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Author Topic: de Fermat's last theorem  (Read 879 times)
Description: A trigonometric approach
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jehovajah
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« on: February 11, 2011, 10:55:54 AM »

x^n+y^n \neq z^n
Reduces to \frac{x^n}{ z^n} +  \frac{y^n}{ z^n} \neq 1^n in the unit circle.

This is the same as  cos^n\theta +sin^n\theta\neq 1^n

Which can be written as  cos^n\theta +sin^n\theta\neq (cos^2\theta +sin^2\theta)^n

Which by de Moivre reduces to  cos^n\theta +sin^n\theta\neq (cos n\theta +i*sin n\theta)(cos n\theta - i*sin n\theta)

You can explore these relations graphically here.

Post what you find out, think, render, or create.

Have fun dancing banana dancing chilli
« Last Edit: April 06, 2012, 06:57:38 AM by jehovajah » Logged

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jehovajah
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« Reply #1 on: February 15, 2011, 11:02:21 PM »

x^n+y^n \neq z^n

This is the same as  cos^n\theta +sin^n\theta\neq 1^n

Which can be written as  cos^n\theta +sin^n\theta\neq (cos^2\theta +sin^2\theta)^n

Which  can be expressed   (e^{i\theta}-isin\theta)^n +(-ie^{i\theta}+icos\theta)^n\neq (e^{in\theta})(e^{-in\theta})
which can be arranged as
 (e^{i\theta}-isin\theta)^n +i^n(cos\theta-e^{i\theta})^n\neq (e^{in\theta})(e^{-in\theta})
or
 (e^{i\theta}-isin\theta)^n +i^n(cos\theta+e^{-i\theta})^n\neq (e^{in\theta})(e^{-in\theta})
You can explore these relations graphically using any fractalgenerator.

Indulge me, someone?

Post what you find out, think, render, or create.

Have fun dancing banana dancing chilli
« Last Edit: November 12, 2011, 10:13:39 AM by jehovajah » Logged

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jehovajah
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« Reply #2 on: February 17, 2014, 09:23:20 AM »

I think I have made a mistake in one of the conversions.

Anyone care to find it? embarrass
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jehovajah
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« Reply #3 on: March 01, 2014, 10:36:54 AM »

z = x + y

Can be understood as  z is a binomial form.
Then zn can be expanded by the binomial theorem.

We find then that zn can never equal xn + yn.

However we do have a special case, proved by Pythagoras

z2 = x2 + y2  + 2xy by the binomial expansion
 = c2  + 2xy by the Pythagoras theorem , using the proof of embedding in a larger square.

What this means is, by Pythagoras we can always find some c2 that sums the 2 squares . We have no such identity for n>2.
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May a trochoid of hh iteratively entrain your Logos Response transforming into iridescent fractals of orgasmic delight and joy, with kindness, peace and gratitude at all scales within your experience. I beg of you to enrich others as you have been enriched, in vorticose pulsations of extravagance!
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