Inscribe(inner) and circumscribe (outer) two similar "n-gon"s
For any $n \in \mathbb{N}$ and $n \geq 3$ (The diagram shows n=6)
Define $\pi_n^{up} = P_{big\ shape} = nL_n$, $\pi_n^{low} = nl_n$ (note that $\pi \in [nL_n,nl_n]$)
Therefore $\pi_{est} = \frac{\pi_n^{up} + \pi_n^{low}}{2}$, $\pi_{error} = \frac{\pi_n^{up} - \pi_n^{low}}{2}$, Hence.
$$
\pi = \pi_{est} \pm \pi_{error}
$$
Retain: If we wish to estimate a quantity $x$ from lower bound $x_{low}$ and upper bound $x_{up}$, if for $n$ sufficiently LARGE, the bounds "approach" together, then we have a way to compute $x$ with arbitary accuracy.
Finite sums of powers of integers and proof by induction.
Objective: Prove that$\sum\limits_{i=1}^n i = \frac{n(n+1)}{2}$
By induction:
Let P(n) be a statement about counting numbers with the following properties:
Base case: P(1) is true
Induction Hypothesis: forall $k \geq 1$, if P(k) is true then $P(k+1)$ is all true.
Conclusion: P(n) is true forall $n \geq 1$
Applying to Objective
P(1): 1 = $\frac{1 \times (1+1)}{2}$ - Holds true
P(k) infer P(k+1):
Does $\sum\limits^{k}_{i = 1} + (k+1) = \frac{(k+1)(k+2)}{2}$?
We already know $\sum\limits^k_{i = 1} i = \frac{k(k+1)}{2}$, so
$$
\begin{matrix}
\frac{k(k+1)}{2} + (k+1)\\
= (k+1)[\frac{k}{2}+1]\\
= (k+1) \frac{k+2}{2}\\
= \frac{(k+1)(k+2)}{2}
\end{matrix}
$$
P(k) does infer P(k+1)
Therefore, by principle of matematical induction, it holds true for all P(n) where $n \geq 1$
The amazing power of rectangle: Finding the Area under a curve
Apply Archimedes principle.
$\Delta x = \frac{b-a}{n} = \frac{b}{n} $ by a = 0 and $n \geq 2$
$x_{i+1} = x_i + \Delta x = (i - 1) \times \Delta x$
For any $n \in \mathbb{N}$ and $n \geq 3$ (The diagram shows n=6)
Define $\pi_n^{up} = P_{big\ shape} = nL_n$, $\pi_n^{low} = nl_n$ (note that $\pi \in [nL_n,nl_n]$)
Therefore $\pi_{est} = \frac{\pi_n^{up} + \pi_n^{low}}{2}$, $\pi_{error} = \frac{\pi_n^{up} - \pi_n^{low}}{2}$, Hence.
$$
\pi = \pi_{est} \pm \pi_{error}
$$
Retain: If we wish to estimate a quantity $x$ from lower bound $x_{low}$ and upper bound $x_{up}$, if for $n$ sufficiently LARGE, the bounds "approach" together, then we have a way to compute $x$ with arbitary accuracy.
$\Delta x = \frac{b-a}{n} = \frac{b}{n} $ by a = 0 and $n \geq 2$
$x_{i+1} = x_i + \Delta x = (i - 1) \times \Delta x$