Pascal's Triangle

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

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.

Pascal triangle extended to negative values

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 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 (1+x)^n} , where 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 n} 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, and calculated the areas following Wallis and Fermat method that was known:

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 = &&\to && A(y_0) = x = \frac11 x \\ y_1 &= (1-x^2)^{1/2} \\ y_2 &= (1-x^2)^{2/2} = 1-x^2 &&\to && A(y_2) = x - \frac13 x^3 = \frac11 x - \frac13 x^3 \\ y_3 &= (1-x^2)^{3/2} \\ y_4 &= (1-x^2)^{4/2} = (1-x^2)^2 = 1 - 2x^2 + x^4 &&\to && A(y_4) = x - \frac23 x^3 + \frac15 x^5 = \frac11 x - \frac23 x^3 + \frac15 x^5 \\ 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 &&\to && A(y_6) = x - x^3 + \frac35 x^5 - \frac17x^7 = \frac11 x - \frac33 x^3 + \frac35 x^5 - \frac17x^7 \end{align} }

Newton noted that

• the first term is always $x$. He assumed that that is true also for half-integer numbers
• The denominator is always an odd integer: $1,3,5,7,\dots$
• the second term is ${\frac {0}{3}}x^{3}$, ${\frac {1}{3}}x^{3}$, ${\frac {2}{3}}x^{3}$, ${\frac {3}{3}}x^{3}$, etc. Thus, because the numerator of the second term is separated by $1$ he assumed that when adding the half-integers into the list, the separation will be $1/2$, also $0,{\frac {1}{2}},1,{\frac {3}{2}},2,{\frac {5}{2}},3\dots$ So, this will give the the first and second term half-integer to be

${\begin{aligned}A(y_{1})&=A[(1-x^{2})^{1/2}]=x-{\frac {\tfrac {1}{2}}{3}}x^{3}+{\frac {a_{12}}{5}}x^{5}-\cdots =x-{\frac {1}{6}}x^{3}+{\frac {a_{12}}{5}}x^{5}-\cdots \\A(y_{3})&=A[(1-x^{2})^{3/2}]=x-{\frac {\tfrac {1+1}{2}}{3}}x^{3}+{\frac {a_{32}}{5}}x^{5}-\cdots =x-{\frac {1}{3}}x^{3}+{\frac {a_{32}}{5}}x^{5}-\cdots \\A(y_{5})&=A[(1-x^{2})^{5/2}]=x-{\frac {\tfrac {1+2}{2}}{3}}x^{3}+{\frac {a_{5_{2}}}{5}}x^{5}-\cdots =x-{\frac {1}{2}}x^{3}+{\frac {a_{52}}{5}}x^{5}-\cdots \end{aligned}}$

Newton argued that the coefficient $a_{12}$ must be $a_{12}={\frac {1}{2}}\times {\frac {{\tfrac {1}{2}}-1}{2}}=-{\frac {1}{8}}$ and by multiplying continously, he got the next term in $A(y_{1})$ to $a_{13}=-{\frac {1}{8}}\times {\frac {{\tfrac {1}{2}}-2}{2}}={\frac {1}{16}}$ and next $a_{14}={\frac {1}{16}}\times {\frac {{\tfrac {1}{2}}-3}{2}}=-{\frac {5}{128}}$ and thus he got $A(y_{1})=A[(1-x^{2})^{1/2}]=x-{\frac {1}{6}}x^{3}-{\frac {1}{40}}x^{5}-{\frac {1}{112}}x^{7}-\cdots$

Furthermore, Newton realised that 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_1} also can be presented as a power series: 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_1 = (1-x^2)^{1/2} = 1 - \frac12 x^2 - \frac18 x^4 - \frac1{16}x^6 - \cdots } and Newton multiplied it by itself, and got the 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 1-x^2} .

Numbers

• Natural numbers
• Triangular numbers
• Tetrahedral numbers
• Pentatope numbers
• 5-simplex numbers
• 6-simplex numbers
• 7-simplex numbers