Pascal's Triangle: Difference between revisions

Latest revision as of 22:20, 19 October 2022

Introduction

Binomial expansion

${\begin{aligned}(a+b)^{0}&=1\\(a+b)^{1}&=a+b\\(a+b)^{2}&=a^{2}+2ab+b^{2}\\(a+b)^{3}&=(a+b)(a+b)^{2}=(a+b)(a^{2}+2ab+b^{2})\\&=a^{3}+2a^{2}b+ab^{2}+ba^{2}+2ab^{2}+b^{3}\\&=a^{3}+3a^{2}b+3ab^{2}+b^{3}\\(a+b)^{4}&=\cdots \end{aligned}}$

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 $(a+b)^{n}$ , we will use $(1+x)^{n}$ , where $n$ is a negative integer. The exponent of each terms grows when going to left. We get according to the Pascal triangle

${\begin{aligned}(1+x)^{-1}&=1-x+x^{2}-x^{3}+\cdots \\(1+x)^{-2}&=1-2x+3x^{2}-4x^{3}+\cdots \\(1+x)^{-3}&=1-3x+6x^{2}-\cdots \end{aligned}}$

And by Taylor series (expansion at $x=-1$ Laurent series) we get

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 $y={\sqrt {1-x^{2}}}=(1-x^{2})^{1/2}$ , because it is a quarter of a unit circle $A={\frac {\pi }{4}}$ . He couldn't do that, so he took some other powers, and calculated the areas following Wallis and Fermat method that was known:

${\begin{aligned}y_{0}&=(1-x^{2})^{0/2}=(1-x^{2})^{0}=1=&&\to &&A(y_{0})=x={\frac {1}{1}}x\\y_{1}&=(1-x^{2})^{1/2}\\y_{2}&=(1-x^{2})^{2/2}=1-x^{2}&&\to &&A(y_{2})=x-{\frac {1}{3}}x^{3}={\frac {1}{1}}x-{\frac {1}{3}}x^{3}\\y_{3}&=(1-x^{2})^{3/2}\\y_{4}&=(1-x^{2})^{4/2}=(1-x^{2})^{2}=1-2x^{2}+x^{4}&&\to &&A(y_{4})=x-{\frac {2}{3}}x^{3}+{\frac {1}{5}}x^{5}={\frac {1}{1}}x-{\frac {2}{3}}x^{3}+{\frac {1}{5}}x^{5}\\y_{5}&=(1-x^{2})^{5/2}\\y_{6}&=(1-x^{2})^{6/2}=(1-x^{2})^{3}=1-3x^{2}+3x^{4}-x^{6}&&\to &&A(y_{6})=x-x^{3}+{\frac {3}{5}}x^{5}-{\frac {1}{7}}x^{7}={\frac {1}{1}}x-{\frac {3}{3}}x^{3}+{\frac {3}{5}}x^{5}-{\frac {1}{7}}x^{7}\end{aligned}}$

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 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 $a_{12}={\frac {1}{2}}\times {\frac {{\tfrac {1}{2}}-1}{2}}=-{\frac {1}{8}}$ 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 $y_{1}$ also can be presented as a power series: $y_{1}=(1-x^{2})^{1/2}=1-{\frac {1}{2}}x^{2}-{\frac {1}{8}}x^{4}-{\frac {1}{16}}x^{6}-\cdots$ and Newton multiplied it by itself, and got the $1-x^{2}$ .

Numbers

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

Fibonacci numbers

@@ Line 64: / Line 64: @@
 \begin{align}
 y_0 &= (1-x^2)^{0/2}
-= (1-x^2)^0 = 1
+= (1-x^2)^0 = 1 =
-&&\to && A(y_0) = x
+&&\to && A(y_0) = x = \frac11 x
 \\
 y_1 &= (1-x^2)^{1/2}
@@ Line 71: / Line 71: @@
 y_2 &= (1-x^2)^{2/2}
 = 1-x^2
-&&\to && A(y_2) = x - \frac13 x^3
+&&\to && A(y_2) = x - \frac13 x^3 = \frac11 x - \frac13 x^3
 \\
 y_3 &= (1-x^2)^{3/2}
@@ Line 77: / Line 77: @@
 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
+&&\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}
@@ Line 84: / Line 85: @@
 = (1-x^2)^3 = 1 - 3x^2 + 3x^4 -x^6
 &&\to && A(y_6) = x - x^3 + \frac35 x^5 - \frac17x^7
-= \x - \frac33 x^3 + \frac35 x^5 - \frac17x^7
+= \frac11 x - \frac33 x^3 + \frac35 x^5 - \frac17x^7
 \end{align}
 </math>
@@ Line 90: / Line 91: @@
 Newton noted that
 * the first term is always <math>x</math>. He assumed that that is true also for half-integer numbers
-* The denominator is always an odd integer
+* The denominator is always an odd integer: <math>1, 3, 5, 7, \dots</math>
-* the second term is <math>\frac03 x^3</math>, <math>\frac13 x^3</math>, <math>\frac23 x^3</math>, <math>\frac33 x^3</math>, etc. Thus, because the numerator of the second term is separated by <math>1</math> he assumed that when adding the half-integers into the list, the separation will be <math>1/2</math>, also <math>0, \frac12, 1, \frac 32, 2, \frac52, 3\dots</math>. So, this will give the the first and second term half-integer to be
+* the second term is <math>\frac03 x^3</math>, <math>\frac13 x^3</math>, <math>\frac23 x^3</math>, <math>\frac33 x^3</math>, etc. Thus, because the numerator of the second term is separated by <math>1</math> he assumed that when adding the half-integers into the list, the separation will be <math>1/2</math>, also <math>0, \frac12, 1, \frac 32, 2, \frac52, 3\dots</math> So, this will give the the first and second term half-integer to be
 <math>
 \begin{align}
-A(y_1) &= (1-x^2)^{1/2}
+A(y_1) &= A[(1-x^2)^{1/2}]
-= x - \frac {\tfrac12}3 x^3 + \cdots
+= 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) &= (1-x^2)^{3/2}
+A(y_3) &= A[(1-x^2)^{3/2}]
-= x - \frac {\tfrac{1+1}2}3 x^3 + \cdots
+= 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) &= (1-x^2)^{5/2}
+A(y_5) &= A[(1-x^2)^{5/2}]
-= x - \frac {\tfrac{1+2}2}3 x^3 + \cdots
+= 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}
 </math>
+Newton argued that the coefficient <math>a_{12}</math> must be
+<math>
+a_{12} = \frac12 \times \frac{\tfrac12 -1}{2} = -\frac18
+</math>
+and by multiplying continously, he got the next term in <math>A(y_1)</math> to
+<math>
+a_{13} = -\frac18 \times \frac{\tfrac12 -2}{2} = \frac1{16}
+</math>
+and next
+<math>
+a_{14} = \frac1{16} \times \frac{\tfrac12 -3}{2} = -\frac5{128}
+</math>
+and thus he got
+<math>
+A(y_1) = A[ (1-x^2)^{1/2} ]
+= x - \frac16 x^3 - \frac1{40}x^5 - \frac1{112}x^7 - \cdots
+</math>
+Furthermore, Newton realised that <math>y_1</math> also can be presented as a power series:
+<math>
+y_1 = (1-x^2)^{1/2} = 1 - \frac12 x^2 - \frac18 x^4 - \frac1{16}x^6 - \cdots
+</math>
+and Newton multiplied it by itself, and got the <math>1-x^2</math>.
+=== Numbers ===
+[[File:Pascaltriangle integers triangular .svg|thumb]]
+* Natural numbers
+* Triangular numbers
+* Tetrahedral numbers
+* Pentatope numbers
+* 5-simplex numbers
+* 6-simplex numbers
+* 7-simplex numbers
+=== Fibonacci numbers ===
+[[File:Pascaltriangle fibonacci.svg|thumb]]

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Latest revision as of 22:20, 19 October 2022

Contents

Introduction

Binomial expansion

Pascal's triangle

Pascal's triangle: Negative right

Pascal's triangle: Negative left

Pascal's triangle: half-integers

Numbers

Fibonacci numbers

Navigation

Navigation

Wiki tools

Wiki tools

Anonymous

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Pascal's Triangle: Difference between revisions

Latest revision as of 22:20, 19 October 2022

Introduction

Binomial expansion

Pascal's triangle

Pascal's triangle: Negative right

Pascal's triangle: Negative left

Pascal's triangle: half-integers

Numbers

Fibonacci numbers

Navigation

Wiki tools

Page tools

Categories