Find t to Satisfy e^iat=e^ia_0, e^ibt=e^ib_0, e^ibct=e^ic_0

[maka89]

If there is no upper limit on t, can you find a t such that: $e^{iat} = e^{ia_0}$, $e^{ibt} = e^{ib_0}$ and $e^{ibct} = e^{ic_0}$ at the same time?

No matter what a,b and c is, though given a != b , a!=c, b!=c and a!= 0, b!= 0, c!=0

Or maybe rather:
$at=a_0 +k_12\pi$, $bt=b_0 +k_22\pi$ and $ct=c_0 +k_32\pi$, where the k's are integers

I think it seems reasonable that you can, or at least come arbitrarily close to the equations being satisfied... But don't know how to prove it, or if I am right... Any pointers?

 

[mjc123]

Are you really talking factorials up there, or were you trying to type "not equals" (≠)?

 

[mfb]

Not in general.

Consider a=2, b=1, a₀=0 and b₀ = 0.1. Clearly t=kπ solves the first equation, but does not solve or even approximate the second one.

If you require that a,b,c do not have a pair which have a common multiple, there should be approximations to arbitrary precision. In general there won't be an exact solution here either.

 

[maka89]

Not equals, mjc123 ;)

Thanks mfb! Forgot to mention, I assumed a is not a multiple of b etc.

 

[mfb]

I assumed a is not a multiple of b etc.

That is not general enough.
a=2 and b=3 still lead to a common multiple of 6, and the same result.

If we don't have that case, in general all you get is an approximation.