novaBBS - sci.math - UCLA Dr Tao & Kibo Parry Moroney fail at never able to do a geometry proof of Fundamental Theorem of Calculus. So why should Dr. Tao win any math award when he fails at calculus?

We know Kibo Parry Moroney fails at math for that bozo still thinks 938 is 12% short of 945. We also know that Terry Tao can do percentages correctly, but what is worrying is that Dr. Tao was never able to give a geometry proof of Fundamental Theorem of Calculus, and Terry knows well that calculus is geometry. And so Terry is satisfied as the other loose marbled mathematicians of a "limit analysis" for FTC. But that is failure math. We need Dr. Tao to actually do a geometry proof of FTC.

11th published book

World's First Geometry Proof of Fundamental Theorem of Calculus// Math proof series, book 2 Kindle Edition
by Archimedes Plutonium (Author)

Last revision was 15May2021. This is AP's 11th published book of science.
Preface:
Calculus needs a geometry proof of Fundamental Theorem of Calculus. But none could ever be obtained in Old Math so long as they had a huge mass of mistakes, errors, fakes and con-artist trickery such as the "limit analysis". To give a Geometry Proof of Fundamental Theorem of Calculus requires math be cleaned-up and cleaned-out of most of math's mistakes and errors. So in a sense, a Geometry FTC proof is a exercise in Consistency of all of Mathematics. In order to prove a FTC geometry proof, requires throwing out the error filled mess of Old Math. Can the Reals be the true numbers of mathematics if the Reals cannot deliver a Geometry proof of FTC? Can the functions that are not polynomial functions allow us to give a Geometry proof of FTC? Can a Coordinate System in 2D have 4 quadrants and still give a Geometry proof of FTC? Can a equation of mathematics with a number that is _not a positive decimal Grid Number_ all alone on the right side of the equation, at all times, allow us to give a Geometry proof of the FTC?

Cover Picture: Is my hand written, one page geometry proof of the Fundamental Theorem of Calculus, the world's first geometry proof of FTC, 2013-2015, by AP.

Length: 85 pages

Product details
ASIN : B07PQTNHMY
Publication date : March 14, 2019
Language : English
File size : 1272 KB
Text-to-Speech : Enabled
Screen Reader : Supported
Enhanced typesetting : Enabled
X-Ray : Not Enabled
Word Wise : Not Enabled
Print length : 85 pages
Lending : Enabled
Amazon Best Sellers Rank: #128,729 Paid in Kindle Store (See Top 100 Paid in Kindle Store)
#2 in 45-Minute Science & Math Short Reads
#134 in Calculus (Books)
#20 in Calculus (Kindle Store)

Archimedes Plutonium<plutonium.archimedes@gmail.com>
May 15, 2021, 11:05 AM
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Alright, let us prove the two Power Rules derivative Y= x^n is n(x^(n-1)) and integral (1/(n+1))x^(n+1), using the AP Geometry proof of Fundamental Theorem of Calculus.

Reiterating what I said last night-- that the Cell of the function graph proves the two power rules on polynomials. I need not make a elaborate argument of splitting a polynomial into segments such as x^4 is x^3*(x). Instead, the cell mechanically delivers the Power Rules.

STATEMENT: The two power rules of Calculus on Polynomials are proven true from any arbitrary Cell of a particular Decimal Grid System. Power Rules derivative Y= x^n is n(x^(n-1)) and integral (1/(n+1))*(x^(n+1)).

Proof: It is the arbitrary picked cell that gives us a proof in general. I am not restricted to function graphs of upward slopes like / but can also handle downward slopes such as \ as for example an upward slope is Y= 2x + 3 and a downward slope in 10 Grid is Y = 10 - 2x over interval 0 to 5. The proof handles both upward and downward slopes where the left wall of the rectangle integral does upward slopes and the right wall of rectangle integral does downward slopes.

My arbitrary pick is 10 Grid of function Y= x^4 in Cell 2.0 to 2.1. Remember in true math, there is only empty space from 2.0 to 2.1 in 10 Grid. But we need a midpoint so we borrow from 100 Grid and obtain 2.05 as midpoint of cell.

__A__C
| |
|B |
| |
---------
2.0 2.1

B
/|
/ |
A /----| C
/ |
| |
|____|
2.0 2.1

Now we compute 2^4 and 2.1^4 for the function graph where B= 2.1^4.

And we compute A, the midpoint as 2.05^4.

2^4 = 16
2.1^4 = 19.4481
2.05^4 = 17.6610

The derivative is dy/dx and of x^4 is 4x^3.
The integral is rectangle area under function graph x^4 and is (1/5)x^5.

In the proof I must show that a line segment from (2, 16) to (2.1, 19.4481) is the derivative 4x^3, and must show that the rectangle area of .1 width by 17.6610 height comes from (1/5)x^5.

The length AC is 0.05, and what is the length CB?

The perpendicular length from x-axis to B is 19.4481.
The perpendicular length from x-axis to midpoint A is 17.6610.

Our function graph in Cell 2.0 to 2.1 looks like this.

B
/|
R , |
| |
| |
| |
|____|
2.0 2.1

R= 16
B= 19.4481

At this moment in the proof, we are troubled in seeing whether the line segment from R to B is going to pass through the midpoint A of the cell.

To be continued.

Archimedes Plutonium<plutonium.archimedes@gmail.com>
May 15, 2021, 11:40 AM
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On Saturday, May 15, 2021 at 11:05:08 AM UTC-5, Archimedes Plutonium wrote in sci.math:
> Alright, let us prove the two Power Rules derivative Y= x^n is n(x^(n-1)) and integral (1/(n+1))x^(n+1), using the AP Geometry proof of Fundamental Theorem of Calculus.

In this proof, I am also going to check whether Apostol's Calculus of starting with integral rather than derivative is best logic wise.

So I have the cell from 2.0 to 2.1 and have the midpoint of 2.05 whose lengths on y-axis are these:

16
17.6610
19.4481

Now, let me just take the integral from Power Rule and see if it comes close to being the rectangle of width 0.1 and length 17.6610.

The integral is (1/5)x^5
(1/5)(2.1)^5 subtract (1/5)*(2.0)^5 = (1/5)(40.84101) subtract (1/5)(32) = 8.168202 - 6.4 = 1.768202

Now, comparing the midpoint rectangle of width 0.1 and length 17.6610 and multiply for area would be 1.766. I would say that we have equality match, for the Sigma Error is so small. 1.768/1.766 = 0.1% error. So yes, our power rule integral matches the midpoint rectangle area.

Now we need to see if the power rule derivative is a right triangle carved out of the left wall of the midpoint rectangle and when lift up at the hinge of the midpoint will span across so that the vertex of the right triangle coincides with the point (2.1, 19.4481).

I have a feeling of many Pythagorean theorem has to be applied here.

To be continued.

Archimedes Plutonium<plutonium.archimedes@gmail.com>
May 15, 2021, 12:27 PM
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Derivative Power Rule proof

Alright, we have proven the Integral Power Rule, and now we need to prove the Derivative Power Rule from the Cell. Our arbitrary picked cell was 2.0 to 2.1 for the arbitrary picked function Y=x^4. Remember in New Math all functions are polynomials and only polynomials. If you bring me a trig function, it is not a function until you transformed it into a polynomial over a given interval.

So now we prove the AP geometry proof of FTC delivers the Derivative Power Rule of n*(x^(n-1)) given the function x^n. Basically the Power Rules are subtract 1 from exponent for the derivative and add 1 to exponent for integral as we ignore the prefix factor.

And essentially the adding of 1 to exponent in a cell is the procedure of making a square from starting line segment and the prefix factor is carving out a rectangle area from a square. So it looks as though Apostol's choice of starting with a integral vice derivative was a wise choice, a logical choice.

Now we do the derivative, and I am almost certain it is a application of the Pythagorean theorem. But what I was hesitant about when starting this proof, is how much the midpoint of the cell from 2.0 to 2.1, how that midpoint is always on the derivative slope line and is part of the 4x^3 derivative of Y=x^4. So let us do the derivative proof and see how that midpoint is always on the derivative line slope.

We have the y lengths of
16
17.6610 midpoint
19.4481

/|
/ |
| |
| |
| |

We want the lengths of the right triangle that is folded down and forms the rectangle at the midpoint 17.6610 in length and 0.1 in width.

So we must have one leg be that of length 0.05 and the other leg must be 19..4481 subtract 17.6610 = 1.7871 length.

Now looking at what the Power Rule derivative gives as 4x^3 and we use the midpoint x of 2.05, where 4(2.05^3) = 34.4605. And according to our right-triangle we have tangent of 1.7871/0.05 = 35.742. And we view this derivative number as a angle of trigonometry. And here is the tangent function of trigonometry as 88.5degree with complament of 1.5 degree.

Is it close enough in Sigma Error for agreement? 35.74/34.46 = 3.7%. A bit high. And makes me wonder if the midpoint has a role in that error.

So I did not have to apply the Pythagorean theorem for it was applied in the tangent function of trigonometry.

Archimedes Plutonium<plutonium.archimedes@gmail.com>
May 15, 2021, 12:47 PM
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The accuracy of the Integral Power Rule and the area of the rectangle at midpoint as being sigma error of 0.1% cannot be ignored. Having said that, the Derivative Power Rule proof is accepted. And thus we conclude the derivative slope line segment always passes through the midpoint of the Cell.

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Subject	Author
UCLA Dr Tao & Kibo Parry Moroney fail at never able to do a geometry proof of Fu	Archimedes Plutonium
Re: Archimedes "ibozo" Plutonium flunked the math test of a	Michael Moroney

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tech / sci.math / UCLA Dr Tao & Kibo Parry Moroney fail at never able to do a geometry proof of Fundamental Theorem of Calculus. So why should Dr. Tao win any math award when he fails at calculus?