Re: extensions of Bezout's theorem to non-integer powers?

In article <1133807736.072787.270880@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
c.rowat@xxxxxxxxxx <c.rowat@xxxxxxxxxxxxx> wrote:

> I am working with
>
> a*x^b - c*x^d + f*y^b - g*y^d
>
> where x is a real variable, and y a (real) linear function of x; a-g
> are all real;b may be negative but the rest are positive.
>
> Is there a good reference for results on the number of zeros that such
> functions have?

I don't know about a reference but I think we can count the zeros.
Your subject line refers to non-integer powers. I don't know what
x^d will mean for e.g. d = pi if x<0, so probably you mean to
restrict your attention to _positive_ values of x and y.
If you didn't mean that, you'll have to adapt what I say below;
the principles don't change.


I think the answer is that there can be at most 7 (x,y) pairs in
the first quadrant where the curve meets the line, irrespective
of the values of a-g. A sharper bound may also exist though it's
easy to see that there can be 4 such points for some choices of
the line, at least for some a-g.


I'll assume for simplicity that b is also positive.

Your description of the problem is to count the (first-quadrant)
solutions to a pair of equations { F(y) = G(x) , y = L(x) } where
L is linear and each of F and G have the form
F(z) = a z^b - c z^d.
Such functions are smooth on the positive reals and have one critical
point (at z = (ab / cd)^1/(d-b) ). In particular, for any constant
k there are either no solutions to F(z) = k or else there are two,
one on either side of the critical point. The two cases are distinguished
by the sign of k - F( crit.pt. ) ; whether "too high" or "too low" is
the problem giving no solutions depends only on which of { b, d } is
larger, and in particular will be the same for F and for G .
For concreteness, let me assume that d > b so that F''( crit. pt.) < 0
and so F has a local max there; F(z) = k has solutions iff k is
smaller than some bound.

Now it's easy to draw the curve F(y) = G(x) in the xy plane.
We simply plot the points (x,y) for which F(y) = G(x) = k, for
one value of k at a time. As above, there are no points to plot
until k is below the lower of the two peaks (of F and G); at
that value of k we will typically get two points (e.g. we must
have y = critical value for F but can have two solutions to G(x) = k).
Qualitatively this curve must have the following form: there is an arc
which starts at the origin, increases both coordinates until one of
them has reached a critical point (say, for F(y)) , then doubles
back to the (in this case, y-) axis. There is another branch of
the curve that passes from far away on the other (x-) axis,
decreases x and increases y again until the critical value of y,
then turns back and increases both coordinates without bound.

Now that you've got that curve drawn, just superimpose the other
curve, which you state is a straight line. This line can easily intersect
each of the two components of the prior graph twice, giving up to
four points of intersection.


The qualititative picture is not enough to bound the number of
points of intersection -- yes, one can prove that the slopes of the curve
are positive or negative where they appear to be, but it is consistent
with that information for the curve to wiggle gently around the line
and so to have many points of intersection. That requires the existence
of inflection points, whose locations can be found by setting
d^2y/dx^2 = 0, that second derivative computed via the Implicit Function
Theorem as a sum of multiples of a few powers of x and y. But
then we have an even more complicated kind of intersection to worry
about than the one initially given, so this doesn't seem like a
productive way to count the number of intersections without knowledge
of the specific values of the constants.


But we can find an absolute upper bound on the number of intersections.


Lemma 1: There is at most one positive solution to a binomial equation
A1 x^r1 + A2 x^r2 = 0
Proof: find it!


Lemma 2: There are at most two positive solutions to a trinomial equation
A1 x^r1 + A2 x^r2 + A3 x^r3 = 0
Proof: Divide by the lowest power of x, then apply Rolle's theorem
and Lemma 1.


Cor.: In the same way we show (by induction on k ) that the equation
A1 x^r1 + A2 x^r2 + ... + A_k x^{r_k} = 0
has at most k positive real solutions.


Lemma 3: There are at 7 most positive solutions to
A1 (x^a + A2) = ( x^b + A3 )( x - A4 )^c
Proof: Write this product equation multiplicatively: we are counting the
locations where F(x) = 1 where F(x) =
[ (A1) ( x^a + A2 ) ] / [ ( x^b + A3 ) ( x - A4 ) ]
By Rolle's theorem, between any two lies a zero of the derivative, which
we can compute with the quotient rule; equivalently (using "logarithmic
differentiation") we know the derivative is F(x) itself times
a x^(a-1) / (x^a + A2) - b x^(b-1) / (x^b + A3) - c / (x - A4)
whose terms may be collected into a fraction with numerator
a x^(a-1) (x^b +A3)(x-A4) - b x^(b-1)(x^a +A2)(x-A4) - c (x^a +A2)(x^b +A3)
which is a linear combination of these seven powers of x:
x^(a+b) x^(a+b-1) x^a x^(a-1) x^b x^(b-1) 1
By the Corollary to lemma 2, this can vanish for at most 7 values of x.

Remark: I am ignoring the possibility that the both sides of the original
equation vanish because that would require specific combinations of signs
of the A_i which don't occur below. Also note that the result holds
equally well with A4 - x replacing x - A4 (only one of which will
make sense for a given fractional value of c ).


Now suppose there are positive values of x,y making both

a*x^b - c*x^d + f*y^b - g*y^d = 0 and y = m*x + h [*]

Writing y = x z, we cancel x^b and (setting k = d-b ) obtain
a - c*x^k + z^b*( f - g z^k* x^k) = 0 and x*(z-m) = h
i.e x^k = ( a + f * z^b )/(c + g * z^d ) and x = h/(z-m).
Thus each of these two curves shows a unique x for each z, and
moreover the points of intersection have z coordinates making
h^k * (c + g*z^d) - (a + f*z^b) * (z-m)^k
vanish. (For fractional k this only makes sense if z > m; if z<m,
write the equation arising from the straight line as x = h' / (m-z)
instead.) With some rearrangements of terms we get an equation of the
form shown in Lemma 3, so there are at most seven possible positive
solutions z to this equation, giving at most seven positive points
where the graphs of x^k = ( a + f * z^b )/(c + g * z^d ) and x = h/(z-m)
meet, i.e. seven points in the first quadrant where [*] holds.


I didn't check to see whether this bound is sharp --- I was mostly
pleased to note the existence of an upper bound independent of the
exponents (a sort of Descartes' Rule of Signs, if you like).


dave
.

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