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README.md
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# NoteNextra # NoteNextra
static note sharing site for minmum care static note sharing site for minmum care

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next.config.mjs
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import nextra from 'nextra' import nextra from 'nextra'
const withNextra = nextra({ const withNextra = nextra({
theme: 'nextra-theme-blog', theme: 'nextra-theme-docs',
themeConfig: './theme.config.jsx' themeConfig: './theme.config.jsx',
latex: {
renderer: 'katex',
// options: {
// macros: {
// '\\RR': '\\mathbb{R}'
// }
// }
}
}) })
export default withNextra() export default withNextra()
// If you have other Next.js configurations, you can pass them as the parameter: // If you have other Next.js configurations, you can pass them as the parameter:
// export default withNextra({ /* other next.js config */ }) // export default withNextra({ /* other next.js config */ })

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package.json
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}, },
"dependencies": { "dependencies": {
"next": "^15.0.3", "next": "^15.0.3",
"react": "^18.3.1",
"react-dom": "^18.3.1",
"nextra": "^3.2.3", "nextra": "^3.2.3",
"nextra-theme-blog": "^3.2.3" "nextra-theme-docs": "^3.2.3",
"react": "^18.3.1",
"react-dom": "^18.3.1"
} }
} }

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pages/Math4111/Math4111_E2.md → pages/Math4111/Exam_reviews/Math4111_E2.md
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# Math 4111 Exam 2 review # Math 4111 Exam 2 review
$E$ is open if $\forall x\in E$, $x\in E^\circ$ ($E\subset E^\circ$) $E$ is open if $\forall x\in E$, $x\in E^\circ$ ($E\subset E^\circ$)
$E$ is closed if $E\supset E'$ $E$ is closed if $E\supset E'$
Then $E$ closed $\iff E^c$ open $\iff \forall x\in E^\circ, \exists r>0$ such that $B_r(x)\subset E^c$ Then $E$ closed $\iff E^c$ open $\iff \forall x\in E^\circ, \exists r>0$ such that $B_r(x)\subset E^c$
$\forall x\in E^c$, $\forall x\notin E$ $\forall x\in E^c$, $\forall x\notin E$
$B_r(x)\subset E^c\iff B_r(x)\cap E=\phi$ $B_r(x)\subset E^c\iff B_r(x)\cap E=\phi$
## Past exam questions ## Past exam questions
$S,T$ is compact $\implies S\cup T$ is compact $S,T$ is compact $\implies S\cup T$ is compact
Proof: Proof:
Suppose $S$ and $T$ are compact, let $\{G_\alpha\}_{\alpha\in A}$ be an open cover of $S\cup T$ Suppose $S$ and $T$ are compact, let $\{G_\alpha\}_{\alpha\in A}$ be an open cover of $S\cup T$
(NOT) $\{G_\alpha\}$ is an open cover of $S$, $\{H_\beta\}$ is an open cover of $T$. (NOT) $\{G_\alpha\}$ is an open cover of $S$, $\{H_\beta\}$ is an open cover of $T$.
... ...
EOP EOP
## K-cells are compact ## K-cells are compact
We'll prove the case $k=1$ and $I=[0,1]$ (This is to simplify notation. This same ideas are used in the general case) We'll prove the case $k=1$ and $I=[0,1]$ (This is to simplify notation. This same ideas are used in the general case)
Proof: Proof:
That $[0,1]$ is compact. That $[0,1]$ is compact.
(Key idea, divide and conquer) (Key idea, divide and conquer)
Suppose for contradiction that $\exists$ open cover $\{G_a\}_{\alpha\in A}$ of $[0,1]$ with no finite subcovers of $[0,1]$ Suppose for contradiction that $\exists$ open cover $\{G_a\}_{\alpha\in A}$ of $[0,1]$ with no finite subcovers of $[0,1]$
**Step1.** Divide $[0,1]$ in half. $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$ and at least one of the subintervals cannot be covered by a finite subcollection of $\{G_\alpha\}_{\alpha\in A}$ **Step1.** Divide $[0,1]$ in half. $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$ and at least one of the subintervals cannot be covered by a finite subcollection of $\{G_\alpha\}_{\alpha\in A}$
(If both of them could be, combine the two finite subcollections to get a finite subcover of $[0,1]$) (If both of them could be, combine the two finite subcollections to get a finite subcover of $[0,1]$)
Let $I_1$ be a subinterval without a finite subcover. Let $I_1$ be a subinterval without a finite subcover.
**Step2.** Divide $I_1$ in half. Let $I_2$ be one of these two subintervals of $I_1$ without a finite subcover. **Step2.** Divide $I_1$ in half. Let $I_2$ be one of these two subintervals of $I_1$ without a finite subcover.
**Step3.** etc. **Step3.** etc.
We obtain a seg of intervals $I_1\subset I_2\subset \dots$ such that We obtain a seg of intervals $I_1\subset I_2\subset \dots$ such that
(a) $[0,1]\supset I_1\supset I_2\supset \dots$ (a) $[0,1]\supset I_1\supset I_2\supset \dots$
(b) $\forall n\in \mathbb{N}$, $I_n$ is not covered by a finite subcollection of $\{G_\alpha\}_{\alpha\in A}$ (b) $\forall n\in \mathbb{N}$, $I_n$ is not covered by a finite subcollection of $\{G_\alpha\}_{\alpha\in A}$
(c) The length of $I_n$ is $\frac{1}{2^n}$ (c) The length of $I_n$ is $\frac{1}{2^n}$
By (a) and **Theorem 2.38**, $\exists x^*\in \bigcap^{\infty}_{n=1} I_n$. By (a) and **Theorem 2.38**, $\exists x^*\in \bigcap^{\infty}_{n=1} I_n$.
Since $x^*\in [0,1]$, $\exists \alpha_0$ such that $x^*\in G_{\alpha_0}$ Since $x^*\in [0,1]$, $\exists \alpha_0$ such that $x^*\in G_{\alpha_0}$
Since $G_{\alpha_0}$ is open, $\exist r>0$ such that $B_r(x^*)\subset G_{\alpha_0}$ Since $G_{\alpha_0}$ is open, $\exist r>0$ such that $B_r(x^*)\subset G_{\alpha_0}$
Let $n\in \mathbb{N}$ be such that $\frac{1}{2^n}<r$. Then by $(c)$, $I(n)\subset B_r(x^*)\subset G_{\alpha_0}$ Let $n\in \mathbb{N}$ be such that $\frac{1}{2^n}<r$. Then by $(c)$, $I(n)\subset B_r(x^*)\subset G_{\alpha_0}$
Then $\{G_{\alpha_0}\}$ is a cover of $I_n$ which contradicts with (b) Then $\{G_{\alpha_0}\}$ is a cover of $I_n$ which contradicts with (b)
EOP EOP
## Redundant subcover question ## Redundant subcover question
$M$ is compact and $\{G_\alpha\}_{\alpha\in A}$ is a "redundant" subcover of $M$. $M$ is compact and $\{G_\alpha\}_{\alpha\in A}$ is a "redundant" subcover of $M$.
$\exists \{G_{\alpha_i}\}_{i=1}^n$ is a finite subcover of $M$. $\exists \{G_{\alpha_i}\}_{i=1}^n$ is a finite subcover of $M$.
We define $S$ be the $x\in M$ that is only being covered once. We define $S$ be the $x\in M$ that is only being covered once.
$$ $$
S=M\backslash\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right) S=M\backslash\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right)
$$ $$
We claim $S$ is a closed set. We claim $S$ is a closed set.
$G_{\alpha_i}\cap G_{\alpha_j}$ is open. $G_{\alpha_i}\cap G_{\alpha_j}$ is open.
$\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right)$ is closed $\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right)$ is closed
$S=M\backslash\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right)$ is closed. $S=M\backslash\left(\bigcup_{i\neq j,i,j\in A} G_{\alpha_i}\cap G_{\alpha_j}\right)$ is closed.
So $S$ is compact, we found another finite subcover yeah! So $S$ is compact, we found another finite subcover yeah!

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pages/Math4111/Math4111_E3.md → pages/Math4111/Exam_reviews/Math4111_E3.md
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# Exam 3 Review session # Exam 3 Review session
## Relations between series and topology (compactness, closure, etc.) ## Relations between series and topology (compactness, closure, etc.)
Limit points $E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\backslash\{x\}\cap E\neq\emptyset\}$ Limit points $E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\backslash\{x\}\cap E\neq\emptyset\}$
Closure $\overline{E}=E\cup E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\cap E\neq\emptyset\}$ Closure $\overline{E}=E\cup E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\cap E\neq\emptyset\}$
$p_n\to p\implies \forall \epsilon>0, \exists N$ such that $\forall n\geq N, p_n\in B_\epsilon(p)$ $p_n\to p\implies \forall \epsilon>0, \exists N$ such that $\forall n\geq N, p_n\in B_\epsilon(p)$
### Some interesting results ### Some interesting results
#### Lemma #### Lemma
$p\in \overline{E}\iff \exists (p_n)\subseteq E$ such that $p_n\to p$ $p\in \overline{E}\iff \exists (p_n)\subseteq E$ such that $p_n\to p$
$p\in E'\iff \exists (p_n)\subseteq E\backslash\{p\}$ such that $p_n\to p$ (you cannot choose $p$ in the sequence) $p\in E'\iff \exists (p_n)\subseteq E\backslash\{p\}$ such that $p_n\to p$ (you cannot choose $p$ in the sequence)
#### Bolzano-Weierstrass Theorem #### Bolzano-Weierstrass Theorem
Let $E$ be a compact set and $(p_n)$ be a sequence in $E$. Then $\exists (p_{n_k})\subseteq (p_n)$ such that $p_{n_k}\to p\in E$. Let $E$ be a compact set and $(p_n)$ be a sequence in $E$. Then $\exists (p_{n_k})\subseteq (p_n)$ such that $p_{n_k}\to p\in E$.
Rudin Proof: Rudin Proof:
Rudin's proof uses a fact from Chapter 2. Rudin's proof uses a fact from Chapter 2.
If $E$ is compact, and $S\subseteq E$ is infinite, then $S$ has a limit point in $E$ ($S'\cap E\neq\emptyset$). If $E$ is compact, and $S\subseteq E$ is infinite, then $S$ has a limit point in $E$ ($S'\cap E\neq\emptyset$).
## Examples of Cauchy sequence that does not converge ## Examples of Cauchy sequence that does not converge
> Cauchy sequence in $(X,d),\forall \epsilon>0, \exists N$ such that $\forall m,n\geq N, d(p_m,p_n)<\epsilon$ > Cauchy sequence in $(X,d),\forall \epsilon>0, \exists N$ such that $\forall m,n\geq N, d(p_m,p_n)<\epsilon$
Let $X=\mathbb{Q}$ and $(p_q)=\{1,1.4,1.41,1.414,1.4142,1.41421,\dots\}$ The sequence is Cauchy but does not converge in $\mathbb{Q}$. Let $X=\mathbb{Q}$ and $(p_q)=\{1,1.4,1.41,1.414,1.4142,1.41421,\dots\}$ The sequence is Cauchy but does not converge in $\mathbb{Q}$.
This does not hold in $\mathbb{R}$ because compact metric spaces are complete. This does not hold in $\mathbb{R}$ because compact metric spaces are complete.
Fact: Every Cauchy sequence is bounded. Fact: Every Cauchy sequence is bounded.
## Proof that $e$ is irrational ## Proof that $e$ is irrational
> $e=\sum_{n=0}^\infty \frac{1}{n!}$ > $e=\sum_{n=0}^\infty \frac{1}{n!}$
Let $s_n=\sum_{k=0}^n \frac{1}{k!}$ Let $s_n=\sum_{k=0}^n \frac{1}{k!}$
So $e-s_n=\left(\sum_{k=n+1}^\infty \frac{1}{k!}\right)<\frac{1}{n!n}$ So $e-s_n=\left(\sum_{k=n+1}^\infty \frac{1}{k!}\right)<\frac{1}{n!n}$
If $e$ is rational, then $\exists p,q\in\mathbb{Z}$ such that $e=\frac{q}{p}$ and $q!s_q\in\mathbb{Z}$, $q!e=q!\frac{p}{q}\in \mathbb{Z}$, so $q!(e-s_q)\in\mathbb{Z}$ If $e$ is rational, then $\exists p,q\in\mathbb{Z}$ such that $e=\frac{q}{p}$ and $q!s_q\in\mathbb{Z}$, $q!e=q!\frac{p}{q}\in \mathbb{Z}$, so $q!(e-s_q)\in\mathbb{Z}$
$0<q!(e-s_q)<\frac{1}{n!n}$ leads to contradiction. $0<q!(e-s_q)<\frac{1}{n!n}$ leads to contradiction.
## $\limsup$ and $\liminf$ ## $\limsup$ and $\liminf$
Let $(a_n)=(-1)^n$ Let $(a_n)=(-1)^n$
$\limsup a_n=1$ and $\liminf a_n=-1$ $\limsup a_n=1$ and $\liminf a_n=-1$
Let $(a_n)\to a$ Let $(a_n)\to a$
$\limsup a_n=\liminf a_n=a$ $\limsup a_n=\liminf a_n=a$
### Facts about $\limsup$ and $\liminf$ ### Facts about $\limsup$ and $\liminf$
#### Convergence of subsequence #### Convergence of subsequence
_$\limsup$ is the largest value that subsequence of $a_n$ can approach to._ _$\limsup$ is the largest value that subsequence of $a_n$ can approach to._
_$\liminf$ is the smallest value that subsequence of $a_n$ can approach to._ _$\liminf$ is the smallest value that subsequence of $a_n$ can approach to._
#### Elements of sequence #### Elements of sequence
$\forall x>s^*,\{n:a_n>x\}$ is finite. $\exists N$ such that $\forall n\geq N, a_n\leq x$ $\forall x>s^*,\{n:a_n>x\}$ is finite. $\exists N$ such that $\forall n\geq N, a_n\leq x$
$\forall x<s^*,\{n:a_n>x\}$ is infinite. $\forall x<s^*,\{n:a_n>x\}$ is infinite.
One example is $(a_n)=(-1)^n\frac{n}{n+1}$ One example is $(a_n)=(-1)^n\frac{n}{n+1}$
$\limsup a_n=1$ and $\liminf a_n=-1$ $\limsup a_n=1$ and $\liminf a_n=-1$
So the size of set of elements of $a_n$ that are greater than any $x<1$ is infinite. and the size of set of elements of $a_n$ that are greater than any $x>1$ is finite. So the size of set of elements of $a_n$ that are greater than any $x<1$ is infinite. and the size of set of elements of $a_n$ that are greater than any $x>1$ is finite.
#### $\limsup(a_n+b_n)\leq \limsup a_n+\limsup b_n$ #### $\limsup(a_n+b_n)\leq \limsup a_n+\limsup b_n$
One example for smaller than is $(a_n)=(-1)^n$ and $(b_n)=(-1)^{n+1}$ One example for smaller than is $(a_n)=(-1)^n$ and $(b_n)=(-1)^{n+1}$
$\limsup(a_n+b_n)=0$ and $\limsup a_n+\limsup b_n=2$ $\limsup(a_n+b_n)=0$ and $\limsup a_n+\limsup b_n=2$
## ($\forall n,s_n\leq t_n$) $\implies \limsup s_n\leq \limsup t_n$ ## ($\forall n,s_n\leq t_n$) $\implies \limsup s_n\leq \limsup t_n$
One example of using this theorem is $(s_n)=\left(\sum_{k=1}^n\frac{1}{k!}\right)$ and $(t_n)=\left(\frac{1}{n}+1\right)^n$ One example of using this theorem is $(s_n)=\left(\sum_{k=1}^n\frac{1}{k!}\right)$ and $(t_n)=\left(\frac{1}{n}+1\right)^n$
## Rearrangement of series ## Rearrangement of series
Will not be tested. Will not be tested.
_infinite sum is not similar to finite sum. For infinite sum, the order of terms matters. But for finite sum, the order of terms does not matter, you can rearrange the terms as you want._ _infinite sum is not similar to finite sum. For infinite sum, the order of terms matters. But for finite sum, the order of terms does not matter, you can rearrange the terms as you want._
## Ways to prove convergence of series ## Ways to prove convergence of series
### n-th term test (divergence test) ### n-th term test (divergence test)
If $\lim_{n\to\infty}a_n\neq 0$, then $\sum a_n$ diverges. If $\lim_{n\to\infty}a_n\neq 0$, then $\sum a_n$ diverges.
### Definition of convergence of series (convergence and divergence test) ### Definition of convergence of series (convergence and divergence test)
If $\sum a_n$ converges, then $\lim_{n\to\infty}\sum_{k=1}^n a_k=0$. If $\sum a_n$ converges, then $\lim_{n\to\infty}\sum_{k=1}^n a_k=0$.
Example: Telescoping series and geometric series. Example: Telescoping series and geometric series.
### Comparison test (convergence and divergence test (absolute convergence)) ### Comparison test (convergence and divergence test (absolute convergence))
Let $(a_n)$ be a sequence in $\mathbb{C}$ and $(c_n)$ be a non-negative sequence in $\mathbb{R}$. Suppose $\forall n, |a_n|\leq c_n$. Let $(a_n)$ be a sequence in $\mathbb{C}$ and $(c_n)$ be a non-negative sequence in $\mathbb{R}$. Suppose $\forall n, |a_n|\leq c_n$.
(a) If the series $\sum_{n=1}^{\infty}c_n$ converges, then the series $\sum_{n=1}^{\infty}a_n$ converges. (a) If the series $\sum_{n=1}^{\infty}c_n$ converges, then the series $\sum_{n=1}^{\infty}a_n$ converges.
(b) If the series $\sum_{n=1}^{\infty}a_n$ diverges, then the series $\sum_{n=1}^{\infty}c_n$ diverges. (b) If the series $\sum_{n=1}^{\infty}a_n$ diverges, then the series $\sum_{n=1}^{\infty}c_n$ diverges.
### Ratio test (convergence and divergence test (absolute convergence)) ### Ratio test (convergence and divergence test (absolute convergence))
> $$ \left|\frac{a_{n+1}}{a_n}\right| \leq \alpha \implies |a_n|\leq \alpha^n$$ > $$ \left|\frac{a_{n+1}}{a_n}\right| \leq \alpha \implies |a_n|\leq \alpha^n$$
Given a series $\sum_{n=0}^{\infty} a_n$, $a_n\in\mathbb{C}\backslash\{0\}$. Given a series $\sum_{n=0}^{\infty} a_n$, $a_n\in\mathbb{C}\backslash\{0\}$.
Then Then
(a) If $\limsup_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| < 1$, then $\sum_{n=0}^{\infty} a_n$ converges. (a) If $\limsup_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| < 1$, then $\sum_{n=0}^{\infty} a_n$ converges.
(b) If $\left|\frac{a_{n+1}}{a_n}\right| \geq 1$ for all $n\geq n_0$ for some $n_0\in\mathbb{N}$, then $\sum_{n=0}^{\infty} a_n$ diverges. (b) If $\left|\frac{a_{n+1}}{a_n}\right| \geq 1$ for all $n\geq n_0$ for some $n_0\in\mathbb{N}$, then $\sum_{n=0}^{\infty} a_n$ diverges.
### Root test (convergence and divergence test (absolute convergence)) ### Root test (convergence and divergence test (absolute convergence))
> $$ \sqrt[n]{|a_n|} \leq \alpha \implies |a_n|\leq \alpha^n$$ > $$ \sqrt[n]{|a_n|} \leq \alpha \implies |a_n|\leq \alpha^n$$
Given a series $\sum_{n=0}^{\infty} a_n$, put $\alpha = \limsup_{n\to\infty} \sqrt[n]{|a_n|}$. Given a series $\sum_{n=0}^{\infty} a_n$, put $\alpha = \limsup_{n\to\infty} \sqrt[n]{|a_n|}$.
Then Then
(a) If $\alpha < 1$, then $\sum_{n=0}^{\infty} a_n$ converges. (a) If $\alpha < 1$, then $\sum_{n=0}^{\infty} a_n$ converges.
(b) If $\alpha > 1$, then $\sum_{n=0}^{\infty} a_n$ diverges. (b) If $\alpha > 1$, then $\sum_{n=0}^{\infty} a_n$ diverges.
(c) If $\alpha = 1$, the test gives no information (c) If $\alpha = 1$, the test gives no information
### Cauchy criterion ### Cauchy criterion
### Geometric series ### Geometric series
### P-series ### P-series
(a) $\sum_{n=0}^{\infty}\frac{1}{n}$ diverges. (a) $\sum_{n=0}^{\infty}\frac{1}{n}$ diverges.
(b) $\sum_{n=0}^{\infty}\frac{1}{n^2}$ converges. (b) $\sum_{n=0}^{\infty}\frac{1}{n^2}$ converges.
### Cauchy condensation test (convergence test) ### Cauchy condensation test (convergence test)
Suppose $(a_n)$ is a non-negative sequence. The series $\sum_{n=1}^{\infty}a_n$ converges if and only if the series $\sum_{k=0}^{\infty}2^ka_{2^k}$ converges. Suppose $(a_n)$ is a non-negative sequence. The series $\sum_{n=1}^{\infty}a_n$ converges if and only if the series $\sum_{k=0}^{\infty}2^ka_{2^k}$ converges.
### Dirichlet test (convergence test) ### Dirichlet test (convergence test)
Suppose Suppose
(a) the partial sum $A_n$ of $\sum a_n$ form a bounded sequence. (a) the partial sum $A_n$ of $\sum a_n$ form a bounded sequence.
(b) $b_0\geq b_1\geq b_2\geq \cdots$ (non-increasing) (b) $b_0\geq b_1\geq b_2\geq \cdots$ (non-increasing)
(c) $\lim_{n\to\infty}b_n=0$. (c) $\lim_{n\to\infty}b_n=0$.
Then $\sum a_nb_n$ converges. Then $\sum a_nb_n$ converges.
Example: $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$ converges. Example: $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$ converges.
### Abel's test (convergence test) ### Abel's test (convergence test)
Let $(b_n)^\infty_{n=0}$ be a sequence such that: Let $(b_n)^\infty_{n=0}$ be a sequence such that:
(a) $b_0\geq b_1\geq b_2\geq \cdots$ (non-increasing) (a) $b_0\geq b_1\geq b_2\geq \cdots$ (non-increasing)
(b) $\lim_{n\to\infty}b_n=0$ (b) $\lim_{n\to\infty}b_n=0$
Then if $|z|=1$ and $z\neq 1$, $\sum_{n=0}^\infty b_nz^n$ converges. Then if $|z|=1$ and $z\neq 1$, $\sum_{n=0}^\infty b_nz^n$ converges.

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export default {
Exam_reviews: "Exam reviews",
Math4111_L1: "Lecture 1",
Math4111_L2: "Lecture 2",
Math4111_L3: "Lecture 3",
Math4111_L4: "Lecture 4",
Math4111_L5: "Lecture 5",
Math4111_L6: "Lecture 6",
Math4111_L7: "Lecture 7",
Math4111_L8: "Lecture 8",
Math4111_L9: "Lecture 9",
Math4111_L10: "Lecture 10",
Math4111_L11: "Lecture 11",
Math4111_L12: "Lecture 12",
Math4111_L13: "Lecture 13",
Math4111_L14: "Lecture 14",
Math4111_L15: "Lecture 15",
Math4111_L16: "Lecture 16",
Math4111_L17: "Lecture 17",
Math4111_L18: "Lecture 18",
Math4111_L19: "Lecture 19",
Math4111_L20: "Lecture 20",
Math4111_L21: "Lecture 21",
Math4111_L22: {
display: 'hidden'
},
Math4111_L23: {
display: 'hidden'
},
Math4111_L24: {
display: 'hidden'
},
Math4111_L25: {
display: 'hidden'
},
Math4111_L26: {
display: 'hidden'
},
Math4111_L27: {
display: 'hidden'
},
Math4111_L28: {
display: 'hidden'
},
index: {
display: 'hidden'
}
}

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import { ThemeProvider } from 'next-themes'
export default function App({ Component, pageProps }) { export default function App({ Component, pageProps }) {
return <Component {...pageProps} /> return (
<ThemeProvider attribute='class'>
<Component {...pageProps} />
</ThemeProvider>
)
} }

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pages/_meta.js
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export default { export default {
index: 'My Homepage', index: {
contact: 'Contact Us', title: 'Home',
about: 'About Us' type: 'menu',
items: {
index: {
title: 'Home',
href: '/'
},
about: {
title: 'About',
href: '/about'
},
contact: {
title: 'Contact Me',
href: '/contact'
}
}
},
Math4111: {
title: 'Math 4111',
type: 'page'
},
about: {
display: 'hidden'
},
contact: {
display: 'hidden'
}
} }

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# About
##

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# contact
##

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pnpm-lock.yaml generated
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nextra: nextra:
specifier: ^3.2.3 specifier: ^3.2.3
version: 3.2.3(@types/react@18.3.12)(acorn@8.14.0)(next@15.0.3(react-dom@18.3.1(react@18.3.1))(react@18.3.1))(react-dom@18.3.1(react@18.3.1))(react@18.3.1)(typescript@5.6.3) version: 3.2.3(@types/react@18.3.12)(acorn@8.14.0)(next@15.0.3(react-dom@18.3.1(react@18.3.1))(react@18.3.1))(react-dom@18.3.1(react@18.3.1))(react@18.3.1)(typescript@5.6.3)
nextra-theme-blog: nextra-theme-docs:
specifier: ^3.2.3 specifier: ^3.2.3
version: 3.2.3(next@15.0.3(react-dom@18.3.1(react@18.3.1))(react@18.3.1))(nextra@3.2.3(@types/react@18.3.12)(acorn@8.14.0)(next@15.0.3(react-dom@18.3.1(react@18.3.1))(react@18.3.1))(react-dom@18.3.1(react@18.3.1))(react@18.3.1)(typescript@5.6.3))(react-dom@18.3.1(react@18.3.1))(react@18.3.1) version: 3.2.3(next@15.0.3(react-dom@18.3.1(react@18.3.1))(react@18.3.1))(nextra@3.2.3(@types/react@18.3.12)(acorn@8.14.0)(next@15.0.3(react-dom@18.3.1(react@18.3.1))(react@18.3.1))(react-dom@18.3.1(react@18.3.1))(react@18.3.1)(typescript@5.6.3))(react-dom@18.3.1(react@18.3.1))(react@18.3.1)
react: react:
@@ -694,6 +694,9 @@ packages:
resolution: {integrity: sha512-e2i4wANQiSXgnrBlIatyHtP1odfUp0BbV5Y5nEGbxtIrStkEOAAzCUirvLBNXHLr7kwLvJl6V+4V3XV9x7Wd9w==} resolution: {integrity: sha512-e2i4wANQiSXgnrBlIatyHtP1odfUp0BbV5Y5nEGbxtIrStkEOAAzCUirvLBNXHLr7kwLvJl6V+4V3XV9x7Wd9w==}
engines: {node: ^12.20.0 || >=14} engines: {node: ^12.20.0 || >=14}
compute-scroll-into-view@3.1.0:
resolution: {integrity: sha512-rj8l8pD4bJ1nx+dAkMhV1xB5RuZEyVysfxJqB1pRchh1KVvwOv9b7CGB8ZfjTImVv2oF+sYMUkMZq6Na5Ftmbg==}
confbox@0.1.8: confbox@0.1.8:
resolution: {integrity: sha512-RMtmw0iFkeR4YV+fUOSucriAQNb9g8zFR52MWCtl+cCZOFRNL6zeB395vPzFhEjjn4fMxXudmELnl/KF/WrK6w==} resolution: {integrity: sha512-RMtmw0iFkeR4YV+fUOSucriAQNb9g8zFR52MWCtl+cCZOFRNL6zeB395vPzFhEjjn4fMxXudmELnl/KF/WrK6w==}
@@ -972,6 +975,9 @@ packages:
fault@2.0.1: fault@2.0.1:
resolution: {integrity: sha512-WtySTkS4OKev5JtpHXnib4Gxiurzh5NCGvWrFaZ34m6JehfTUhKZvn9njTfw48t6JumVQOmrKqpmGcdwxnhqBQ==} resolution: {integrity: sha512-WtySTkS4OKev5JtpHXnib4Gxiurzh5NCGvWrFaZ34m6JehfTUhKZvn9njTfw48t6JumVQOmrKqpmGcdwxnhqBQ==}
flexsearch@0.7.43:
resolution: {integrity: sha512-c5o/+Um8aqCSOXGcZoqZOm+NqtVwNsvVpWv6lfmSclU954O3wvQKxxK8zj74fPaSJbXpSLTs4PRhh+wnoCXnKg==}
format@0.2.2: format@0.2.2:
resolution: {integrity: sha512-wzsgA6WOq+09wrU1tsJ09udeR/YZRaeArL9e1wPbFg3GG2yDnC2ldKpxs4xunpFF9DgqCqOIra3bc1HWrJ37Ww==} resolution: {integrity: sha512-wzsgA6WOq+09wrU1tsJ09udeR/YZRaeArL9e1wPbFg3GG2yDnC2ldKpxs4xunpFF9DgqCqOIra3bc1HWrJ37Ww==}
engines: {node: '>=0.4.x'} engines: {node: '>=0.4.x'}
@@ -1387,6 +1393,14 @@ packages:
react-cusdis: react-cusdis:
optional: true optional: true
nextra-theme-docs@3.2.3:
resolution: {integrity: sha512-kRhnLxbAbD3FgR93yLbu6Iz6XvErka3I5CcVo3VobLuV1mefbZ1T6DfiY6q0KJoHLGRrJESsFSarIqPjKOx00g==}
peerDependencies:
next: '>=13'
nextra: 3.2.3
react: '>=18'
react-dom: '>=18'
nextra@3.2.3: nextra@3.2.3:
resolution: {integrity: sha512-MyNA2kPvDyJK1trjFkwpTdMOKJu/MIueENHtmLoxPnyOi3fxtk9H5k6b5WdMGBibsyFeXqTz9REnz7d1/xL9Hg==} resolution: {integrity: sha512-MyNA2kPvDyJK1trjFkwpTdMOKJu/MIueENHtmLoxPnyOi3fxtk9H5k6b5WdMGBibsyFeXqTz9REnz7d1/xL9Hg==}
engines: {node: '>=18'} engines: {node: '>=18'}
@@ -1570,6 +1584,9 @@ packages:
scheduler@0.23.2: scheduler@0.23.2:
resolution: {integrity: sha512-UOShsPwz7NrMUqhR6t0hWjFduvOzbtv7toDH1/hIrfRNIDBnnBWd0CwJTGvTpngVlmwGCdP9/Zl/tVrDqcuYzQ==} resolution: {integrity: sha512-UOShsPwz7NrMUqhR6t0hWjFduvOzbtv7toDH1/hIrfRNIDBnnBWd0CwJTGvTpngVlmwGCdP9/Zl/tVrDqcuYzQ==}
scroll-into-view-if-needed@3.1.0:
resolution: {integrity: sha512-49oNpRjWRvnU8NyGVmUaYG4jtTkNonFZI86MmGRDqBphEK2EXT9gdEUoQPZhuBM8yWHxCWbobltqYO5M4XrUvQ==}
section-matter@1.0.0: section-matter@1.0.0:
resolution: {integrity: sha512-vfD3pmTzGpufjScBh50YHKzEu2lxBWhVEHsNGoEXmCmn2hKGfeNLYMzCJpe8cD7gqX7TJluOVpBkAequ6dgMmA==} resolution: {integrity: sha512-vfD3pmTzGpufjScBh50YHKzEu2lxBWhVEHsNGoEXmCmn2hKGfeNLYMzCJpe8cD7gqX7TJluOVpBkAequ6dgMmA==}
engines: {node: '>=4'} engines: {node: '>=4'}
@@ -2501,6 +2518,8 @@ snapshots:
commander@9.2.0: {} commander@9.2.0: {}
compute-scroll-into-view@3.1.0: {}
confbox@0.1.8: {} confbox@0.1.8: {}
cose-base@1.0.3: cose-base@1.0.3:
@@ -2817,6 +2836,8 @@ snapshots:
dependencies: dependencies:
format: 0.2.2 format: 0.2.2
flexsearch@0.7.43: {}
format@0.2.2: {} format@0.2.2: {}
get-stream@3.0.0: {} get-stream@3.0.0: {}
@@ -3629,6 +3650,20 @@ snapshots:
react: 18.3.1 react: 18.3.1
react-dom: 18.3.1(react@18.3.1) react-dom: 18.3.1(react@18.3.1)
nextra-theme-docs@3.2.3(next@15.0.3(react-dom@18.3.1(react@18.3.1))(react@18.3.1))(nextra@3.2.3(@types/react@18.3.12)(acorn@8.14.0)(next@15.0.3(react-dom@18.3.1(react@18.3.1))(react@18.3.1))(react-dom@18.3.1(react@18.3.1))(react@18.3.1)(typescript@5.6.3))(react-dom@18.3.1(react@18.3.1))(react@18.3.1):
dependencies:
'@headlessui/react': 2.2.0(react-dom@18.3.1(react@18.3.1))(react@18.3.1)
clsx: 2.1.1
escape-string-regexp: 5.0.0
flexsearch: 0.7.43
next: 15.0.3(react-dom@18.3.1(react@18.3.1))(react@18.3.1)
next-themes: 0.4.3(react-dom@18.3.1(react@18.3.1))(react@18.3.1)
nextra: 3.2.3(@types/react@18.3.12)(acorn@8.14.0)(next@15.0.3(react-dom@18.3.1(react@18.3.1))(react@18.3.1))(react-dom@18.3.1(react@18.3.1))(react@18.3.1)(typescript@5.6.3)
react: 18.3.1
react-dom: 18.3.1(react@18.3.1)
scroll-into-view-if-needed: 3.1.0
zod: 3.23.8
nextra@3.2.3(@types/react@18.3.12)(acorn@8.14.0)(next@15.0.3(react-dom@18.3.1(react@18.3.1))(react@18.3.1))(react-dom@18.3.1(react@18.3.1))(react@18.3.1)(typescript@5.6.3): nextra@3.2.3(@types/react@18.3.12)(acorn@8.14.0)(next@15.0.3(react-dom@18.3.1(react@18.3.1))(react@18.3.1))(react-dom@18.3.1(react@18.3.1))(react@18.3.1)(typescript@5.6.3):
dependencies: dependencies:
'@formatjs/intl-localematcher': 0.5.7 '@formatjs/intl-localematcher': 0.5.7
@@ -3964,6 +3999,10 @@ snapshots:
dependencies: dependencies:
loose-envify: 1.4.0 loose-envify: 1.4.0
scroll-into-view-if-needed@3.1.0:
dependencies:
compute-scroll-into-view: 3.1.0
section-matter@1.0.0: section-matter@1.0.0:
dependencies: dependencies:
extend-shallow: 2.0.1 extend-shallow: 2.0.1

69
theme.config.jsx
View File

@@ -1,21 +1,54 @@
import { useRouter } from 'next/router'
import { useConfig } from 'nextra-theme-docs'
export default { export default {
footer: <p>MIT 2023 © Nextra.</p>, footer: {
head: ({ title, meta }) => ( content: (
<span>
MIT {new Date().getFullYear()} ©{' '}
<a href="https://github.com/Trance-0" target="_blank">
Trance-0
</a>
.
</span>
)
},
head() {
const { asPath, defaultLocale, locale } = useRouter()
const { frontMatter } = useConfig()
const url =
'https://vercel.com/notenextra' +
(defaultLocale === locale ? asPath : `/${locale}${asPath}`)
return (
<> <>
{meta.description && ( <meta property="og:url" content={url} />
<meta name="description" content={meta.description} /> <meta property="og:title" content={frontMatter.title || 'NoteNextra'} />
)} <meta
{meta.tag && <meta name="keywords" content={meta.tag} />} property="og:description"
{meta.author && <meta name="author" content={meta.author} />} content={frontMatter.description || 'A static note sharing site for minimum care'}
/>
</> </>
), )
readMore: 'Read More →', },
postFooter: null,
darkMode: false, logo: (
navs: [ <>
{ <svg width="24" height="24" viewBox="0 0 24 24">
url: 'https://github.com/shuding/nextra', <path fillRule="evenodd" d="M1.114 8.063V7.9c1.005-.102 1.497-.615 1.497-1.6V4.503c0-1.094.39-1.538 1.354-1.538h.273V2h-.376C2.25 2 1.49 2.759 1.49 4.352v1.524c0 1.094-.376 1.456-1.49 1.456v1.299c1.114 0 1.49.362 1.49 1.456v1.524c0 1.593.759 2.352 2.372 2.352h.376v-.964h-.273c-.964 0-1.354-.444-1.354-1.538V9.663c0-.984-.492-1.497-1.497-1.6M14.886 7.9v.164c-1.005.103-1.497.616-1.497 1.6v1.798c0 1.094-.39 1.538-1.354 1.538h-.273v.964h.376c1.613 0 2.372-.759 2.372-2.352v-1.524c0-1.094.376-1.456 1.49-1.456v-1.3c-1.114 0-1.49-.362-1.49-1.456V4.352C14.51 2.759 13.75 2 12.138 2h-.376v.964h.273c.964 0 1.354.444 1.354 1.538V6.3c0 .984.492 1.497 1.497 1.6M7.5 11.5V9.207l-1.621 1.621-.707-.707L6.792 8.5H4.5v-1h2.293L5.172 5.879l.707-.707L7.5 6.792V4.5h1v2.293l1.621-1.621.707.707L9.208 7.5H11.5v1H9.207l1.621 1.621-.707.707L8.5 9.208V11.5z"/>
name: 'Nextra' </svg>
} <span style={{ marginLeft: '.4em', fontWeight: 800 }}>
] NoteNextra
} </span>
</>
),
readMore: 'Read More →',
postFooter: null,
darkMode: false,
navs: [
{
url: 'https://github.com/Trance-0/NoteNextra',
name: 'Source'
}
]
}