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 August 5th, 2016, 08:42 AM #1 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Very Complicate Modulus Algebra I like to frake your mind with my Very Complicate Modulus Algebra: I't possible to define, in the "sense" of Lebesgue integral, so summing integer bars that cover the area of the derivate: $y'=x^n$ from left to right, and having known thickness: $HM_ny = ((Y_{i})^n-(Y_{(i-1)})^n)$ and Base: $BM_nx = (A-x)$ So it's possible to square the area: $\displaystyle A^n= \int_{0}^{A^n} ( y/n)^{(1/(n-1))} dy = \sum_{y \in I} M_{ny}$ where: $I$= { $(x^n-(x-1)^n)|_{(x=1)}, (x^n-(x-1)^n)|_{(x=2)}... , (x^n-(x-1)^n)|_{(x=A)}$} And $M_{ny}$ is the very complicate modulus: $M_{ny} = HM_{ny}* BM_{ny} = (y-(x-1)^n)*(A-x)$ so where $x=$ {1 for $y=I_1$, 2 for $y=I_2$, 3...., A for $y=I_A$ ) Some example: 3^3= 3 + 12 + 12 or 4^3= 4+18+24+18 or 5^3=5+24+36+36+24 or: 6^3=6+30+48+54+48+30 or 7^3 = 7 + 36 + 60 +72 +72 + 60 +36 So I've my Complicate Triangle too ! And my theorem that a square is: $\displaystyle A^2= A+ \sum_{p=1}^{A-1} 2p$ So the Sum of the Root, plus the Sum of the first (A-1) Even... that is the "mirror" formula in the Even numbers of the well known one with the Even ...lot can follow... Thanks ciao Stefano Last edited by complicatemodulus; August 5th, 2016 at 09:12 AM.
 August 5th, 2016, 09:52 AM #2 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 edit1: Sorry is: $\displaystyle A^n= \int_{0}^{A}nx^{(n-1)} dx = \sum_{y \in I} M_{ny}$ the F(y) function I wrote in the integral is wrong, but follows... (still working in office with few minutes to post...) Thanks Ciao Stefano
 August 6th, 2016, 11:48 AM #3 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 edit2: removing typing errors: I like to frake your mind with my Very Complicate Modulus Algebra: I't possible to define, in the "sense" of Lebesgue integral, so summing integer bars that cover the area of the derivate of $y=x^n$ : $y'=nx^{(n-1)}$ from left to right, and having known thickness: $HM_ny = ((Y_{i})^n-(Y_{(i-1)})^n)$ and Base: $BM_nx = (A-x)$ So it's possible to square the area bellow the derivate with orizontal integer bars (Lebesgue like): $\displaystyle A^n= \int_{0}^{A^n} nx^{(n-1)}= \sum_{y \in I} M_{ny}$ where: $I$= { $(x^n-(x-1)^n)|_{(x=1)}, (x^n-(x-1)^n)|_{(x=2)}... , (x^n-(x-1)^n)|_{(x=A)}$} And $M_{ny}$ is the very complicate modulus for orizontal bars integration: $M_{ny} = HM_{ny}* BM_{ny} = (y-(x-1)^n)*(A-x)$ so where $x=$ {1 for $y=I_1$, 2 for $y=I_2$, 3...., A for $y=I_A$ ) Thanks to the properties of the curves of the type $y=x^n$ I already show for the comeplicate modulus, so for the classic Rieman integration, we can perfectly square the derivate still if it's a curve with Integer o Rational bars. So for Powers, Riemann integral works in both directions but there is NO approximation since in the integer the result is the same using the Sum, the Step Sum or the Integral. Some example for n=3 to see that we need to turn the head 180° respect to the classic binomial develope...: 3^3= 3 + 12 + 12 or 4^3= 4+18+24+18 or 5^3=5+24+36+36+24 or: 6^3=6+30+48+54+48+30 or 7^3 = 7 + 36 + 60 +72 +72 + 60 +36 So I've my Complicate Triangle too ! And my theorem that a square is: $\displaystyle A^2= A+ \sum_{p=1}^{A-1} 2p$ So the Sum of the Root, plus the Sum of the first (A-1) Even... that is the "mirror" formula in the Even numbers of the well known one with the Even ...lot can follow... Thanks ciao Stefano
 August 9th, 2016, 08:59 AM #4 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Of course the Sum can also work in the rationals as: $\displaystyle A^n= \sum_{x=1/K}^{A} ((M_{(n,k) |x}-M_{(n,k) |x-1}) (AK-1+Kx))$ where: $\displaystyle M_{(n,k) |x}$ is the well known Complicate Rational Modulus (K dependent) computate for the single value of x and: $\displaystyle M_{(n,k) |x-1}$ is the well known Complicate Rational Modulus (K dependent) computate for the single value of x-1 And of course is possible to go to the limit... $\displaystyle A^n= \lim_{K\to\infty} \sum_{x=1/K}^{A} ((M_{(n,k) |x}-M_{(n,k) |x-1}) (AK-1+Kx))$ Returning to the same area of: $\displaystyle A^n=\int_{0}^{A}n x^n dx$ Here in the integers for n=3: Here it's possible to see that the Delta Area change from a minimum A, to a maximum, then again fells going to the last value... This let us give an answer for some A,B,C triplets (known solution with easy differential equation) Thanks ciao Stefano Last edited by complicatemodulus; August 9th, 2016 at 09:03 AM.
 August 9th, 2016, 09:46 PM #5 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 And here the problem: How can be the area bellow the derivate $y'=nx^{(n-1)}$ from 0 to 1 equal to $1$, while in both case Riemann and Lebesgue all the points $0=< x <1$ the derivate $y'|_x <1$ ?
August 10th, 2016, 03:31 AM   #6
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 Originally Posted by complicatemodulus And here the problem: How can be the area bellow the derivate $y'=nx^{(n-1)}$ from 0 to 1 equal to $1$, while in both case Riemann and Lebesgue all the points $0=< x <1$ the derivate $y'|_x <1$ ?
Ok... no way to get interest still with clear errors in... (the derivate into the picture is too low in y...)

 August 10th, 2016, 05:16 AM #7 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Here the right graph with right curves... That is: $\displaystyle A^n= \sum_{x=1}^{A} ((M_{(n) |x}-M_{(n) |x-1}) (A+1-x))$ where: $\displaystyle M_{(n) |x}$ is the well known Complicate Integer Modulus computate for the single value of x and: $\displaystyle M_{(n) |x-1}$ is the well known Complicate Integer Modulus computate for the single value of x-1 Of course the Sum can also work in the rationals as: $\displaystyle A^n= \sum_{x=1/K}^{A} ((M_{(n,k) |x}-M_{(n,k) |x-1}) (AK+1-Kx))$ where: $\displaystyle M_{(n,k) |x}$ is the well known Complicate Rational Modulus (K dependent) computate for the single value of x and: $\displaystyle M_{(n,k) |x-1}$ is the well known Complicate Rational Modulus (K dependent) computate for the single value of x-1 And of course is possible to go to the limit... $\displaystyle A^n= \lim_{K\to\infty} \sum_{x=1/K}^{A} ((M_{(n,k) |x}-M_{(n,k) |x-1}) (AK-1+Kx))$ Returning to the same area of: $\displaystyle A^n=\int_{0}^{A}n x^n dx$ Here in the graph what happen in the integers for n=3... As we can see we do not square using a linear Delta2Y, where Delta2Y is the difference between two following gnomons $M_{n |x}$ and $M_{n |x-1}$ I hope will not soo hard to prove there is no other way to do that... since the derivate is a curve. So moving left/right around a solution we have NON LINEAR value, while FLT state we have them... $\displaystyle \sum_{1}^{C}M_n = \sum_{1}^{A}M_n + \sum_{1}^{B}M_n$ or: $\displaystyle \sum_{1}^{C}M_n = 2 \sum_{1}^{A}M_n + \sum_{A+1}^{B}M_n$ or: $\displaystyle \sum_{1}^{C}M_n = 2 \sum_{1}^{B}M_n - \sum_{A+1}^{B}M_n$ So we can call: $\displaystyle Delta1 = \sum_{A+1}^{B}M_n$ and $\displaystyle Delta2 = \left( \sum_{A+1}^{B}M_n \right) / (2^{1/n} B -C)$ = costant for each point around the solution Last edited by complicatemodulus; August 10th, 2016 at 05:43 AM.
 August 11th, 2016, 08:43 AM #8 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 I forgot to say that: - with my Step Sum it's possible, under certain condition to work with IRRATIONAL step like: $1/\sqrt(2)$ In case we are working with the double of a Square it's true: $\displaystyle B^2 = \sum_{x= 1/\sqrt(2)}^{2B/\sqrt(2)} 2x/\sqrt(2) - 1/2$ B can be an integer or half an integer. Here the table: X;x= X/sqr(2);2x/sqr(2)-1/2;SUM = 2P^2;P^2; P 1 - 0.707106781 - 0.5 - 0.5 - 0.25 -0.5 2 - 1.414213562 - 1.5 - 2 - 1 -1 3 - 2.121320344 - 2.5 - 4.5 - 2.25 -1.5 4 - 2.828427125 - 3.5 - 8 - 4 -2 5 - 3.535533906 - 4.5 - 12.5 - 6.25 -2.5 6 - 4.242640687 - 5.5 - 18 - 9 -3 7 - 4.949747468 - 6.5 - 24.5 - 12.25 -3.5 8 - 5.656854249 - 7.5 - 32 - 16 -4 9 - 6.363961031 - 8.5 - 40.5 - 20.25 -4.5 10 -7.071067812 - 9.5 - 50 - 25 -5 11 -7.778174593 - 10.5 -60.5 - 30.25 -5.5 12 -8.485281374 - 11.5 -72 - 36 -6 ...I know nobody find that interesting...
 August 19th, 2016, 06:50 AM #9 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Sorry still a label error in the graph that is:
 August 22nd, 2016, 11:49 AM #10 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Remembering that: \[ A^3=\sum_{1}^{A} M_n /] While the Gnomons $M_n$ (x) for n=3 are connected to n! so 3! so 6 and the triangular numbers A000217 - OEIS: $Mn_1 = 0*3!+1= 1$ $Mn_2 = 1*3!+1= 7$ $Mn_3 = 3*3!+1=19$ $Mn_4 = 6*3!+1=37$ $Mn_3 = 10*3!+1=61$ $Mn_4 = 15*3!+1=91$ ... $Mn_P = P*(P+1)/2*3!+1= ...*3!+1$ So a monotone rising function.... the Gnomons Mn(y) are strictly connected to A, and the product of: $3!*(A-k)*\overline {(A-k)} +1$ where in case $K=0$, the product $n!*0 = 1$ (and not by convention). Example $A=5$ then $A^3$ is equal to the sum of: $Mn_{y1} = ( 3!*(5-5=0))=1 *(5-0=5))=5$ $Mn_{y2} = 3!*1*4= 24$ $Mn_{y3} = 3!*2*3= 36$ $Mn_{y4} = 3!*3*2= 36$ $Mn_{y3} = 3!*4*1= 24$ Sum =125 ...so a non simmetric non monotone (non invertible) curve.... So since FLT state that around the solution (+/- $delta$) the function is linear, continuos and for so ...must be invertible.... Last edited by complicatemodulus; August 22nd, 2016 at 12:13 PM.

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