On 26/07/2012 02:58 PM, Invisible wrote:

> So the solutions to one equation are sin and cos, and to the other are
> sinh and cosh. That's really pretty...

Of course, sin and sinh look totally unrelated on the real line. But in 
the complex plane, one is a trivially rotated version of the other. (The 
same goes for cos verses cosh.)

Wolfram|Alpha claims that f''=f actually has "the" solution

   f(t) = A exp(t) + B exp(-t)

I notice the following:

* If A=B=0 then we have f(t)=0.

* With A=1 and B=0, we have f(t)=exp(t).

* Setting A=B=1/2 gives us f(t)=cosh(t).

* Finally, A=1/2 and B=-1/2 gives us f(t)=cosh(t).

More tantalisingly, sinh and cosh are defined as

   sinh(x) = 1/2 [exp(x) - exp(-x)]
   cosh(x) = 1/2 [exp(x) + exp(-x)]

and sin and cos can be similarly defined as

   sin(x) = 1/2i [exp(ix) - exp(-ix)]
   cos(x) = 1/2  [exp(ix) + exp(-ix)]

This is the "trivial rotation in the complex plane" which I mentioned. 
So perhaps the most general solution to f''=-f is given by

   f(t) = A exp(it) + B exp(-it)

It's certainly an interesting idea.



Getting back to physics for a moment, we know that in the case of 
f''=-f, one solution is f(t) = sin(t). But would, say, f(t) = sin(2t) work?

Looking up my table of derivatives, it appears that we have

   f(t) = sin(2t)
   f'(t) = 2cos(2t)
   f''(t) = -4sin(2t)

So that doesn't work at all. It seems our system can oscillate at 
exactly one frequency only. But how able different amplitudes?

   f(t) = 2sin(t)
   f'(t) = 2cos(t)
   f''(t) = -2sin(t)

OK, so that works. It seems our system can oscillate at different 
amplitudes, but only one frequency. So how about different phases? We 
know that both sin and cos work, but how about a combination of them?

   f(t) = sin(t) + cos(t)
   f'(t) = cos(t) - sin(t)
   f''(t) = -sin(t) - cos(t)

OK, so that seems to work. So it appears that we have two basic 
solutions, sin(t) and cos(t), and /any/ linear combination of them is 
also a solution. That's interesting.

Does the same hold for the f''=f case? Let's look.

   f(t) = sinh(2t)
   f'(t) = 2cosh(2t)
   f''(t) = 4sinh(2t)

That doesn't work, as before.

   f(t) = 2sinh(t)
   f'(t) = 2cosh(t)
   f''(t) = 2sinh(t)

That still works.

   f(t) = sinh(t) + cosh(t)
   f'(t) = cosh(t) + sinh(t)
   f''(t) = sinh(t) + cosh(t)

So again, it appears that any linear combination of sinh(t) and cosh(t) 
should be a solution.

Alternatively, we can say that in one case the solutions are exp(x) and 
exp(-x) and in the other they are exp(ix) and exp(-ix), and that any 
linear combination of these works. This is slightly more fiddly, since 
in the second case we need to keep the solutions purely real (so the 
multiplication factors sometimes become imaginary).

Either way, it's a fascinating result...

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