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April 20th, 2017, 11:36 PM   #1
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Fermat's rosebud

I hope this will be the correct end of Fermat the last:

In case $n>2$ the First Derivate is a Curve

So for any Symmetric condition we fix on $Y$

$\displaystyle C^n = 2A^n+\Delta$

&

$\displaystyle C^n = 2B^n-\Delta$

So not just the Fermat's one I've shown in my previous post with:

$\displaystyle \Delta=\sum_{A+1}^{B}(X^n-(X-1)^n))$

We, clearly,

- fix a bigger abscissa $X_{Y_{1/2}}$

- than any Integer or Rational one $X_i =$ $A$, $B$ or $C$, or else (Integer or Rational) we Correctly fix using the property that a Power of an Integer or a Rational, is always equal to a Sum of Integer or Rational Gnomons (I already define here several times) witch the last one as an Abscissa $X_i$ bigger than the Average $(X_i+X_{i-1})/2$, but littlest than the one Fermat (or any other linear definition) fix.

And sice we can say this for any $\Delta$ we choose, and we can easy show we can continue to reduce it (since I prove we can also go rational but the property for genunine powers doesen't change), we always get the same result, so we can be sure Fermat is talking of an irrational Value (or the infinite descent he found, or I was, just, looking for).

We cannot say exactly where our Fermat's Point is (without going infimus) but we can for sure say that it is more Right, so is more bigger than the Right one defined by the last Integer or Rational Gnomon...

Was a very long way... but I hope I get the Full Stop.

Picture asap.

Last edited by complicatemodulus; April 21st, 2017 at 12:03 AM.
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April 21st, 2017, 02:04 AM   #2
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It's little hard to show on the picture since the difference is very little.

Taking as example $n=3$, for A=5, B=6

We define:

$2A^n = 2*5^3 = 250$

$2B^n = 2*6^3 = 432$

So:

$\Delta = (432-250)/2 = 91$

So our $P= 250+91=341$ is, at the moment just an hypotetic $C^3$ but for sure an ordinate on the curve $Y=2X^n$

So it's abscissa will be:

$X_{Y_{1/2}}= (341/2)^{1/3} = 5,545....$

And if we take as reference 341 as the "Linear" Average height beween $(X_i,Y_i)$ and $(X_{i-1},Y_{i-1})$, we get for sure just $5,5$ as Average Abscissa


And now follow that on the First Derivate $Y'=2X^{n-1}=6X^3$

we get:

$Y_{X_{Y_{1/2}}} = 6*5,545..^2 = 184,478...$


While the Double of the Integer Gnomon $M_{n, i}= (X^n-(X-1)^n)$ for $n=3$ and $X=6$ is:

$M3_{6}= 3X^2-3X+1= 3*6^2-3*6+1= 91$

so de double is $91*2=182$

Again if we take as reference the "linear" Average (geometrical mean) on the derivate we get:

$2*nX_{i-1}^{n-1} = 2*3*5^2 = 150$

And

$2*nX_{i}^{n-1} = 2*3*6^2 = 216$

So the medium height on the derivate is: $150+ (216-150)/2 =183$

Since the Power $n$ once bigger than 2 always produce curved derivates and we define an algo that said us that indipendently from the value we take, Fermat's Ordinate on the Derivate it's always bigger than the Geometrical Mean, we can be sure it is also bigger than the one of the Last Good Gnomon (that has a known function $Mn=(X^n-(X-1)^n)$ and satisfy the property that exceding area equal to missing one, or $A^+ = A^-$ (see my previous post), so on any power's derivate curve the medium height has to be littlest than the Geometrical Mean....

OF course once n=2 the derivate it's linear and the equation can find a solution.

Last edited by complicatemodulus; April 21st, 2017 at 02:12 AM.
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April 21st, 2017, 05:23 AM   #3
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Here the first picture. I know I've to work lot to adjust all the labels...

1) How the Symmetric Fermat's condition define the Height on the First Derivate (Curve):




2) While the Ordinate of the last Bigger known Gnomon (that define the Area 2B^3 > C^3), is Littlest than the Height defined by Fermat's Symmetric condition:



We cannot have in the hands an Integer nor a Rational number by this construction, but just the assurance of the fact that still if we change value, or scale (so we reduce going Rational) we continue to have the same condition respect to the Geometric Mean, till the limit when $\Delta$ becomes an infimus and the equation give a result, that is, at this point a "for sure" Irrational (the shown process is exactly the same of how an irrational is defined).
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