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Pythagorean theorem permeates Calculus, thus Physics, thus every science
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Archimedes Plutonium
Jul 1, 2015, 2:57:53 AM7/1/15
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Now it used to be where I thought the Pythagorean theorem in the Maxwell Equations took some effort to find and elucidate, as seen here in this 2014 post of mine:
Newsgroups: sci.math
Date: Sun, 16 Feb 2014 10:47:14 -0800 (PST)
Subject: How the Maxwell Equations derive the Pythagorean Theorem #1558
Correcting Math
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Sun, 16 Feb 2014 18:47:14 +0000
How the Maxwell Equations derive the Pythagorean Theorem #1558 Correcting Math
Alright, I discovered several years back that since all of mathematics is but a tiny subset of physics and that since all of physics is derived from the Maxwell Equations as an axiom set, that all of the theorems and axioms of mathematics comes out of the Maxwell Equations. In other words, the Maxwell Equations are a super-axiom-set.
Every axiom, every theorem in mathematics can be derived by the Maxwell Equations.
So how does the Maxwell Equations derive the Pythagorean theorem? Well, it is not as difficult as what may first appear. The derivation comes from the fact that the Maxwell Equations of the Faraday Law of a moving bar magnet into a stationary closed loop of wire causes a current to flow. That phenomenon is the very same phenomenon as a stationary bar magnet thrust through by a moving closed loop of wire causes a current to flow. In other words, as electrons flow in the wire, they could be from either a stationary bar magnet moving loop or a stationary loop with moving bar magnet. This is familiar to most in physics as special relativity.
Now, consider the proton as a square on the leg of a triangle and the electron as a different square on the leg of a triangle and the bar magnet as a square on the hypotenuse of a triangle. In order for the proton square plus electron square to equal the bar magnet square in Faraday's law, the triangle must be a right-triangle.
--- quoting *in parts* from http://www.berkeleyscience.com/relativity.htm
..
It followed from studies of Maxwell's equations. It reshaped the Newtonian physics we have studied. It explains magnetism - this is the coolest thing! ..
As for math, all you need is the Pythagorean theorem! No kidding.
..
Then both mirrors will have travelled to the right the distance vT, and distance the light will have travelled as observed by O, is cT, and, by the Pythagorean theorem (cT)2 = (vT)2 + L2 so, T2(c2 - v2) = L2 T2 = L2 / (c2 - v2) T2 = L2 / c2(1 - v2/c2)) T = L/c (1/sqrt(1 - v2/c2))
--- end quoting ---
--
AP
That was a 2014 post where I thought that the Pythagorean theorem at work and play in the Maxwell Equations was not all that obvious.
But as for a few days ago, here in 2015, I realized that whereever, and whenever Calculus is used in science, it is permeated by the Pythagorean theorem. So that when asked where is Pythagorean Theorem in the Maxwell Equations, the straightforward answer is that the Maxwell Equations are differential equations or Line-Integral equations, hence, Pythagorean theorem is all over the place.
Now here is a 2015 post of mine that explains the derivative and explains how the derivative is about right-triangles and the Pythagorean Theorem involved in the hypotenuse. So that whenever we have a science that uses the Calculus, that science has the Pythagorean Theorem.
What I am driving at, is the idea that all the axioms of Geometry can be reduced to the statement of the facts and details of a right-triangle and the same goes for the axioms of Numbers for Algebra. What this distills into, is the idea that the whole of mathematics is a tiny subset of the Maxwell Equations.
Newsgroups: sci.math
Date: Wed, 8 Apr 2015 19:56:28 -0700 (PDT)
Subject: Derivative; Introduction True Calculus #134 Correcting Math 4th ed;
(53) High-School & Uni-text 9th
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Thu, 09 Apr 2015 02:56:29 +0000
Derivative; Introduction True Calculus #134 Correcting Math 4th ed; (02) High-School & Uni-text 9th
Introduction to True Calculus, first 5 pages, page 2 Derivative
Alright, I am writing these 5 pages to expose the full spectrum of calculus from Coordinate System to function to derivative to integral all from just 2 pictures. The logic of this is that all are related and to see that relationship in full view is to use just 2 pictures. We end up seeing how the derivative relates to integral as inverse in the Fundamental Theorem of Calculus. We also see how function and the function-graph relates to derivative and integral.
The design of this text is to make it ultra easy for High School students to learn. Unlike books of the past, those authors, in most cases could not teach and their book ended up as silly contraptions. In order to teach, you have to be on the level of comprehension of the students you want to reach. This means, little to no terminology which often gets students lost. Most professors of science and math, in my opinion simply do not know how to teach and this text makes up for that.
On the first page we saw the 2 pictures and we did the Grid Coordinate System and the function and function-graph. Now we move onto the derivative.
Up front and out in the open the Derivative means rate-of-change. The definition is very simple in that of a division dy/dx, and what those symbols mean is the rate of change of y value (dy) divided by rate of change (dx) of x value.
So, in our 2 pictures we have:
THE 2 PICTURES:
The function y= 3 looks like this in Integer Grid
y-axis
^
|
-----------------------------------> y=3
| | | | |
| | | | |
| | | | |
------------------------------------------------> x-axis
0 1 2 3 4
Cell from 0 to 1 is a rectangle as well as cell from 1 to 2, with a flat top roof or ceiling, and same goes for all the other cells in y=3.
The function y=x^2 looks like this in Integer Grid:
y-axis
^
|
|
|
9 /| 9
|
|
|
/ |
|
|
/ |
|
/ |
|
4 4/ | |
| |
/ | |
| |
1 / |1 | |
/ | | |
------------------------------------------------> x-axis
0 1 2 3
Cell from 0 to 1 is a pure right-triangle sides 1 by 1 with hypotenuse sqrt2
Cell from 1 to 2 is a right-angled-trapezoid (picketfence) and the picketfence is composed of a square with a right-triangle atop the square
Cell from 2 to 3 is a right-angled-trapezoid (picketfence) and the picketfence is composed of a rectangle with a right-triangle atop the rectangle
DERIVATIVE
We define derivative as rate of change and write it as dy/dx across two adjoining cells. That is important for in most cases you need two adjoining cells, so we say, in general 2 adjoining cells are required to have a derivative. Notice in the first function above of y=3 as a flat straightline parallel to the x-axis. What is the dy? Well, the y-value always is 3. So the dy is 3-3 =0. And the dx of two adjoining cells is always 2 since the width of a cell is 1 and thus 1+1=2 in the Integer Grid. So in the function y=3 we always have dy/dx as being 0/2 which is 0.
Now let us examine the second function of y= x^2 across two adjoining cells. The first derivative we can compute would be for x=1 because we need two adjoining cells, of that of cell from 0 to 1 and then from 1 to 2 for x. Notice, we hold x=1 in the middle of these two cells, and that is important for to obtain the rate-of-change, we have the x value as the middle of the two adjoining cells. So what is the dx of these two cells? It is of course 1 +1 = 2 since the width of one cell is 1. Now, we ask what is the dy of these two adjoining cells. And we have to take the y-value of the furthest left wall which is y=0 and the y value of the furthest right wall in second cell which is y=4. So we have a dy as that of 4-0 = 4. Notice that x=1 is directly in the middle of the leftward wall and the furthest rightward wall. So the derivative of x=1 for function y=x^2 is that of dy/dx which is 4/2 = 2. The rate of change at x=1 of y=x^2 was 4/2=2.
Later, we will learn that this dy/dx, the derivative is an actual angle of geometry and that 4/2=2 represents angles. But that is later on.
Now, we are not finished with the derivative, for it is more complicated than just dy/dx computing.
Let us draw a straightline segment from the coordinate point (0,0) to (2,4) in the second picture, looking somewhat like this:
function y=x^2 showing vectors, u, v, u+v
//| 4
/ / |
/ / |
u+v / / v |
/ / |
/ /| |
/ / u| |
------------------------------------------------> x-axis
0 1 2 3
VECTORS
We keep it simple, a vector is just a straightline segment that has direction and magnitude. So the straightline segment of the right triangle hypotenuse in cell from 0 to 1 is a vector and let us call it u. The line segment, a hypotenuse of a right triangle atop a picketfence in cell from 1 to 2 is another vector and call it v. The straightline segment from (0,0) to (2,4) is another vector and it is the addition of vectors u+v.
The vector, u+v is the derivative. And the three line segments of u, v, u+v form a triangle. Every adjoining cells have this u, v, u+v and this general triangle unless the function is a flat straightline function like y=3 or a diagonal straightline function like y=x. The u+v changes and varies in adjoining cells. For example the derivative of y=x^2 for x=2 involves the adjoining cells of 1 to 2 and then 2 to 3 so the dx is again 2 but the dy is now furthest leftward wall is 1 and furthest rightward wall is 9 so we have 9-1=8, so the derivative of y=x^2 at x=2 is now dy/dx = 8/2 = 4. When x=1 the derivative was 2, when x=2, the derivative is 4.
So, basically that is it on the derivative. We graph the function in Integer Grid. We partition the x-axis with width of 1 for each cell. We focus on a x-value as the middle of two adjoining cells and then compute the dy/dx.
There is more complexity to the derivative, but all of that can wait. Essentially, if you learned the above, you know what the derivative is, and it is rate of change, some call it velocity, and is dy/dx.
AP
Newsgroups: sci.math
Date: Sun, 16 Feb 2014 10:47:14 -0800 (PST)
Subject: How the Maxwell Equations derive the Pythagorean Theorem #1558
Correcting Math
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Sun, 16 Feb 2014 18:47:14 +0000
How the Maxwell Equations derive the Pythagorean Theorem #1558 Correcting Math
Alright, I discovered several years back that since all of mathematics is but a tiny subset of physics and that since all of physics is derived from the Maxwell Equations as an axiom set, that all of the theorems and axioms of mathematics comes out of the Maxwell Equations. In other words, the Maxwell Equations are a super-axiom-set.
Every axiom, every theorem in mathematics can be derived by the Maxwell Equations.
So how does the Maxwell Equations derive the Pythagorean theorem? Well, it is not as difficult as what may first appear. The derivation comes from the fact that the Maxwell Equations of the Faraday Law of a moving bar magnet into a stationary closed loop of wire causes a current to flow. That phenomenon is the very same phenomenon as a stationary bar magnet thrust through by a moving closed loop of wire causes a current to flow. In other words, as electrons flow in the wire, they could be from either a stationary bar magnet moving loop or a stationary loop with moving bar magnet. This is familiar to most in physics as special relativity.
Now, consider the proton as a square on the leg of a triangle and the electron as a different square on the leg of a triangle and the bar magnet as a square on the hypotenuse of a triangle. In order for the proton square plus electron square to equal the bar magnet square in Faraday's law, the triangle must be a right-triangle.
--- quoting *in parts* from http://www.berkeleyscience.com/relativity.htm
..
It followed from studies of Maxwell's equations. It reshaped the Newtonian physics we have studied. It explains magnetism - this is the coolest thing! ..
As for math, all you need is the Pythagorean theorem! No kidding.
..
Then both mirrors will have travelled to the right the distance vT, and distance the light will have travelled as observed by O, is cT, and, by the Pythagorean theorem (cT)2 = (vT)2 + L2 so, T2(c2 - v2) = L2 T2 = L2 / (c2 - v2) T2 = L2 / c2(1 - v2/c2)) T = L/c (1/sqrt(1 - v2/c2))
--- end quoting ---
--
AP
That was a 2014 post where I thought that the Pythagorean theorem at work and play in the Maxwell Equations was not all that obvious.
But as for a few days ago, here in 2015, I realized that whereever, and whenever Calculus is used in science, it is permeated by the Pythagorean theorem. So that when asked where is Pythagorean Theorem in the Maxwell Equations, the straightforward answer is that the Maxwell Equations are differential equations or Line-Integral equations, hence, Pythagorean theorem is all over the place.
Now here is a 2015 post of mine that explains the derivative and explains how the derivative is about right-triangles and the Pythagorean Theorem involved in the hypotenuse. So that whenever we have a science that uses the Calculus, that science has the Pythagorean Theorem.
What I am driving at, is the idea that all the axioms of Geometry can be reduced to the statement of the facts and details of a right-triangle and the same goes for the axioms of Numbers for Algebra. What this distills into, is the idea that the whole of mathematics is a tiny subset of the Maxwell Equations.
Newsgroups: sci.math
Date: Wed, 8 Apr 2015 19:56:28 -0700 (PDT)
Subject: Derivative; Introduction True Calculus #134 Correcting Math 4th ed;
(53) High-School & Uni-text 9th
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Thu, 09 Apr 2015 02:56:29 +0000
Derivative; Introduction True Calculus #134 Correcting Math 4th ed; (02) High-School & Uni-text 9th
Introduction to True Calculus, first 5 pages, page 2 Derivative
Alright, I am writing these 5 pages to expose the full spectrum of calculus from Coordinate System to function to derivative to integral all from just 2 pictures. The logic of this is that all are related and to see that relationship in full view is to use just 2 pictures. We end up seeing how the derivative relates to integral as inverse in the Fundamental Theorem of Calculus. We also see how function and the function-graph relates to derivative and integral.
The design of this text is to make it ultra easy for High School students to learn. Unlike books of the past, those authors, in most cases could not teach and their book ended up as silly contraptions. In order to teach, you have to be on the level of comprehension of the students you want to reach. This means, little to no terminology which often gets students lost. Most professors of science and math, in my opinion simply do not know how to teach and this text makes up for that.
On the first page we saw the 2 pictures and we did the Grid Coordinate System and the function and function-graph. Now we move onto the derivative.
Up front and out in the open the Derivative means rate-of-change. The definition is very simple in that of a division dy/dx, and what those symbols mean is the rate of change of y value (dy) divided by rate of change (dx) of x value.
So, in our 2 pictures we have:
THE 2 PICTURES:
The function y= 3 looks like this in Integer Grid
y-axis
^
|
-----------------------------------> y=3
| | | | |
| | | | |
| | | | |
------------------------------------------------> x-axis
0 1 2 3 4
Cell from 0 to 1 is a rectangle as well as cell from 1 to 2, with a flat top roof or ceiling, and same goes for all the other cells in y=3.
The function y=x^2 looks like this in Integer Grid:
y-axis
^
|
|
|
9 /| 9
|
|
|
/ |
|
|
/ |
|
/ |
|
4 4/ | |
| |
/ | |
| |
1 / |1 | |
/ | | |
------------------------------------------------> x-axis
0 1 2 3
Cell from 0 to 1 is a pure right-triangle sides 1 by 1 with hypotenuse sqrt2
Cell from 1 to 2 is a right-angled-trapezoid (picketfence) and the picketfence is composed of a square with a right-triangle atop the square
Cell from 2 to 3 is a right-angled-trapezoid (picketfence) and the picketfence is composed of a rectangle with a right-triangle atop the rectangle
DERIVATIVE
We define derivative as rate of change and write it as dy/dx across two adjoining cells. That is important for in most cases you need two adjoining cells, so we say, in general 2 adjoining cells are required to have a derivative. Notice in the first function above of y=3 as a flat straightline parallel to the x-axis. What is the dy? Well, the y-value always is 3. So the dy is 3-3 =0. And the dx of two adjoining cells is always 2 since the width of a cell is 1 and thus 1+1=2 in the Integer Grid. So in the function y=3 we always have dy/dx as being 0/2 which is 0.
Now let us examine the second function of y= x^2 across two adjoining cells. The first derivative we can compute would be for x=1 because we need two adjoining cells, of that of cell from 0 to 1 and then from 1 to 2 for x. Notice, we hold x=1 in the middle of these two cells, and that is important for to obtain the rate-of-change, we have the x value as the middle of the two adjoining cells. So what is the dx of these two cells? It is of course 1 +1 = 2 since the width of one cell is 1. Now, we ask what is the dy of these two adjoining cells. And we have to take the y-value of the furthest left wall which is y=0 and the y value of the furthest right wall in second cell which is y=4. So we have a dy as that of 4-0 = 4. Notice that x=1 is directly in the middle of the leftward wall and the furthest rightward wall. So the derivative of x=1 for function y=x^2 is that of dy/dx which is 4/2 = 2. The rate of change at x=1 of y=x^2 was 4/2=2.
Later, we will learn that this dy/dx, the derivative is an actual angle of geometry and that 4/2=2 represents angles. But that is later on.
Now, we are not finished with the derivative, for it is more complicated than just dy/dx computing.
Let us draw a straightline segment from the coordinate point (0,0) to (2,4) in the second picture, looking somewhat like this:
function y=x^2 showing vectors, u, v, u+v
//| 4
/ / |
/ / |
u+v / / v |
/ / |
/ /| |
/ / u| |
------------------------------------------------> x-axis
0 1 2 3
VECTORS
We keep it simple, a vector is just a straightline segment that has direction and magnitude. So the straightline segment of the right triangle hypotenuse in cell from 0 to 1 is a vector and let us call it u. The line segment, a hypotenuse of a right triangle atop a picketfence in cell from 1 to 2 is another vector and call it v. The straightline segment from (0,0) to (2,4) is another vector and it is the addition of vectors u+v.
The vector, u+v is the derivative. And the three line segments of u, v, u+v form a triangle. Every adjoining cells have this u, v, u+v and this general triangle unless the function is a flat straightline function like y=3 or a diagonal straightline function like y=x. The u+v changes and varies in adjoining cells. For example the derivative of y=x^2 for x=2 involves the adjoining cells of 1 to 2 and then 2 to 3 so the dx is again 2 but the dy is now furthest leftward wall is 1 and furthest rightward wall is 9 so we have 9-1=8, so the derivative of y=x^2 at x=2 is now dy/dx = 8/2 = 4. When x=1 the derivative was 2, when x=2, the derivative is 4.
So, basically that is it on the derivative. We graph the function in Integer Grid. We partition the x-axis with width of 1 for each cell. We focus on a x-value as the middle of two adjoining cells and then compute the dy/dx.
There is more complexity to the derivative, but all of that can wait. Essentially, if you learned the above, you know what the derivative is, and it is rate of change, some call it velocity, and is dy/dx.
AP
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