Pascal's Triangle: Difference between revisions
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A(y_1) &= A[ (1-x^2)^{1/2} ] | A(y_1) &= A[ (1-x^2)^{1/2} ] | ||
= x - \frac16 x^3 - \frac1{40}x^5 - \frac1{112}x^7 - \ | = x - \frac16 x^3 - \frac1{40}x^5 - \frac1{112}x^7 - \cdots | ||
</math> | </math> | ||
Revision as of 21:55, 19 October 2022
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 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 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} . He assumed that that is true also for half-integer numbers
- The denominator is always an odd integer: 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, 3, 5, 7, \dots}
- the second term is 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 \frac03 x^3} , 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 \frac13 x^3} , 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 \frac23 x^3} , 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 \frac33 x^3} , etc. Thus, because the numerator of the second term is separated by 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} he assumed that when adding the half-integers into the list, the separation will be 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 1/2} , also 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 0, \frac12, 1, \frac 32, 2, \frac52, 3\dots} So, this will give the the first and second term half-integer to be
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(y_1) &= A[(1-x^2)^{1/2}] = x - \frac {\tfrac12}3 x^3 + \frac{a_{12}}5 x^5 - \cdots = x - \frac16 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 - \frac13 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 - \frac12 x^3 + \frac{a_{52}}5 x^5 - \cdots \end{align} }
Newton argued that the coefficient 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_{12}} must be 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_{12} = \frac12 \times \frac{\tfrac12 -1}{2} = -\frac18 } and by multiplying continously, he got the next term in 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(y_1)} to 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_{13} = -\frac18 \times \frac{\tfrac12 -2}{2} = \frac1{16} } and next 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_{14} = \frac1{16} \times \frac{\tfrac12 -3}{2} = -\frac5{128} } and thus he got 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(y_1) &= A[ (1-x^2)^{1/2} ] = x - \frac16 x^3 - \frac1{40}x^5 - \frac1{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