December 13th, 2016, 10:25 PM  #1 
Senior Member Joined: Dec 2012 Posts: 979 Thanks: 24  Tessellation problem
Answer to this: 1) How many (contiguos, not interfering with each other) tessel of area 10 (measure unit don't care so can be square feet or meter, don't care) can be arranged bellow the first derivate of $Y=X^2$ from 0 to 10 ? 2) Write the list of their Base x Height dimensions Thanks Ciao Stefano 
December 15th, 2016, 12:50 AM  #2 
Senior Member Joined: Dec 2012 Posts: 979 Thanks: 24 
As shown in other post I've found that all the derivates of: $\displaystyle Y=X^n$ For $X\in \mathbb{Q}$ Are tessellable with 2 kinds of Rectangular Tessels: T1: The unitary area [1/K x 1/K] or TM: Rectangular Columns called: Gnomon's, of area BxH: $Base = 1/K$ $\displaystyle Height = M_{n,x,K}= {n \choose 1}x^{n1}/K + {n \choose 2}x^{n2}/K^2 + {n \choose 3}x^{n3}/K^3 +... +/ \frac{1}{K^n} $ Starting with the most simple case $n=2$, $K=1$ so mesauring for the moment the INTEGER areas only, bellow the first derivate: From the well known relation: $\displaystyle A^n=\int_{X=0}^{A} nX^{n1} dX$ From the non well before investigated Telescoping Sum property we can write if $A\in \mathbb{N}$: $\displaystyle A^n=\sum_{X=1}^{A} (X^n(X1)^n)*1$ Infact developing the Sum we have: $\displaystyle A^n=\sum_{X=1}^{A} (X^n(X1)^n)*1= A^n(A1)^n+(A1)^n ..... +11= A^n$ So it's clear that the $Base=1$ was forgotten in the computation, so not well investigated, also because of two, now discovered wrong, "math statments" that:  The index of the sum is aways "mute" (this it's true, but too penalizing definition let many thiks it's unusefull) and for so called with a insignificant "i" or else name, while as shown it's important to connect it to the abscissa $X$ on the graph.  The $Base$ can be scaled under certain condition so it can be, in this case: $dx < Base \leq A$ Infact if $A\in N^{+}$ we can write $A^n$ as a Sum of Gnomons (Integer Rectangular strips), or as a Step Sum of Rational Gnomons (Rational Rectangular strips) or, at the limit as integral as Sum of Rectangles base dx, so Lines: $\displaystyle A^n = \sum_{X=1}^{A} {M_n} = \sum_{x=\frac{1}{K}}^{A} M_{n,K} = \lim_{K\to\infty} \sum_{x=\frac{1}{K}}^{A} M_{n,K} = \int_{0}^{A} n x^{(n1)} dx $ A proof can be easy found remembering some property of the SUM: For example $n=2$ Given: $a, k \in {\mathbb{N^{+}}} \implies$ $\displaystyle \sum_{X=1}^{ak}\left ( \dfrac{2X}{k^2}  \dfrac{1}{k^2} \right ) = a^2 = \int_0^a\ 2X\ dX.$ Proof: $ \sum_{X=1}^{ak} \left ( \dfrac{2X}{k^2}  \dfrac{1}{k^2} \right ) = \dfrac{1}{k^2} *\left \{ \left ( 2 * \sum_{x=1}^{ak}X \right )  \sum_{X=1}^{ak}1 \right \}$ $= \dfrac{1}{k^2} \left \{ \left (\cancel 2 * \dfrac{ak (ak + 1)}{\cancel 2} \right ) ak \right \} \implies$ $ \sum_{X=1}^{ak} \left ( \dfrac{2X}{k^2}  \dfrac{1}{k^2} \right ) = \dfrac{1}{k^2} * \{(ak)(ak) + \cancel {ak}  \cancel {ak)} \} = \dfrac{1}{\cancel {k^2}} * (a^2 \cancel {k^2}) = a^2 = \int_0^a\ 2X\ dX.$ But to understand better the fact that there is continuity between the Integer Sum and the Integral you've to digest that: due to the $mute$ property of the Index: Nothing change in the result of the Sum if: We make a change of variable calling $x=X/K$ so we have K times scaled variable and we multiply by $K$ the number of the times we count the new Index $1/K$, adjusting the Lower and the Upper limit as necessary: $\displaystyle A^2= \sum_{X=1}^{A=aK}\left ( \dfrac{2X}{K^2}  \dfrac{1}{K^2} \right ) = \sum_{x=1/K}^{A=aK}\left ( \dfrac{2x}{K}  \dfrac{1}{K^2} \right ) = \lim_{K\to\infty} = \sum_{x=1/K}^{A=aK}\left ( \dfrac{2x}{K}  \dfrac{1}{K^2} \right ) = \int_0^A\ 2x\ dx$ Now it's clear that if we are talking of the area till an integer $A$ we can Tesselate it with:  Unit tessel  Integer Gnomons  Rational Gnomons  Integral Without changing the result. This simple adjustment can give us the base to debate some of the most "difficoult" number theory problems like Fermat and Beal, using the simple connection between the continuos function $Y=X^n$ and it's derivates and what i call the Integer or the Rational Derivate. Since I prove they are Tessellable (so modular) one we go into Fermat problem we can say immediately that:  If we state a such function must be: $f: \mathbb{N^{+}}\to\mathbb{N^{+}}$ and that all the involved quantities are (or must be) Power of Integers: (1) $\displaystyle C^n=A^n+B^n$ with $A<B\in \mathbb{N^{+}}$ Than we immediately see we can reduce the problem to the question: in what is our analitical extention where our problem lay, so in $\mathbb{R}$ it's always true that: (3) $\displaystyle P= C^n=A^n+B^n$ with $A<B\in \mathbb{N^{+}}$ Considering $ \mathbb{X} = \mathbb{N}$ the set of the integer abscissa $X$ (I let you add here the property of the Domain set $ \mathbb{X}$ and of the CoDomain set $ \mathbb{Y}$ ) Rewriting the (3) with the well known formulation I already present several times here: $\displaystyle P= 2*\sum_{X=1}^{A} (X^n(X1)^n)*1 + Delta$ $\displaystyle P= 2*\sum_{X=1}^{B} (X^n(X1)^n)*1  Delta$ with: $\displaystyle Delta= \sum_{X=A+1}^{B} (X^n(X1)^n)*1 $ We can immediately see on the cartesian plane what this imply: We are asking if given an height $P$ defined by the above equations (symetric condition), we ask if it can be a Power of an integer, so $P=? C^n$ has to be one element of the set we are talking of, so must lay on our analitic extention $Y=2X^n$ and to it correspond an abscissa $X_F$ that is equal to: $X_F = C/{2^{(1/n}}$ that is for sure an irrational value. Now the end: We know that if our Fermat's function must be of the type: $\displaystyle f:\mathbb{N} \to \mathbb{N}$ So if: $\displaystyle C^n=A^n+B^n$ than we can tessellate all the 3 value with the Unit tessel, or the Integer (and I add will, "or Rational") Gnomon Tessel, But since our Right Border of the area bellow $Y'=2nX^{n1}$, so our upper limit is an IRRATIONAL We have NO WAY TO PLAY with our INTEGER TESSEL SET (some is double so if necessary we can dismount it in Unit Tessel, but not littlest, remembering Unitary or Integer or Rational Gnomons, makes no difference) to try to Tessellate such AREA. Proof: Trying to FULLFILL the area bellow $Y'=2nX^{n1}$, till the irrational $Xf = C/{2^{(1/n}}$ We can immediately see that with Integer Gnomons we stop to: $X_I <X_F$ and also going with our Rationa Gnomons ww stop to: $X_I < X_Q <X_F$ And since we are in a well ordered SET and we knwo that with the Rational Step Sum going to the limit for $K\to\infty$ we can for sure write: \[ P= C^n = \lim_{K\to\infty} \sum_{x=1/K}^{C/{2^{(1/n}}} M_{n,K,x} = \int_{0}^{C/{2^{(1/n}}} M_{n,K,x} dx = C^n = A^nB^n\] and: $\displaystyle \sum_{X=1/}^{X_i} M_{n,X} < \sum_{x=1/K}^{X_Q} M_{n,K,x} < \int_{0}^{C/{2^{(1/n}}} M_{n,K,x} dx $ Remembering that for $n=2$ it's clear why we can have a solution (the linear derivate can be squared using the (Base x Height/2) forumla, so talking just of derivate that are curve $n>2$ We have no other way to square such area except going Infimus Tessel, so via integral, so we violate the condition Fermat's equation is a function $\displaystyle f:\mathbb{N} \to \mathbb{N}$ So, I hope, I prove in the forgotten elegant way Fermat is Right. The proof for all $n$ comes both by Induction, and by the transfinte law of induction, I hope is well shown here (still if I not write both the easy proof). I just remember we have this relation between different $nth Gnomons$: A property of the Gnomons $M_n$ is that: \[ M_{n+1} =(n+1) \int ( M_{n}) + C \] Where $C$ is the Integration Constant For example: \[ M_{2} = 2X1 \] then: \[ M_{3} =(n+1) \int ( M_{n}) + C = 3 \int (2X1) = 3* ( 2/2 X^2  X ) + C \] $C$ will be $+1$ in case $n$ is ODD, $C$ will be $1$ in case $n$ is EVEN So: \[ M_{3} =3X^2  3X + 1 \] And so on. The Proof it's easy and follow the well known integration rules. Another property of the Gnomons $M_n$ is that: \[ M_{n1} = \frac{1}{n} * \frac{d}{dx} ( M_{n}) \] So \[ M_{2} = \frac{1}{n} * \frac{d}{dx} ( M_{n}) = \frac{1}{3} * \frac{d}{dx} ( 3X^23X+1) = 2X1 \] Pls not say it's self reference, or looping it is not, and we can claim Set theory recognising my Ordinal as the property of all the derivates to be squared in the integers and in the rationals with Integer or Rational Gnomons and the fact that the $\displaystyle f:\mathbb{N} \to \mathbb{N}$ Imply we can play just with a minimum tessel that it's the unit [1x1], but I hope, it's clear I go over proving fermat is right also if we go rational with the unit tessel Rational as [1/K x 1/K]. I know will be a long way to understand and digest because I involve several math fields, so the error can be not so easy to be found, or not present. Last edited by complicatemodulus; December 15th, 2016 at 01:26 AM. 
December 15th, 2016, 03:45 AM  #3 
Senior Member Joined: Dec 2012 Posts: 979 Thanks: 24 
What I don't write, but I hope will be well clear is that we MUST check what the symmetric condition in $Y$ means also on all the following curves we find moving right:  $Y=X^n$ first, where on $X$ so in $\mathbb{X}$ still if $C$ was an integer, the other 2 limits $A/{2^{1/n}}$ and $B/{2^{1/n}}$ are Irrationals And all that will rest in irrational ratio till the first linear derivate that is the second last (ans the last it's linear too). So we immediately see, viceversa, that for n=2 the only 2 derivates are both linears. 
December 15th, 2016, 06:39 AM  #4 
Senior Member Joined: Dec 2012 Posts: 979 Thanks: 24 
This ordinal works as follow: So there cannot be another law to sort the last difference (and in case of FLT it will no longer be $n!$ if $n>2$) 
December 29th, 2016, 06:43 AM  #5 
Senior Member Joined: Feb 2010 Posts: 621 Thanks: 98 
Greg1313 ... Didn't you close this thread on the number theory forum?

December 29th, 2016, 08:55 AM  #6 
Senior Member Joined: Dec 2012 Posts: 979 Thanks: 24 
...it was many post after. If for you it's a problem, pls just discard it, also because now you're just wasting the time of interested readers (if any). Ciao Stefano 

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