We can show LaTeX display math by using either the \\[ ... \\] or $$ ... $$
environments:
$$ \frac{d}{dx} \left( \int_{0}^{x} f(u) , du \right) = f(x) $$
\[ \frac{d}{dx} \left( \int_{g(x)}^{h(x)} f(u) , du \right) = f(h(x)) h'(x) - f(g(x)) g'(x) \]
LaTeX can be written inline as well using the \\( ... \\) or $ ... $
environments: $x = 25$ or \(y = \pi\)
Another cool thingy
We will prove this statement by induction on $h$.
Base Case: $h = 0$
The AVL tree of height 0 has no nodes, so $N_0 = 0$. The Fibonacci number $F_{0+2} = F_2 = 1$. Therefore,
$$ N_0 = 0 \geq F_2 - 1 = 1 - 1 = 0 $$
which holds true.
Base Case: $h = 1$
The AVL tree of height 1 has one node, so $N_1 = 1$. The Fibonacci number $F_{1+2} = F_3 = 2$. Therefore,
$$ N_1 = 1 \geq F_3 - 1 = 2 - 1 = 1 $$
which holds true.
Inductive Step:
Assume the claim holds true for $h = k$. That is,
$$ N_{k} \geq F_{k+2} - 1 \tag{I.H.} $$
For $h = k+1$, we know:
$$ N_{k+1} = N_{k} + N_{k-1} + 1 \tag{def. of $N_h$} $$
From the inductive hypothesis, $N_{k} \geq F_{k+2} - 1$ and $N_{k-1} \geq F_{k+1} - 1$. Substituting:
$$ N_{k+1} = N_{k} + N_{k - 1} + 1 \geq (F_{k+2} - 1) + (F_{k+1} - 1) + 1 $$
Thus, we get:
$$ N_{k+1} \geq F_{k+2} + F_{k+1} - 1 $$
By the definition of Fibonacci numbers:
$$ N_{k+1} \geq F_{k+3} - 1 $$
By the principle of mathematical induction, the statement holds for all $h \geq 0$.
Different $\LaTeX$ Fonts
$$ \text{Normal text.}\ \textit{Italic text.}\ \texttt{Typewriter text.}\ $$
Escape Characters
To disambiguate the use of the dollar sign, you can use the backslash
character to escape it: \\$.
Example:
> To write the dollar sign, use \\$ in your text. That way, if there is another
> dollar sign in your text like this \\$, it will not be confused with the start of a math
> environment.This will render as:
To write the dollar sign, use \$ in your text. That way, if there is another dollar sign in your text like this \$, it will not be confused with the start of a math environment.