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November 26th, 2016, 08:31 PM   #21
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Quote:
Originally Posted by complicatemodulus View Post
$\displaystyle A^2= \sum_{X=1}^{A}(2X-1)$
Granted

Quote:
Originally Posted by complicatemodulus View Post
It's A WAY TO SQUARE THE DERIVATE $y'=2X$, till A ?
What does this mean? As JeffM1 says, we have acquired a function $y$ which appears to be $y=x^2$, but what does "square the [derivative] until A" mean? From the diagram, you appear to mean that you want to make a square having the same area as the area underneath $f(x)=2x$ between $0$ and $A$. That is, $\int_0^A 2x \, \mathrm dx$.

Quote:
Originally Posted by complicatemodulus View Post
With rectangular integer columns: $M_{2,X} = 1 * (2X-1)$
So, in your diagram, you have $Mnx$ is there any significance in the $x$ here being in lower case, but in upper case $X$ in the formula? You state in the diagram $Mnx = X^n - (X-1)^n$ which agrees with $M_{2,X} = 1 * (2X-1)$, but why do we have a "1*" here? Is that supposed to represent the width of the strip? Also, what does that have to do with the green areas indicated in the diagram? The best guess I have here is that you are forming a Riemann sum here with 5 ordinals forming the right hand side of rectangular strip. The talk about $y'=2x$ and the line drawn in the second diagram suggests that we are taking the centre of each rectangle as our ordinal and so $$R=\sum_{k=1}^A 2\left(k-\tfrac12\right)$$
The summand here is obviously equal to the $2k-1$ you are looking for, but it will not always be so, and that is likely to cause us some problems later.
Quote:
Originally Posted by complicatemodulus View Post
Now it's clear that the upper integer limit A, it's just a quesion of the SCALE WE CHOOSE ?
I think that here you are just saying that $A$ is arbitrary. You seem to want to limit it to rational numbers though, although there is no obvious reason why this should be. I have a feeling that it's because you insist on have slices of unit thickness, but then later you are going to make them aribitrarily thin anyway. And again we will hit trouble.

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Originally Posted by complicatemodulus View Post
Than we have just to see that when we multiply THE NUMBER OF INDEX by $10^m$, we obtain as result an area $10^{2m}$ bigger (in case of a square), in general $10^{n*m}$ bigger.
The underlined part may well cause problems because of the impending confusion over what function we are dealing with.

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Originally Posted by complicatemodulus View Post
Staring from:

$\displaystyle A^n = \sum_{X=1}^{A}{[ X^n-(X-1)^n]} $

We keep as example the case n=3,
Here comes the confusion: here, you want to talk about the function $y=x^3$ or $y'=3x^2$.

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Originally Posted by complicatemodulus View Post
the term $M_n= {[ X^n-(X-1)^n]}$ become: $\displaystyle M_3= {( 3X^2-3X+1 )} $
And here you want to talk about $3x^2-3x+1$. It's far from clear that we can put all this right as we did before by talking about using a different ordinal in our Riemann sum.

As it is, the area beneath the derivative of $x^3$ between zero and $a$ is $a^3$, but the area beneath $3x^2-3x+1$ is $a^3-\tfrac32a^2+a$. And if we multiply $a$ by some factor $k$, we can't just divide that area by $k^3$ to get back to an answer of $a^3$.

I think I'd need to see your diagram for $n=3$ to have some idea of what is happening here. And it's rather important I think, because you have previously claimed that this all turns into a Riemann sum.

Since you claim that this is all about areas and what happens on the plane, you have to be clear about what area you are talking about on the plane. I think you've been making some unjustified generalisations.
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November 26th, 2016, 11:12 PM   #22
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Quote:
Originally Posted by v8archie View Post
Granted


What does this mean? As JeffM1 says, we have acquired a function $y$ which appears to be $y=x^2$, but what does "square the [derivative] until A" mean? From the diagram, you appear to mean that you want to make a square having the same area as the area underneath $f(x)=2x$ between $0$ and $A$. That is, $\int_0^A 2x \, \mathrm dx$.


So, in your diagram, you have $Mnx$ is there any significance in the $x$ here being in lower case, but in upper case $X$ in the formula? You state in the diagram $Mnx = X^n - (X-1)^n$ which agrees with $M_{2,X} = 1 * (2X-1)$, but why do we have a "1*" here? Is that supposed to represent the width of the strip? Also, what does that have to do with the green areas indicated in the diagram? The best guess I have here is that you are forming a Riemann sum here with 5 ordinals forming the right hand side of rectangular strip. The talk about $y'=2x$ and the line drawn in the second diagram suggests that we are taking the centre of each rectangle as our ordinal and so $$R=\sum_{k=1}^A 2\left(k-\tfrac12\right)$$
The summand here is obviously equal to the $2k-1$ you are looking for, but it will not always be so, and that is likely to cause us some problems later.

I think that here you are just saying that $A$ is arbitrary. You seem to want to limit it to rational numbers though, although there is no obvious reason why this should be. I have a feeling that it's because you insist on have slices of unit thickness, but then later you are going to make them aribitrarily thin anyway. And again we will hit trouble.


The underlined part may well cause problems because of the impending confusion over what function we are dealing with.


Here comes the confusion: here, you want to talk about the function $y=x^3$ or $y'=3x^2$.


And here you want to talk about $3x^2-3x+1$. It's far from clear that we can put all this right as we did before by talking about using a different ordinal in our Riemann sum.

As it is, the area beneath the derivative of $x^3$ between zero and $a$ is $a^3$, but the area beneath $3x^2-3x+1$ is $a^3-\tfrac32a^2+a$. And if we multiply $a$ by some factor $k$, we can't just divide that area by $k^3$ to get back to an answer of $a^3$.

I think I'd need to see your diagram for $n=3$ to have some idea of what is happening here. And it's rather important I think, because you have previously claimed that this all turns into a Riemann sum.

Since you claim that this is all about areas and what happens on the plane, you have to be clear about what area you are talking about on the plane. I think you've been making some unjustified generalisations.
Thanks, but I'm stucked from your answers.

This let me understand that I'm 8 light years ahead here, and I've to come back to the beginning to show what I discover, and probably my mech mind find very easy, but it is not, also for thoose as you are very familiar with high level of math (I'm not able to rise for sure!).

Ok, my fault. From the very beginning:

Yes, right I've found a very interesting property of Power curves $Y=X^n$

Any of their derivate $y'=nx^{(n-1)}$ (and also all the followings) can be SQUARED in the integers, or in the Rationals too

Via Sum of Gnomons (grey rectangular strips):



The case n=5 follows in the same way:



(sorry in blue labels there are several erros: $M_3$ insthead of $M_5$, since I left there what it's true for the n=3 case... I've no time to fix now all that graphical problems...)

I think this was very simple to be understood by mathematician, but now from your answer it's clearly not.

The property I discover is that for such derivates, between two integers (or 2 rationals as follows) the exceding area fo the gnomons respect to the derivate it's always equal to the missing one, respect to the point in X where:

Derivate height = Gnomon heigh

I know you think here there is the loop let my proof fail, but this it's not true.

So if you are pushing me to prove this step by step, pls no, it's very easy and this is not the place.

I will write a book for all those are not able to arrive themself till here,

but now I claimed I've solved 2 of the biggest problems math has till now.

I cannot return each time from this beginning.

V8 I'm sure you can arrive there yourself extending this from n=2 to n=3 to all n using Newton's develope and induction.

Are able to do that ? If yes we can go on, else, sorry I was in the wrong place, and I understand I'm in the wrong planet too...

I'm now very close to Riemann problem, still if I want stay far from it... what I'm discovering it's like an avalance falling itself in that direction...

Ok, we can for so start again giving for good all my sums in the integers so the general telescopic sum:

$\displaystyle A^n=\sum_{1}^{A} M_n$

?

I'm sure you are not trying to lose my time to let you (other) have time to publish on Arxiv or elsewhere.

Pls don't adbandone me since your is the only voice it's answering me.

Thanks
Ciao
Stefano

p.s. Yes, Lower/Uppercase "x" / "X" it's the core of the question...

Last edited by complicatemodulus; November 26th, 2016 at 11:26 PM.
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November 27th, 2016, 12:06 AM   #23
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On your $x^5$ picture, how do you prove that $A_i=A_{i+1}$? Because I'd be surprised if it were (assuming that the blue vertical line bisects the strip).
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November 27th, 2016, 12:38 AM   #24
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Quote:
Originally Posted by v8archie View Post
On your $x^5$ picture, how do you prove that $A_i=A_{i+1}$? Because I'd be surprised if it were (assuming that the blue vertical line bisects the strip).
Ok, it's clear you're pushing me to prove... I did it when it's simple like now, I'll skip in other case...

For n=2 JUST, it bisect !

For all the other $n>2$ it's clear that if you wanna computate what I start to call $X_m$ (after I made this picture several years ago, so sorry you'll not find $X_m$ on that picture), it's necessary to computate the intersection between the Derivate height, and the Gnomon's heigh $M_n$

So $X_m$ (Local Balancing Point) it's the abscissa where for such integer $X = Xi$ :

$\displaystyle Y'|_{Xm} = M_n|_{Xi} $

so:

$\displaystyle n X_m ^{(n-1)} = {n \choose 1}X^{n-1} + {n \choose 2}X^{n-2} + {n \choose 3}X^{n-3} +... +/- 1 $

You can also have that making the integral of the two areas, than finding where equal.

BUT THIS IT'S NOT NECESSARY TO PROVE FLT / BEAL ! Or we was in a clear loop...

p.s. Regardin one of your question in the previous post it's clear that talking of $Y=X^n$ it's the same (when correctly applied) to talk of the area bellow $Y'= n X ^{(n-1)}$ and this is the trick on what Fermat's probably play, and for sure on what I'm playing from 8 years...

Last edited by complicatemodulus; November 27th, 2016 at 12:51 AM.
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November 27th, 2016, 01:29 AM   #25
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hope more clear now:


Last edited by complicatemodulus; November 27th, 2016 at 01:31 AM.
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November 27th, 2016, 07:24 AM   #26
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So now you have a huge whole in your work that you have dimply decided not to bother proving. If you are going to rely on $A_i=A_{i+1}$, you need to show what effect that has on the ordinal in the Riemann sum. This is particularly necessary as you are now saying (I think) that the height of that ordinal is governed by the function $3x^2-3x+1$. This suggests that the integral you are looking to form is no longer that of the derivative of $x^3$ and thus the equality with $a^3$ won't hold.
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November 27th, 2016, 08:02 AM   #27
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Quote:
Originally Posted by v8archie View Post
So now you have a huge whole in your work that you have dimply decided not to bother proving. If you are going to rely on $A_i=A_{i+1}$, you need to show what effect that has on the ordinal in the Riemann sum. This is particularly necessary as you are now saying (I think) that the height of that ordinal is governed by the function $3x^2-3x+1$. This suggests that the integral you are looking to form is no longer that of the derivative of $x^3$ and thus the equality with $a^3$ won't hold.
What ???

I think you're trying to put me in a corner...

You are again jumped to the wrong conclusion I'm in wrong...

Be sure, not now.

First, I know what is the problem and how to solve it, but I have some problem with english and math too, so pls you have to make a table where I show you how and what works how... and you translate it in your Math english... after you try to delete as possible non usefull definitions.

Ok ?

Back on point:

- It's clear that Fermat's problem and $A_{i+} = A_{i-}$ are very similar problems ?

If you wanna understand how I've solved:

$\displaystyle C^n = 2 \sum_{1}^{A}M_n + Delta $

and

$\displaystyle C^n = 2 \sum_{1}^{B}M_n - Delta $

with


$\displaystyle Delta = \sum_{A+1}^{B}M_n $


$\displaystyle \implies C\in (R-Q) $ if $n>2$

And, problem number 2:

$A_{i+} = A_{i-} \implies X_m \in (R-Q)$ for any $n>2$

You've to understand what I discover.... and the process to let it works from Integers, than in the Rationals, than to the limit.

You accept the generalized Integer Sum ?

If yes, You're not alone, and we can go on...

I'm quite sure I'm not wasting our time since I've found a ex-mathematician here in Italy that already follow me till the integral, but I lost him once was time to put all together.

To answer you: both conditions are proved by the limit, so what it's realy an infinte descent, never stop.

So I've proved that $X_m \in (R-Q)$ since the only way to rise it it's to go infinity with a divisor I've called $K$


We have to enter in a field where the Induction let all proved, but where at the moment you can't ask me to say or go more deep, I cannot say you what I've in mind with the mountain of definition you study in several years in a field that it's not mine...

You've to follow me than all mathematician can be free to well study any implication of what I discover...

I think that there will be work for generations, and evreybody can be free to start from here to go over using his proper math language.

Sorry I need to see think at works.


Again on the point: I understand now you call Ordinal what I call Modulus.

I think it's not good to stick a new label to each think we discover... it's better to use what we know in the most general way possible.

Here the first problem of math: you call Modulus the Operator, while (I think the inventor, as me whant to say) the Modulus, in mechanic, is a "standardized part" that can be easily duplicated, using the same dimensions ( your broken Clock, with one hand), or also easy scaled (as in my 2 hands clock...) making thinks always equal, just bigger or smaller...

From here my definition "Complicate Modulus" means a way to call just apparently different object, we discover they have different dimensions, but that depends on the same function.

So we have a mesuring device capable to mesure Naturals in terms of an n-th Power, we decide, pus a rest.

1 : 1$M_2$ + 0
2 : 1$M_2$ + 1
3 : 1$M_2$ + 2
4 : 1$M_2$ + 0
5 : 2$M_2$ + 1
6 : 2$M_2$ + 2
7 : 2$M_2$ + 3

etc...

Or the same with any $M_n$ we choose.


You call this bijection... right ?

..and now ho to go to the "transfinite induction" ? Be sure I will have no time to go there with you with your crazy Abstract sings... I wanna go there following my clear graph and std induction... I hope...

Sorry I just discover I can say I add "integration" and "derivative process" to this ordinals...

It's clear we can find $M_{n+1}$ via integration from $M_n$ (+/-1 as constant term) ; or deriving we can go down from $M_n$ n$M_{n-1}$

... there is no limit to go over and over... Fermat and Beal are just an invisible cherry on the cake...

Ok, I stop editing... I've to left you for now.

Last edited by complicatemodulus; November 27th, 2016 at 08:49 AM.
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November 27th, 2016, 09:43 AM   #28
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I think you're trying to put me in a corner...
If you aren't interested in addressing my concerns, how do you expect me to change my mind. Do you think I want to spend 8 years reproducing all your work in English and standard mathematical notation?

Quote:
Originally Posted by complicatemodulus View Post
you have to make a table where I show you how and what works how... and you translate it in your Math english...
I don't have to do anything of the sort. I've never been good at listening to people making unproven claims, especially when they expect me to do all the hard work of providing a proof.

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Originally Posted by complicatemodulus View Post
I'm quite sure I'm not wasting our time since I've found a ex-mathematician here in Italy that already follow me till the integral,
So why isn't he writing it out for you?

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Originally Posted by complicatemodulus View Post
To answer you: both conditions are proved by the limit, so what it's realy an infinte descent, never stop.
That's not an answer and it certainly isn't a proof. Even if I thought you understood limits, I'd expect more detail in response to the query I've raised.

Quote:
Originally Posted by complicatemodulus View Post
So I've proved...
You haven't. You are waving your hands around trying to get me to buy into your belief. That's religion, not mathematics. I gave that up when I was 15 or 16 after an in-depth discussion with my local priest and Master of Divinity.
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November 27th, 2016, 12:20 PM   #29
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I really can't understand if you wanna joke with me or you/ others really not understand my method...

You've to do nothing, just translate in your language, if you wanna understand... of forgot it and use mine that also a joung student with a free mind can understand.

Unfortunately we know from history what happen when someone collect the necessary information from the work of others...

I won't be remembered as who made just stupid think or several stupid errors I did ... nor for who stole the work of others, nor the one who lived 8 years in a secret bunker to ask back for the million dollar prize...

I've addressed you a question: it's clear that using induction we can prove the Sum into the Integers for any $n$?

I can't believe you're not able to do that.

As we prove it for $M_n$ we can do the same for $M_n,K$ and/or for the $y'=nX^n$ formula.

If you can't follow me till there, I can't say you to follow me in the conclusion, you already said 2 times, it's for sure wrong.

I'm starting to think that Fermat understood that nobody can understand his proof, since too deep for that time, but now we have all well known...

Pls don't missunderstand my "low-profile" approach, with ignorance...

I won't play with you, simply jump what already written here 100 times...

Very sorry, but still here.

Thanks
Ciao
Stefano

Last edited by complicatemodulus; November 27th, 2016 at 12:25 PM.
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November 27th, 2016, 01:26 PM   #30
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I really can't understand if you wanna joke with me or you/ others really not understand my method...

You've to do nothing, just translate in your language, if you wanna understand... of forgot it and use mine that also a joung student with a free mind can understand.

Unfortunately we know from history what happen when someone collect the necessary information from the work of others...

I won't be remembered as who made just stupid think or several stupid errors I did ... nor for who stole the work of others, nor the one who lived 8 years in a secret bunker to ask back for the million dollar prize...

I've addressed you a question: it's clear that using induction we can prove the Sum into the Integers for any $n$?

I can't believe you're not able to do that.

As we prove it for $M_n$ we can do the same for $M_n,K$ and/or for the $y'=nX^n$ formula.

If you can't follow me till there, I can't say you to follow me in the conclusion, you already said 2 times, it's for sure wrong.

I'm starting to think that Fermat understood that nobody can understand his proof, since too deep for that time, but now we have all well known...

Pls don't missunderstand my "low-profile" approach, with ignorance...

I won't play with you, simply jump what already written here 100 times...

Very sorry, but still here.

Thanks
Ciao
Stefano
Please understand I'm being serious here and not trying to "tear you down."

Has it occurred to you that the reason why no one seems to be following you all that well is because you are wrong and there isn't an elementary proof of FLT? Note that Fermat claimed he had one but never went back to write it down. Best guess is that he didn't have one either. Examples of simple looking theorems that have highly non-trivial proofs abounds in Math. My favorite is Tychonoff theorem in Topology that says that the product of any collection of compact spaces is compact in the product topology. Once you get the idea of what a compact space is the theorem seems obvious. However I've been trying for years on my own and I still can't get the details of the proof in order. Sometimes there is no simple way out.

-Dan
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