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September 26th, 2015, 12:35 PM   #81
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Quote:
 Originally Posted by v8archie Actually, in the most simple form of a Riemann sum, the $\displaystyle x$ interval (or step) is constant and the $\displaystyle y=f(x)$ is variable. But we can build a Riemann sum where the $\displaystyle y$ interval is constant and the $\displaystyle x=f^{-1}(y)$ interval is variable, and this is very closely related to the sum I think you are looking for. In essence I am saying that it doesn't matter which (if either) are constant. In practice, for Riemann sums we pick whichever step regime makes the calculations easier.
Thanks lot, for sure I've to cooperate.

You are the first, in 7 years, that put time and mind to understand my crazy math.

I know just little think about RH integration so I kindly ask you if is clear enought that all this can be applied to the Powers and as I've written in the other thread I open, to some polynomials (or all, I don't know now) because both can use the binomial develope or the telescopic sum property that is:

in this curves the primitive can be squared with a sum of BARS, of eight can be described with a function comes from the binomial develope, I call the Integer Primitive, or the from the modified one I call Rational Primitive or, at the limit, from the known Primitive that is just the first term that rest alive from what happen in Q once passing to the limit.

It means that the roof of the Bars, don't care if we are in N,Q or R always cut his primitive, leaving at his left the same missing area of the one exceeds at his right.

I don't know if there are other curves or else in some "space" that has this property ...

Definition is not my job, but I build all that to escape the long known way of abstract algebra, group theory etc... that already explain all that in a soo complicated manner that at the end little mind as mine, forgot what and where is the beginning...

For sure there are soo big mind that will see all that as "trivial" and unlikely because is an attack to the magnificent abstract building they built...

But find a way to figure very complex math by simple formula or simple drawing (if possible) rest (for me) the most important way to teach it to the fresh mind of childs, to set them free to discover more....

Well, after this boring boiling soup, that can be my speech at the ignobel prize, I'm here to answer to your questions.

I was here for long time in this days thanks to the flu... but from tomorrow I'll be again busy, so answers will be slowly and probably again (more) messy ....

Thanks lot,
Ciao
Stefano

p.s. I've a long .PDF /latex on all that with picture and more, but is far from presentability...

Last edited by complicatemodulus; September 26th, 2015 at 12:40 PM. September 27th, 2015, 10:18 PM #82 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Here a picture that show my Riemann modified integration method. Using the same Riemann's concept is possible to square a curve (here the primitive of ALL the functions of tipe y=x^n n>=2) in 3 ways: 1) Useing just integers: how the Integer Primitive y'=3x^2-3x+1 perfectly square the Primitive y=3x^2. 2) Useing Rationals: The same happen if we use the Rational Primitive: y'=3x^2/K-3x/K^2+1/K^3 with x1= 1/K, 2/K ... till our upper limit that can be a Rational that again perfectly square the Primitive y=3x^2. That is what I call my "Step Sum" method where I'm also able to show that when $\displaystyle K\to\infty$ we come back to known Riemann Integral /primitive: (Infimus) Primitive y=3x^2.  September 27th, 2015, 10:21 PM #83 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 All that is just the begnning of the way to understand why Fermat & Beal are both right... September 29th, 2015, 07:01 AM #84 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 I accept in my step sum: $\displaystyle K\to\infty$ 1/K = 0 While when we have a variable over K we accept it is: $\displaystyle x/K = x dx$ So we have to accept that each time we have a variable over soethink that goes to $\displaystyle \infty$ we have to consider not zero, so also if we have a variable or else in the form: $\displaystyle x/K^n$ also with n>2 we assume: $\displaystyle x/K^n <>0$ This is because in R we accept what we know as "continuity". So for example we have to accept that on the line of R numbers we can have an irrational "distance" like: $\displaystyle \infty^2 * A$ that can be SCALED to A, since from 0 to A whe can immagine the same number of infimus "dx"... there is in the apprently bigger "distance" All that (sorry you can find more beautifull words) allow me to explain why my Step Sum that works in N, and in Q works also in R with what we can imagine is a "distance" with an infinite number of digits, so for example if we go at the limit, with an upper limit with an infinite number of digits... So still if right hand is exactly equal to right hand terms using the right hands and changeing the variable from X to X= x K we have: $\displaystyle A^2 = \sum_{X=1}^{KA} 2X/K^2 -1/K^2 = \sum_{x=1/K}^{A} 2x/K -1/K^2$ and this rest true for any K with $\displaystyle K\in N$ if we put my definition on the left hand terms we rest stucked and we can say nothing when $\displaystyle K\to \infty$ while in the right hand one, (accepting what said) we can write and understand: $\displaystyle A^2 = \lim_{K\to\infty} \sum_{x=1/K}^{A} 2x/K -1/K^2 = \int_{0}^{A} 2x dx$ and our motto is that we can prove $\displaystyle 1 / \infty^2 = 0$ while we can't say nothing of SURE about $\displaystyle 2x / \infty$ so we consider it as $\displaystyle 2x dx$ Note that in this way the initial property of power rest true for $\displaystyle K\to\infty$ The integral is, as known good for any A, so also for: A = Rational with infinite digit, or A = Irrational ...hope some will understand.... and translate in math.... Last edited by complicatemodulus; September 29th, 2015 at 07:04 AM. September 30th, 2015, 12:42 AM #85 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Just little note: In case we have $\displaystyle K\to\infty$ and terms like: $\displaystyle (x/K)^n$ also with n>2 we assume: $\displaystyle (x/K)^n <>0$ This is because in R we accept what we know as "continuity". And is of course known that in case we have more than one term we can vanish the bigger order terms so in: $\displaystyle M3K= 3x^2/K - 3x/K^2 +1/K^3$ We can forgot: - 3x/K^3 +1/K^3 and put : 1/K = dx All that still if we well know thar our x is: x = X/K (with $\displaystyle K=\infty$). This is because we have to let the variable has a chance to be of the same (or bigger) order of $\displaystyle \infty$... Hope now it can be acceptable... numbers match (till possible)... Thanks Ciao Stefano September 30th, 2015, 02:16 AM #86 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 I'm rewriting all with known notation where the INDEX is NOT connected to the ABSCISSA x... This, I hope, will let think much clear... But at the end the problem will rest: my notation (that will need another symbol) let several think more "evident" and any kind of trick can be done on the "parameter" so both lower and upper limit, jump of the variable, that can follow now another variable so let it be also PARTIAL SUM where just some of the possible terms in the Step SUm can be picked-up, ...etc... September 30th, 2015, 08:32 AM #87 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 In a known notation where the Index DO NOT COINCIDE with the Abscissa, but is just the position of the columns: 1st, 2th etc... We wanna prove that (simple example for n=2): $\displaystyle A^2 = \sum_{p=1}^{A/p} 2p/K^2 -1/K^2 = \lim_{K\to\infty}\sum_{p=1}^{A/p} 2p/K^2 -1/K^2$ We start from: $\displaystyle A^2 = \sum_{p=1}^{A} 2p -1 = \sum_{p=1}^{KA} 2p/K^2 -1/K^2$ Now we push $\displaystyle K\to\infty$ $\displaystyle A^2 = \lim_{K\to\infty}\sum_{p=1}^{KA} 2p/K^2 -1/K^2$ Since both the limit of the term exists, the limit of the difference is equal to difference of the limits or: $\displaystyle A^2 = \lim_{K\to\infty}\sum_{p=1}^{KA} 2p/K^2 -\lim_{K\to\infty}\sum_{p=1}^{KA} 1/K^2$ or changing the variable putting K= 1/h: $\displaystyle A^2 = \lim_{h\to\ 0}\sum_{1}^{A/h} 2p h^2 = \int_{0}^{A} 2p * dp$ What is not explicit in this notation is the connection between the abscissa x and his height f(x), that is what I try to make with my new notation I call StepSum... In the known notation the reason why the index pass from 1 in the sum to 0 in the inthegral is not "immediately" shown as when we put in my Step Sum the lower limit equal to 1/K and we rise $\displaystyle K\to\infty$ So now is clear that the Sum works till: $\displaystyle KA=A/h \in N$ and for any n-th power since the general term is Mn= (p-(p-1)^n), that must be divided by K^n I hope this time is clear the definition, I let you work on... Thanks Ciao Stefano Last edited by complicatemodulus; September 30th, 2015 at 08:36 AM. October 2nd, 2015, 12:41 AM #88 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 I forgot to write in math here the property of Powers of integers (and not just). - Be f'(X^n) the primitive of X^n - Be the Gnomons of what I call Complicate Modulus Mn = (X^n-(X-1)^n) Always exist an abscissa x_{e}= (X-d) that satisfy: $\displaystyle f'(x^n)|_{X=x_{e}} = (X^n-(X-1)^n)$ (read f'(X) in X=xe) for n=2 is easy to prove that d is fixed and is: d=1/2 $\displaystyle f'(x^2)|_{X-d} = (X^2-(X-1)^2)$ $\displaystyle f'((X-d)^2) = 2X-1$ $\displaystyle 2X -2d = 2X-1$ $\displaystyle d = 1/2$ For n=3 we can easy computate d that is a function of X with clear property: $\displaystyle d \not\in N$ The same rising n (odds) while for n=even is weel known why Fermat is right. Sorry no time today to develope... Thanks Ciao Stefano Pls just see the picture discarding the note that are old and wrongs: Last edited by complicatemodulus; October 2nd, 2015 at 12:46 AM. October 2nd, 2015, 02:45 AM #89 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 So thinking to a balance of the momentum given by C^3-B^3 and the one given by A^3 (that can be done also in 1700) is that - In eachxe can be reduced the area of one BARR of "weight" (3X^2-3X+1) - Each Area reduced to its center of gravity, multiplied by it's distance xe Must be equal to the complete area of the primitive (of A^3 in this example) multiplied by the distance of its "center of gravity" from the fulcrum. Now the property of the powers (n>2) is that still if EACH Xe is irrational the sum of the single momentum become an integer ODD (or half an Odd)... So what FLT said is that this equilibrium (property) vanish is we remove (from the ones more close to the origin one) or more elements (so Barr Area reduced to its C.G. multiplied for the distance from the fulcrum)... Thanks Ciao Stefano October 5th, 2015, 09:48 PM #90 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Forgot my last post... I will send picture and somethink of more similar to math asap... Tags beal, flt Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post complicatemodulus Number Theory 0 June 10th, 2015 05:30 AM lwgula Number Theory 18 November 4th, 2014 10:26 AM bigbuck Number Theory 20 August 2nd, 2014 05:01 AM akhil verma Number Theory 29 July 21st, 2013 04:51 AM johnr Number Theory 8 April 1st, 2013 10:38 PM

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