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Can Infinity border predict that the quartic is the last solvable equation #98 Specialness of 10 and 1, 10, 100, 10^3, . . 8th ed.: TRUE CALCULUS
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Can Infinity border predict that the quartic is the last solvable equation #98 Specialness of 10 and 1, 10, 100, 10^3, . . 8th ed.: TRUE CALCULUS
Posted:
Dec 22, 2013 4:56 PM


Well the borderline of infinity is 1*10^603 where pi has those 3 zero digits in a row.
Now so far I have been saying that in a Grid, a specific Grid like the 10 Grid we can use numbers of the 10^2 and 10^3 Grids to develop a Calculus. However, if the quartic has general solutions would suggest that we must go 4 levels.
Isaac Newton did some excellent work on the cubic, showing many of the curves.
Quartic curves are much more difficult.
Quintic curves follow much like cubic curves.
Can the Infinity borderline of 1*10^603 predict that quartic solutions are the last general solutions?
Well, that would mean 10^600, 10^601, 10^602, 10^603.
It would mean that 4 levels are required for each specific given level, rather than 3. And for some reason, a 5th level is unacceptable.
Now yesterday I started from 0 and worked forwards, and today I am starting from macroinfinity working backwards. With 0 and microinfinity we see that 1.1^5 is equal to phi = 1.61 in the 10 Grid. And that would tell us the quartic has general solutions but the quintic may not.
So what is the reasoning at macroinfinity? Is it that a logarithmic spiral cannot have 5 squares for a single revolution? Take a look at the Fibonacci Golden Mean Logarithmic Spiral of its 8, 5, 3, 2 squares then look at its 55, 34, 21, 13 squares. It takes 4 squares or 4 levels to cover a Revolution of the Spiral.
Is that the reason the general quartic has solutions but the general quintic is unsolvable.
 Recently I reopened the old newsgroup of 1990s and there one can read my recent posts without the hassle of mockers and hatemongers.
https://groups.google.com/forum/?hl=en#!forum/plutoniumatomuniverse
Archimedes Plutonium



