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July 7[edit]

Does this math theory make any sense or should I just drop it?[edit]

Warning: if you look at this your brain will explode. Looie496 (talk) 18:21, 7 July 2013 (UTC)[reply]

Pi is an infinite cubic variable that represents the time it takes for an electron to travel over an amount of space over any period of time.

5 1 4 1 . 3 . 1 4 1 5

Now picture the circle created at the decimal point of Pi on a 2d graph.

Pi is just a cubic palendrome that extends infinitely to create space over time.

So if e- = electron velocity and π = an infinitely and equally expanding amount of space

π = (e-)√(V)³ π³e = (e-)√V(e) π³e = (π)^(-3)√ V So time = π³e and e represents the time is takes for an electron to travel to all points on the circle over a variable amount of time. e(x) = (π)-3√V(x)

e(x!)-3 = (π)√V(x)

(π)-1 = e(-x!)-3√V(x)

π = e((-x!))³(V(x))

π² = e((-x!))(-³)(V(x))√π

if x = 1

π = e((-1!))(-³)(V(1))√π

π=e^(-3)√π

π² = -1√π

π = π (Palendrome)

(x!)³ = π^5

Which means in 5 1 4 1 . 3 . 1 4 1 5 the . represents the new variable, e. And each point on the graph is an exponential variable of x. The centre 3 is Pi, the first point on the graph is π(1!x).

5 1 4 1(e) 3 1 4 1 5

— Preceding unsigned comment added by 99.255.193.121 (talk • contribs) 12:49, 7 July 2013‎

Drop it. Looie496 (talk) 18:21, 7 July 2013 (UTC)[reply]

Yeah, drop it. It makes no sense. —SeekingAnswers (reply) 12:59, 8 July 2013 (UTC)[reply]

As it stands, what you've written doesn't make sense. However, if you'd like to discuss your ideas and try and clarify them down to what you are wanting to express, I'd be happy to help on my talk page. Even if this particular conception doesn't work out to anything, I'm sure there are some great articles and books we can find on the topics that lead you to it (and this could help you refine what you want to say into actual mathematical language). Just drop me a line on my talk page:-)Phoenixia1177 (talk) 06:23, 9 July 2013 (UTC)[reply]

Like Very charitable of you! SemanticMantis (talk) 14:09, 9 July 2013 (UTC)[reply]