Pascal's Triangle
Introduction
Binomial expansion
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} (a+b)^0 &= 1 \\ (a+b)^1 &= a + b \\ (a+b)^2 &= a^2 + 2ab + b^2 \\ (a+b)^3 &= (a+b)(a+b)^2=(a+b)(a^2 + 2ab + b^2) \\ &= a^3 + 2a^2 b + ab^2 +ba^2 + 2ab^2+b^3 \\ &= a^3 + 3a^2 b + 3ab^2+b^3 \\ (a+b)^4 &= \cdots \end{align} }
Pascal's triangle

The coefficients of binomial expansion can be easily seen from the Pascal triangle. The number is a sum of the two numbers above it.
Pascal's triangle: Negative right
This can be extended to negative numbers easily.

Now, instead of expanding Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a+b)^n} , we will use , where is a negative integer. The exponent of each terms grows when going to left. We get according to the Pascal triangle
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} (1+x)^{-1} &= 1 - x + x^2 - x^3 + \cdots \\ (1+x)^{-2} &= 1 - 2x + 3x^2 - 4x^3 + \cdots \\ (1+x)^{-3} &= 1 - 3x + 6x^2 - \cdots \end{align} }
And by Taylor series (expansion at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-1} Laurent series) we get
- https://www.wolframalpha.com/input?i=series%28+%281%2Bx%29%5E%28-1%29+%29
- https://www.wolframalpha.com/input?i=series%28+%281%2Bx%29%5E%28-2%29+%29
- https://www.wolframalpha.com/input?i=series%28+%281%2Bx%29%5E%28-3%29+%29
Pascal's triangle: Negative left
The triangle can be extended to the left also, but it is symmetric to the earlier.
Pascal's triangle: half-integers
Newton: Find the area of the curve Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = \sqrt{1-x^2}=(1-x^2)^{1/2}} , because it is a quarter of a unit circle Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A= \frac\pi/4} . He couldn't do that, so he took some other powers:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} y_0 &= (1-x^2)^{0/2} = (1-x^2)^0 = 1 \\ y_1 &= (1-x^2)^{1/2} \\ y_2 &= (1-x^2)^{2/2} = 1-x^2 \\ y_3 &= (1-x^2)^{3/2} \\ y_4 &= (1-x^2)^{4/2} = (1-x^2)^2 = 1 - 2x + x^4 \\ y_5 &= (1-x^2)^{5/2} \\ y_6 &= (1-x^2)^{6/2} \\ = (1-x^2)^3 = 1 - 3x^2 + 3x^4 -x^6 \end{align} }