Continued Fractions are Trig ID's within Waves

Ok, you guy's were right about these convergents. Thanks, I can now not only explain it much clearer but have tested it and it all checks out. I've just sent an email to a maths professor friend at Sydney University, John Mack, asking him to have a look at it.

Take the ratio a/b where a > b and a/b is either coprime or irrational. If [1/1, p/q,....a/b] represents the convergents of the continued fraction for a/b then the following identity holds:

f(t) = sin(2piat) + sin(2pibt) =

sin(2pi (p((a + b)/(p + q))t + ((aq - pb)/(p + q))t) +
sin(2pi (q((a + b)/(p + q))t - ((aq - pb)/(p + q))t).

Since p and q are integers, then the original wave can be seen as the composite of two 'simpler' waves, the pth and qth harmonics of fundamental GCD = ((a + b)/(p + q)), with an equal and opposite time-dependent phase shift in each component. (Observe that the term (aq - pb) in the 'remainder' equals the determinant. I call them the remainder because they are the difference between the p((a + b)/(p + q)) and q((a + b)/(p + q)) original a, b frequencies. Observe also that we could probably rewrite this more simply in exponential form as

f(t) = exp[i2pi((aq - pb)/(p + q))t]*exp[i2pi (p((a + b)/(p + q))t] + exp[-i2pi((aq - pb)/(p + q))t]*exp[i2pi (q((a + b)/(p + q))t].

But I haven't checked this out properly yet).

Using the standard trigonometric identity sinA + sinB = 2sin((A + B)/2)cos((A - B)/2), we obtain

f(t) = 2sin(pi(a + b)t)cos(pi[(p - q)(a + b)/(p + q) + 2((aq - pb)/(p + q))]t).

With p/q = 1/1, (p - q) = 0, (aq - pb) = (a - b) and (p + q) = 2 giving

f(t) = 2sin(pi(a + b)t)cos(pi(a - b)t).

Thus, this standard trig ID appears to be a special case of a more general set of identities. Interestingly, since the sine part is always (a + b) then we now have many more amplitude modulations. If there are N convergents, then there will be N - 1 such amplitude mods. Since N -> infinity for irrationals then there are, theoretically speaking, infinitely many.

One more thing; The extrema of each (p, q) follows what seems to be a new type of envelope

+/- 2cos(2pi((aq - pb)/(p + q)]t +/- pi((2k + 1)/(2(p + q))).

This I suspect will be an important part of any proof.

I haven't found the proof yet but I'm sure it's only a matter of time. I've also been looking around for previous work but haven't found anything, so please let me know if you know of someone who has been here before me or done similar work.

Thanks again,

Rick

PS: Given f(t) = 2sin(pi(a + b)t)cos(pi[(p - q)(a + b)/(p + q) + 2((aq - pb)/(p + q))]t), the cos values here are all equal to (a - b) (of course). I meant to say that the slow moving 2((aq - pb)/(p + q)) term, which can be taken out, represents a type of modulated amplitude frequency in itself.

These equations are becoming hard to keep track of when formatted this
way and could use some simplification. Does anyone know of any handy
online "pretty print" software or something similar?

I also recommend that we just drop the 2*pi terms here, there's no
point. All of these expressions:

sin(2*pi*4*t) + sin(2*pi*5*t)
sin(4t) + sin(5t)
sin(one million * 4 * t) + sin(one million * 5 * t)
cos(4t) + cos(5t)

are pretty much equivalent in terms of what you're trying to do here,
since they're just denoting the same interval being transposed up and
down the frequency spectrum.

Also, what exactly do you need proven here? It looks like you have it
pretty much worked out.

-Mike

On Fri, May 21, 2010 at 1:58 AM, rick <rick_ballan@yahoo.com.au> wrote:
>
>
>
> Ok, you guy's were right about these convergents. Thanks, I can now not only explain it much clearer but have tested it and it all checks out. I've just sent an email to a maths professor friend at Sydney University, John Mack, asking him to have a look at it.
>
> Take the ratio a/b where a > b and a/b is either coprime or irrational. If [1/1, p/q,....a/b] represents the convergents of the continued fraction for a/b then the following identity holds:
>
> f(t) = sin(2piat) + sin(2pibt) =
>
> sin(2pi (p((a + b)/(p + q))t + ((aq - pb)/(p + q))t) +
> sin(2pi (q((a + b)/(p + q))t - ((aq - pb)/(p + q))t).
>
> Since p and q are integers, then the original wave can be seen as the composite of two 'simpler' waves, the pth and qth harmonics of fundamental GCD = ((a + b)/(p + q)), with an equal and opposite time-dependent phase shift in each component. (Observe that the term (aq - pb) in the 'remainder' equals the determinant. I call them the remainder because they are the difference between the p((a + b)/(p + q)) and q((a + b)/(p + q)) original a, b frequencies. Observe also that we could probably rewrite this more simply in exponential form as
>
> f(t) = exp[i2pi((aq - pb)/(p + q))t]*exp[i2pi (p((a + b)/(p + q))t] + exp[-i2pi((aq - pb)/(p + q))t]*exp[i2pi (q((a + b)/(p + q))t].
>
> But I haven't checked this out properly yet).
>
> Using the standard trigonometric identity sinA + sinB = 2sin((A + B)/2)cos((A - B)/2), we obtain
>
> f(t) = 2sin(pi(a + b)t)cos(pi[(p - q)(a + b)/(p + q) + 2((aq - pb)/(p + q))]t).
>
> With p/q = 1/1, (p - q) = 0, (aq - pb) = (a - b) and (p + q) = 2 giving
>
> f(t) = 2sin(pi(a + b)t)cos(pi(a - b)t).
>
> Thus, this standard trig ID appears to be a special case of a more general set of identities. Interestingly, since the sine part is always (a + b) then we now have many more amplitude modulations. If there are N convergents, then there will be N - 1 such amplitude mods. Since N -> infinity for irrationals then there are, theoretically speaking, infinitely many.
>
> One more thing; The extrema of each (p, q) follows what seems to be a new type of envelope
>
> +/- 2cos(2pi((aq - pb)/(p + q)]t +/- pi((2k + 1)/(2(p + q))).
>
> This I suspect will be an important part of any proof.
>
> I haven't found the proof yet but I'm sure it's only a matter of time. I've also been looking around for previous work but haven't found anything, so please let me know if you know of someone who has been here before me or done similar work.
>
> Thanks again,
>
> Rick
>
>

Yeah I agree Mike, it would be very handy being able to simplify the expressions. I'm just used to thinking in terms of the real frequencies with 2pi, but as you said it's really superfluous.

I also think I have it covered already, come to think of it. Thanks for saying so. However, I did make one mistake in the last message. The envelopes these curves are following were missing a (p - q) term. The correct value is (in my old 'pi' language, and with +/- meaning + or -):

+/- 2cos[2pi(aq - pb)t/(p + q) +/- pi((p - q)(2k + 1)/2(p + q))).

So, for first term, p/q = 1/1, the angle here is 0 and the formula reduces to the usual +/-2cos(pi(a - b)t). As I said it's interesting that for N convergents there are N - 1 of these envelopes, the usual amp mod just being the first.

-Rick

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> These equations are becoming hard to keep track of when formatted this
> way and could use some simplification. Does anyone know of any handy
> online "pretty print" software or something similar?
>
> I also recommend that we just drop the 2*pi terms here, there's no
> point. All of these expressions:
>
> sin(2*pi*4*t) + sin(2*pi*5*t)
> sin(4t) + sin(5t)
> sin(one million * 4 * t) + sin(one million * 5 * t)
> cos(4t) + cos(5t)
>
> are pretty much equivalent in terms of what you're trying to do here,
> since they're just denoting the same interval being transposed up and
> down the frequency spectrum.
>
> Also, what exactly do you need proven here? It looks like you have it
> pretty much worked out.
>
> -Mike
>
>
> On Fri, May 21, 2010 at 1:58 AM, rick <rick_ballan@...> wrote:
> >
> >
> >
> > Ok, you guy's were right about these convergents. Thanks, I can now not only explain it much clearer but have tested it and it all checks out. I've just sent an email to a maths professor friend at Sydney University, John Mack, asking him to have a look at it.
> >
> > Take the ratio a/b where a > b and a/b is either coprime or irrational. If [1/1, p/q,....a/b] represents the convergents of the continued fraction for a/b then the following identity holds:
> >
> > f(t) = sin(2piat) + sin(2pibt) =
> >
> > sin(2pi (p((a + b)/(p + q))t + ((aq - pb)/(p + q))t) +
> > sin(2pi (q((a + b)/(p + q))t - ((aq - pb)/(p + q))t).
> >
> > Since p and q are integers, then the original wave can be seen as the composite of two 'simpler' waves, the pth and qth harmonics of fundamental GCD = ((a + b)/(p + q)), with an equal and opposite time-dependent phase shift in each component. (Observe that the term (aq - pb) in the 'remainder' equals the determinant. I call them the remainder because they are the difference between the p((a + b)/(p + q)) and q((a + b)/(p + q)) original a, b frequencies. Observe also that we could probably rewrite this more simply in exponential form as
> >
> > f(t) = exp[i2pi((aq - pb)/(p + q))t]*exp[i2pi (p((a + b)/(p + q))t] + exp[-i2pi((aq - pb)/(p + q))t]*exp[i2pi (q((a + b)/(p + q))t].
> >
> > But I haven't checked this out properly yet).
> >
> > Using the standard trigonometric identity sinA + sinB = 2sin((A + B)/2)cos((A - B)/2), we obtain
> >
> > f(t) = 2sin(pi(a + b)t)cos(pi[(p - q)(a + b)/(p + q) + 2((aq - pb)/(p + q))]t).
> >
> > With p/q = 1/1, (p - q) = 0, (aq - pb) = (a - b) and (p + q) = 2 giving
> >
> > f(t) = 2sin(pi(a + b)t)cos(pi(a - b)t).
> >
> > Thus, this standard trig ID appears to be a special case of a more general set of identities. Interestingly, since the sine part is always (a + b) then we now have many more amplitude modulations. If there are N convergents, then there will be N - 1 such amplitude mods. Since N -> infinity for irrationals then there are, theoretically speaking, infinitely many.
> >
> > One more thing; The extrema of each (p, q) follows what seems to be a new type of envelope
> >
> > +/- 2cos(2pi((aq - pb)/(p + q)]t +/- pi((2k + 1)/(2(p + q))).
> >
> > This I suspect will be an important part of any proof.
> >
> > I haven't found the proof yet but I'm sure it's only a matter of time. I've also been looking around for previous work but haven't found anything, so please let me know if you know of someone who has been here before me or done similar work.
> >
> > Thanks again,
> >
> > Rick
> >
> >
>

--- In tuning-math@yahoogroups.com, "rick" <rick_ballan@...> wrote:
>
> Yeah I agree Mike, it would be very handy being able to simplify the expressions. I'm just used to thinking in terms of the real frequencies with 2pi, but as you said it's really superfluous.

You can write your identities in term of S(x) = sin(2 pi x) and C(x) =
cos(2 pi x) if you prefer. I think it would help some.