{d_k}^{\text{ any digit but 4 }}
\underbrace{d_{k+1}\ldots d_{k+100}}_{\text{ 4s }}\overbrace{d_{k+101}\ldots }^{\text{ anything }}
$$
Thus it is easy to check that
$$
\lambda (B_k)=\frac{8a_{k-1}[k\geqslant 1]+[k=0]}{10^{k+100}}
$$
(a) The set that we are searching for is $\bigcup_{k\geqslant 0}B_k$, so it is clearly Borel.
(b) Because $B_k\cap B_j=\emptyset$ whenever $k \neq j$ we find that
$$
\begin{align*}
\lambda \left(\bigcup_{k\geqslant 0}B_k\right)&=\frac1{10^{100}}+\frac{8}{10^{101}}\sum_{k\geqslant 1}\frac{a_{k-1}}{10^{k-1}}\\
&=\frac1{10^{100}}+\frac{8}{10^{101}}\left[\sum_{k\geqslant 0}a_kx^k\right]_{x=1/10}\\
&=\frac1{10^{100}}+\frac{8}{10^{101}}A(1/10)\\
&\approx 9\cdot 10^{-100}
\end{align*}
$$
---
## 2D.23
Note that $\Bbb R /\Bbb Q$ is uncountable, and so there is a bijection $g:\Bbb R /\Bbb Q \to \Bbb R$. Now let $f(x):=g([x])$.∎
---
## 2E.8
It is enough to note that, because $\sum_{k\geqslant }2^{-k}=2$ then for any chosen $\varepsilon >0$ there is some $N\in \Bbb N$ such that $0\leqslant 2-\sum_{k=0}^{N}2^{-k}<\varepsilon$, so choosing the finite set defined by $N$ we have finished.
Interesting. In your example, you get an error only on 2D.1. Perhaps it has something to do with \begin{align*} ?
Interesting. In your example, you get an error only on 2D.1. Perhaps it has something to do with
\begin{align*}?
I get errors in the whole document, not just in 2D.1, in many places there is no \begin{align* }.
@Masacroso did you find a way to fix it (not manually I mean)? I also have structure $$ \mathbb{R} $$ and not $$\mathbb{R}$$ And it does not render
