# I Infinity question

1. Mar 24, 2017

### 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?

2. Mar 24, 2017

### mjc123

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

3. Mar 24, 2017

### Staff: Mentor

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.

4. Mar 24, 2017

### maka89

Not equals, mjc123 ;)

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

5. Mar 24, 2017

### Staff: Mentor

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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted