Markdown and MDX

This is an MDX test page that showcases various MDX elements and code blocks.

MDX

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test

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Etiam lobortis volutpat ligula, a facilisis purus. Sed vel felis blandit, sodales urna ac, varius mi. Pellentesque habitant morbi tristique senectus et netus et malesuada fames ac turpis egestas.

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Ordered List

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Blockquote

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Nullam luctus felis vel risus lacinia, eu fringilla urna mattis.

Horizontal Rule

Code Blocks

Inline Code

This is an example of inline code.

Fenced Code Blocks

const add = (a, b) => a + b;
console.log(add(2, 3)); // Output: 5

Math

Lift( $L$ ) can be determined by Lift Coefficient ( $C_L$ ) like the following equation.

L = \frac{1}{2} \rho v^2 S C_L

x = a_0 + \frac{1}{\displaystyle a_1 + \frac{1}{\displaystyle a_2 + \frac{1}{\displaystyle a_3 + a_4}}}

Proof of the Binomial Theorem

The Binomial Theorem states that for any real numbers $a$ and $b$ , and any non-negative integer $n$ ,

(a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k

We will prove this by induction.

Base Case

For $n = 0$ , the theorem states that $(a + b)^0 = {0 \choose 0} a^0 b^0 = 1$ , which is true.

Inductive Step

Assume the theorem holds for some $n = k$ , i.e.,

(a + b)^k = \sum_{i=0}^{k} {k \choose i} a^{k-i} b^i

We need to show that the theorem holds for $n = k + 1$ , i.e.,

(a + b)^{k+1} = \sum_{i=0}^{k+1} {k+1 \choose i} a^{k+1-i} b^i

Expanding $(a + b)^{k+1}$ , we get $(a + b)(a + b)^k$ .

Using the inductive hypothesis, we can write this as

(a + b)\sum_{i=0}^{k} {k \choose i} a^{k-i} b^i

Expanding this, we get

\sum_{i=0}^{k} {k \choose i} a^{k+1-i} b^i + \sum_{i=0}^{k} {k \choose i} a^{k-i} b^{i+1}

By shifting the index of the second sum by 1, we can write this as

\sum_{i=0}^{k} {k \choose i} a^{k+1-i} b^i + \sum_{i=1}^{k+1} {k \choose i-1} a^{k+1-i} b^i

By the property of binomial coefficients, we know that ${k \choose i} + {k \choose i-1} = {k+1 \choose i}$ .

Therefore, we can write the above expression as

\sum_{i=0}^{k+1} {k+1 \choose i} a^{k+1-i} b^i

which completes the inductive step.

Therefore, by mathematical induction, the Binomial Theorem is proven.

Proof of the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if a function $f$ is continuous over the interval $[a, b]$ and $F$ is an antiderivative of $f$ on $[a, b]$ , then

\int_{a}^{b} f(x) dx = F(b) - F(a)

We will prove this theorem.

Let $F(x)$ be the area under the curve of $f$ from $a$ to $x$ , i.e.,

F(x) = \int_{a}^{x} f(t) dt

We want to find the derivative of $F(x)$ , i.e., $F'(x)$ .

By the definition of the derivative, we have

F'(x) = \lim_{h \to 0} \frac{F(x + h) - F(x)}{h}

Substituting the definition of $F(x)$ , we get

F'(x) = \lim_{h \to 0} \frac{\int_{a}^{x + h} f(t) dt - \int_{a}^{x} f(t) dt}{h}

By the properties of integrals, we can write this as

F'(x) = \lim_{h \to 0} \frac{\int_{x}^{x + h} f(t) dt}{h}

As $h$ approaches 0, the integral $\int_{x}^{x + h} f(t) dt$ approaches the area of a rectangle with height $f(x)$ and width $h$ . Therefore, we can write this as

F'(x) = \lim_{h \to 0} \frac{f(x) \cdot h}{h} = f(x)

Therefore, $F'(x) = f(x)$ , i.e., $F$ is an antiderivative of $f$ .

Substituting this into the equation $\int_{a}^{b} f(x) dx = F(b) - F(a)$ , we get

\int_{a}^{b} f(x) dx = F(b) - F(a) = \int_{a}^{b} F'(x) dx

which completes the proof of the Fundamental Theorem of Calculus.