Little problem I created for myself

• Jul 22nd 2013, 02:13 PM
uperkurk
Little problem I created for myself
I want to have some algebra expression so that once it is solved the answer becomes $\displaystyle E=mc^2$. I want it to be a little complicated though where the answer isn't obvious and you do actually have to reduces, cancel ect. Something like this for example. I know it doesn't mean anything but to give you an idea, once solved and everything cancels ect the solution is $\displaystyle E=mc^2$

$\displaystyle ac^2+\frac{E^2}{6-m^2} = -6-a \frac{1}{E^2}$

Thanks!
• Jul 22nd 2013, 05:26 PM
topsquark
Re: Little problem I created for myself
Quote:

Originally Posted by uperkurk
I want to have some algebra expression so that once it is solved the answer becomes $\displaystyle E=mc^2$. I want it to be a little complicated though where the answer isn't obvious and you do actually have to reduces, cancel ect. Something like this for example. I know it doesn't mean anything but to give you an idea, once solved and everything cancels ect the solution is $\displaystyle E=mc^2$

$\displaystyle ac^2+\frac{E^2}{6-m^2} = -6-a \frac{1}{E^2}$

Thanks!

Well, you could start with E = mc^2 and add a term to both sides, divide both sides by something, square both sides, etc. If you want to have real fun you could do something like...
$\displaystyle E = mc^2$

$\displaystyle E + 3ac = mc^2 + 10ac^3$
or something and make them solve for a such that E = mc^2.

Of course you could have some real fun and use $\displaystyle E^2 = p^2c^2 + (mc^2)^2$, where p is the momentum. That's the general form for E = mc^2.

-Dan
• Jul 23rd 2013, 02:12 AM
uperkurk
Re: Little problem I created for myself
Thanks topsquark, I had something like that in mind but I wanted to make it a little more complicated and not make it apparent that the answer is $\displaystyle E=mc^2$
• Jul 23rd 2013, 03:41 AM
ChessTal
Re: Little problem I created for myself
Quote:

Originally Posted by uperkurk
I want to have some algebra expression so that once it is solved the answer becomes $\displaystyle E=mc^2$. I want it to be a little complicated though where the answer isn't obvious and you do actually have to reduces, cancel ect. Something like this for example. I know it doesn't mean anything but to give you an idea, once solved and everything cancels ect the solution is $\displaystyle E=mc^2$

$\displaystyle ac^2+\frac{E^2}{6-m^2} = -6-a \frac{1}{E^2}$

Thanks!

As an exercise for someone you can ask him to solve for E and x(both real number) and for $\displaystyle m \ne 0$ and $\displaystyle c \ne 0$ the following:

$\displaystyle 2{E^2} + {(mxc)^2} + {m^2}{c^4} - 2mxcE - 2{c^2}Em \le 0$

or even better the equivalent inequality:

$\displaystyle 2{\left( {\frac{E}{c}} \right)^2} + {(mc)^2} + {(mx)^2} - 2\frac{{mxE}}{c} - 2mE \le 0$
• Jul 23rd 2013, 05:15 AM
uperkurk
Re: Little problem I created for myself
Quote:

Originally Posted by ChessTal
As an exercise for someone you can ask him to solve for E and x(both real number) and for $\displaystyle m \ne 0$ and $\displaystyle c \ne 0$ the following:

$\displaystyle 2{E^2} + {(mxc)^2} + {m^2}{c^4} - 2mxcE - 2{c^2}Em \le 0$

or even better the equivalent inequality:

$\displaystyle 2{\left( {\frac{E}{c}} \right)^2} + {(mc)^2} + {(mx)^2} - 2\frac{{mxE}}{c} - 2mE \le 0$

When I paste these into Wolfram it tells me it doesn't understand the equation and the closest answer it gives is $\displaystyle c^4m^2$
• Jul 23rd 2013, 10:18 AM
ChessTal
Re: Little problem I created for myself
Quote:

Originally Posted by uperkurk
When I paste these into Wolfram it tells me it doesn't understand the equation and the closest answer it gives is $\displaystyle c^4m^2$

Well Wolfram Alpha is wrong since this equation is solvable and the solution is what you ask.... Just solve it yourself and forget Wolfram A. :)