I was just kinda playing around in my brain with some other thoughts. Then something kinda strange popped out that looks like an equation. I would guess for the velocity of an energy relative to its mass maybe??? \({C^2}= \frac{{2}^{M}}{E!}\)

1) Always post your source -- context is essential to understanding any equation. If this is all you have then you have nothing. 2) Your equation doesn't look like any type of physics, since both M and E are required to be dimensionless (without units) and thus C is without units. If we assume natural units where \(G = c = \hbar = 1\) (which is a wild assumption, based on nothing, and not the only possible choice) then we may assume E and M are to represent energy and mass in natural units and expand the right side. As Special Relativity is the first physical theory to have a concrete energy associated with a free body \(E = \sqrt{(Mc^2)^2 + (pc)^2}\) then we may attempt to extract information from the right side: \(M_0 = \sqrt{\frac{\hbar c}{G}} = 1\) gives us: \(\frac{2^{M}}{E!} = \frac{2^{\tiny \frac{M}{M_0} } }{ \Gamma \left( 1 + \frac{E}{c^2 M_0} \right) } = \frac{2^{\tiny \frac{M}{M_0} } }{ \Gamma \left( 1 + \frac{\sqrt{(Mc^2)^2 + (pc)^2}}{c^2 M_0} \right) }\) Now if \(M=0\) this gives us \( \frac{ 1 }{ \Gamma \left( 1 + \frac{ \left|p \right| }{M_0 c} \right) } \approx 1 + \gamma \frac{ \left| p \right| }{M_0 c} - \left( \frac{1}{12}\pi^2 - \frac{1}{2} \gamma^2 \right) \left( \frac{ \left| p \right| }{M_0 c} \right)^2 \approx 1 + \frac{\omega}{3.2147706 \, \times \, 10^{43} \, \textrm{s}^{\tiny -1} }\) -- this says gamma rays from annihilating electrons differ (in some way that I can't guess at) from radio waves by some 24 parts per million, billion, billion. This is unevidenced. Being unevidenced, it is required to have a chain of persuasive logic to be asserted, but you admit you have nothing of the sort. So posting things like this are just going to get you labeled (perhaps unfairly) as a useless person and (completely accurately) a person incompetent at math and physics who doesn't understand the subjects he plays at. It looks like you started with \(E = Mc^2\) (which is not by itself physics) and invented rules of symbol manipulation that don't preserve the equality of both sides (which makes it useless as a system of logic).

I started with my brain and now I'm working backwards to try and figure out what I did.... It should be based off the basic definition of factorial. It should be a summation that included both the transcendental pi and the natural log hidden by the definition of a factored sum? Odd form of summation?! \( e^x= (\pi_{k=1}^{n}) K \) (=) \( e^x= (\sum_{k=1}^{n}) \frac{x^2}{n!} \) \(E=m{c}^{2}\) \({c}^{2}={\frac_{m}^{e}}\) There isn't a way for me to explain this but I can feel this next transformation is correct... In order to introduce a factorial we have to also introduce pi which could be where I messed up on the last one. \({c}^{2}=\pi {\frac_{m!}^{E}}\) Last but not least to also accomplish this transformation to include both pi and factorial we have to bring with us our natural log given by the definition of factorial in summation form. So we can put that over our E. \({c}^{2}=\pi {\frac_{m!}^{E^e}}\) Summation form to reg is apparently not my strong suit...

\( e^x = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n = \lim_{n \to \infty} \prod_{k=1}^n \left( 1 + \frac{x}{n} \right) = \lim_{n \to \infty} \sum_{k=0}^n \frac{x^k}{k!} = \sum_{k=0}^{\infty}\frac{x^k}{k!}\) Your biggest math problem is that you ignore the rules of math in favor of your abuse of mathematical notation, specifically you cannot add meters to square meters and you cannot take the exponential or factorial or logarithm of a quantity with units attached. You also wrote down the wrong definition of exponential. Your biggest physics problem is that you ignore the fact that physics has to describe the universe which requires you to justify your work, not for me to justify it. You also don't seem to understand the physics content of \(E = mc^2\) which is a consequence of certain assumptions that you toss out when you substitute your "transformations" for rigorous algebra.

Those are two different definitions of Factorial, no where did I mention a definition of exponential, but I think the equation would represent that notion except where exponents occur. You solve two problems of equality by taking our imaginative ideas in science and smashing them with reality.

Lets start over from the correct perspective If I may re-gather my thoughts for a moment. \(e^x = \sum\frac{{x}^n}{n!} \) We then redistribute our summation into a variable and two equal circumferences. \(Ee^x = \frac{{\pi x}^n}{\pi n!}\) select our variables in such a way that it bring balance immediately: \( E=MC^2 =( \frac{C^2}{E} = \frac{1}{m} )\) integrate: \((\frac{C^2}{E})e^x = \frac{{\pi x}^M}{\pi M!}\) X being the amount.

This is not how you do science, starting with some conclusion you assume and then trying to justify it. It's how 'creation science' works and its how many hacks here work. You should start with the evidence and data and see where it takes you. This is just incoherent. Not only are you not doing science properly you cannot even do the maths properly. Why shoe horn in pi? Because it is transcendental? Most numbers are transcendental, the algebraic numbers are staggeringly rare (they are as set of measure zero within the Reals). As it happens e is transcendental too, so if you want an transcendental number you already have one. In fact it is a proven theorem that if x is algebraic then \(e^{x}\) is transcendental unless x=0. Which is actually only a specific case of \(E^{2} = (mc^{2})^{2} + (pc)^{2}\), the true mass-energy-momentum relationship in relativity. There isn't a way to explain it because it is nonsense. Please tell me this is an attempt at a joke. Do you really think this is what physicists do, just randomly tape together different symbols because we feel like it? Personally I prefer \(n! \equiv \Gamma(n+1)\) where \(\Gamma(z) = \int_{0}^{\infty}e^{-t}t^{z-1}dt\). And the definition of e comes from finding the y which satisfies \(\int_{1}^{y}\frac{dt}{t} = 1\). From that you find a series expansion for \(e^{x}\). No, you solve science problems by examining reality and then drawing conclusions from that, not making up your conclusions and then trying to justify them.

The evidence is all around. We are trying to understand energy by testing all its properties. So I messed up the definitions on the linked page because I suck at latex. How many other people not in the field try some other field? Ah so... From both \( Ee^x= \frac{\pi x^m}{\pi m!} . and . (mc^2)^2 + (pc^2)^2 . we can get. . E(mc^2)^2e^x = -(pC^2)^2\frac{\pi x^m}{\pi M!} \) don't fault me too much for my first attempt to ever work a transformation. No I think I can do it cause I feel... Which is something you lack in your own life and reflect online. If you lack patience for a student of another matter it is your own reflection when you fail to teach, but by learning what we have done wrong we find our faults. I am just a person. I can think some peoples thoughts for themselves. But who can't. That is how you end science. Beginning another starts mentally where I reside. I can do this because the equations represent equivalence and the only strange thing I have done is bring Pi back into the definition and create a variable open to balancing the equation naturally.

Oh here is where I messed up instead of \( Ee^x= \frac{\pi x^m}{\pi m!} . it ...should ...be: Circumference... over... diameter... to... represent...pi ..in.. this.... exponential.. .. \frac{C}{D}e^x= \frac{\pi x^m}{\pi m!} . and . (mc^2)^2 + (pc^2)^2 . we can get. . \frac{C}{D}(mc^2)^2e^x = -(pc^2)^2\frac{\pi x^m}{\pi M!} \) C=circumference c=speed of light D=diameter e= natural log X= the amount of our element M= mass number of an element P= position?

Am I really that hopeless in maths??? Maybe if I try to explain someone can do maths better.... Or... \(\frac{C}{D}e^x+ (mc^2)^2 = \pi\frac{ x^m}{ M!} -(pc^2)^2 . (\frac{C}{D}e^x+ (mc^2)^2) - (\pi\frac{ x^m}{ M!} +(pc^2)^2)=E^2 \)