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Dickfore said:What about:
<br /> \sin^{\frac{1}{\sqrt{2}}}{(\alpha)}<br />
Just as in the Newtons Law example, if we had F^{\sqrt{1/2}}=m^{\sqrt{1/2}}a^{\sqrt{1/2}}, it would be something that could be "fixed". If not, then its not a good physics equation.
Dickfore said:Me not finding an example does not constitute a proof of your claims. It is up to you to show that there is no such law. Also, you still had not addressed the question of:
<br /> B(p, q) = B(q, p)<br />
True, but if do find an example, let me know. About the B(p,q)=B(q,p), the p and q are exponents of the trig functions in the integral, so its very similar to the \sin^x(\theta) problem. Just because the exponents are "bad" doesn't mean they cannot be "fixed". Take a definition of the Beta function:
B(p,q)=\int_0^{\pi/2} \sin^{2p+1}(\theta)\cos^{2q+1}(\theta)\,d\theta
If we treat theta as a directed quantity, we write it as \theta 1_x and the sine is directed the same way, the cosine is dimensionless, the d\theta is directed the same way and the direction of the Beta function is 1_x^{2p+2}=(1_x^2)^{p+1} which is dimensionless. Defining \theta =\phi +\pi/2 we get \sin(\theta 1_x)=1_x \cos(\phi 1_x) and \cos(\theta 1_x)=-1_x \sin(\phi 1_x) (i.e. still dimensionless) so that B(p,q) is equal in value to B(q,p) and is again directed as 1_x^{2p+2} (i.e. dimensionless). But if we take your original definition (replacing p with (p+1)/2 etc.)
B\left(\frac{p+1}{2},\frac{q+1}{2}\right)=2\int_0^{\pi/2} \sin^p(\theta)\cos^q(\theta)\,d\theta
And now its directed as 1_x^{p+1} which is not well defined unless p+1 takes on certain values (like 1/3, but not 1/2, as in the sin(alpha) example). I have not checked, but I bet the directions are not consistent for the second case either. But the point is, a "bad" statement like the second case can be "fixed" to be good, like the first case. If Siano's extension is valid, then the Beta function will not occur in any physically meaningful equation in such a way that it cannot be "fixed" somehow, just like the Newtons law case.
Dickfore said:I just noticed you `invented' a new dimension - cycle.
A cycle is 2 pi radians, so if radians are not dimensionless, then neither are cycles. Not an invention, just a consequence of saying radians are not dimensionless.