Formal expression required

[Jack]

I am trying to say the following, in formal terms:

" Any two functions, each of whose graph curve is exclusively convex*,
intersect at most twice".

* meaning that the curve starts off steep then gradually becomes flatter, at
no point violating this trend.

Can anyone help me with the mathematical language I need?

BTW I take it this is an identity that doesn't need a proof...?

With thanks in advance.

[Jussi Piitulainen]

Jack writes:

> I am trying to say the following, in formal terms:
>
> " Any two functions, each of whose graph curve is exclusively
> convex*, intersect at most twice".
>
> * meaning that the curve starts off steep then gradually becomes
> flatter, at no point violating this trend.
>
> Can anyone help me with the mathematical language I need?

It seems clear enough to me. I think it would be standard to call them
decreasing and convex instead of defining a new term.

If you required differentiability, you could define a real function f
as "exclusively convex" if its first derivative is everywhere negative
(f'(x) < 0 for all x) and monotonically increasing (f'(x) < f'(x + d)
for d > 0, say), and then claim that for two exclusively convex
functions f and g there are at most two such real x that f(x) = g(x).

> BTW I take it this is an identity that doesn't need a proof...?

I think it's false. It seems easy to sketch counterexamples.

[Jack]

"Jussi Piitulainen" <jpii...@ling.helsinki.fi> wrote in message
news:qot8v5v...@ruuvi.it.helsinki.fi...

> Jack writes:
>
>> I am trying to say the following, in formal terms:
>>
>> " Any two functions, each of whose graph curve is exclusively
>> convex*, intersect at most twice".
>>
>> * meaning that the curve starts off steep then gradually becomes
>> flatter, at no point violating this trend.
>>
>> Can anyone help me with the mathematical language I need?
>
> It seems clear enough to me.

Is convex a recognised mathematical term for a function, as described above?

I think it would be standard to call them
> decreasing and convex instead of defining a new term.
>
> If you required differentiability, you could define a real function f
> as "exclusively convex" if its first derivative is everywhere negative
> (f'(x) < 0 for all x) and monotonically increasing (f'(x) < f'(x + d)
> for d > 0, say), and then claim that for two exclusively convex
> functions f and g there are at most two such real x that f(x) = g(x).
>
>> BTW I take it this is an identity that doesn't need a proof...?
>
> I think it's false. It seems easy to sketch counterexamples.

I think it must be true. For the two functions, f and g where the gradient
of f(k) is greater than that of g(k) at lower k and less than that of g(k)
at higher k, there exists x for which, for all y < x,
f(x) - g(x) > f(y) - g(y)
and f(y) -g(y) is an increasing function of y, there being one such y for
which f(y) = g(y), and for all y > x, similarly
f(x) -g(x) > f(y) - g(y)
but f(y) - g(y) is a decreasing function of y, there being, again, one such
y for which f(y) = g(y). Can you falsify this?

With thanks.

[Frederick Williams]

f = 1/x and g roughly following f but meandering a bit so that g crosses
f more than twice. By 'a bit' I mean so as to ensure that g' < 0.

--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting

[Frederick Williams]

Jack wrote:
>
> I am trying to say the following, in formal terms:
>
> " Any two functions, each of whose graph curve is exclusively convex*,
> intersect at most twice".
>
> * meaning that the curve starts off steep then gradually becomes flatter, at
> no point violating this trend.

What does that mean? Do you want both second derivatives to be
constant?

[Stan Brown]

On Sun, 10 Mar 2013 15:48:52 -0000, Jack wrote:
>
> I am trying to say the following, in formal terms:
>
> " Any two functions, each of whose graph curve is exclusively convex*,
> intersect at most twice".
>
> * meaning that the curve starts off steep then gradually becomes flatter, at
> no point violating this trend.

I don't know "convex" as you are using it. Isn't the real issue that
the first derivative is either monotonically increasing but never
positive, or monotonically decreasing but never negative?

> BTW I take it this is an identity that doesn't need a proof...?

It _seems_ right (if I understand your language), but even if it is I
think it would need to be proved.

(It wouldn't be an identity in any case. An identity is an equation
that is true for all values of the variables, like (x+y)² = x² + 2xy
+ y². I think he word you want is "axiom", but I don't think it
qualifies at that level of obviousness, even if I'm not overlooking a
counterexample.)

--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com
Shikata ga nai...

[Jack]

"Stan Brown" <the_sta...@fastmail.fm> wrote in message
news:MPG.2ba6c1d71...@news.individual.net...

> On Sun, 10 Mar 2013 15:48:52 -0000, Jack wrote:
>>
>> I am trying to say the following, in formal terms:
>>
>> " Any two functions, each of whose graph curve is exclusively convex*,
>> intersect at most twice".
>>
>> * meaning that the curve starts off steep then gradually becomes flatter,
>> at
>> no point violating this trend.
>
> I don't know "convex" as you are using it. Isn't the real issue that
> the first derivative is either monotonically increasing but never
> positive, or monotonically decreasing but never negative?
>

I can't very well speak derivative-speak. By 'convex' or 'concave' I mean
that no two points on the curve have the same gradient.

What I am trying to say is that both curves follow a gradient that decreases
at a steadily decreasing rate; and for any x<y, the d(f(x))/d(f(y)) <
d(g(x))/d(g(y)), where d is the function denoting the gradient.

What do you think the best way to prove it would be? I would rather not use
calcukus, if that's possible.

>> BTW I take it this is an identity that doesn't need a proof...?
>
> It _seems_ right (if I understand your language), but even if it is I
> think it would need to be proved.
>
> (It wouldn't be an identity in any case. An identity is an equation
> that is true for all values of the variables, like (x+y)² = x² + 2xy
> + y².

I had imagined it would be OK to use this definition

"An identity is a relation which is tautologically true. This means that
whatever the number or value may be, the answer stays the same"
http://en.wikipedia.org/wiki/Identity_%28mathematics%29
-- which looks pretty much the same as definition of the word 'analytic' in
logic.

I think he word you want is "axiom", but I don't think it
> qualifies at that level of obviousness, even if I'm not overlooking a
> counterexample.)

I was once rebuked by a mathematcian for using the word 'axiom' in that
sense. I told him I thought it meant something that is self-evident, and he
said 'well it might have meant that *once*, but not any more! Not in
mathematics!'. My current understanding of the mathematical use of 'axiom'
is that it means an assumption.

[Frederick Williams]

Jack wrote:
>
> "Stan Brown" <the_sta...@fastmail.fm> wrote in message
> news:MPG.2ba6c1d71...@news.individual.net...
> > On Sun, 10 Mar 2013 15:48:52 -0000, Jack wrote:
> >>
> >> I am trying to say the following, in formal terms:
> >>
> >> " Any two functions, each of whose graph curve is exclusively convex*,
> >> intersect at most twice".
> >>
> >> * meaning that the curve starts off steep then gradually becomes flatter,
> >> at
> >> no point violating this trend.
> >
> > I don't know "convex" as you are using it. Isn't the real issue that
> > the first derivative is either monotonically increasing but never
> > positive, or monotonically decreasing but never negative?
> >
>
> I can't very well speak derivative-speak. By 'convex' or 'concave' I mean
> that no two points on the curve have the same gradient.

That is non standard. Convex functions may not have a derivative at
every point.

If it's true you can just call it a fact. (And if it isn't I guess you
won't need to call it anything.)

[Jussi Piitulainen]

Frederick Williams writes:

> f = 1/x and g roughly following f but meandering a bit so that g
> crosses f more than twice. By 'a bit' I mean so as to ensure that
> g' < 0.

That's a good description of the sketches I drew. It appears that Jack
wants to rule out such meandering by "gradient decreases in a steadily
decreasing rate" so I guess there is an improved statement that holds.

[Jussi Piitulainen]

Jack writes:
> Jussi Piitulainen wrote

> > Jack writes:
> >
> >> I am trying to say the following, in formal terms:
> >>
> >> " Any two functions, each of whose graph curve is exclusively
> >> convex*, intersect at most twice".
> >>
> >> * meaning that the curve starts off steep then gradually becomes
> >> flatter, at no point violating this trend.
> >>
> >> Can anyone help me with the mathematical language I need?
> >
> > It seems clear enough to me.
>
> Is convex a recognised mathematical term for a function, as
> described above?

Usually it means that you can connect any two points of the function
with a line segment that stays above the function.

[Jack]

I think this can be proven through the fact that the curve for g(x) will cut
through a circle at most twice. I am imagining the circle being broken at a
point beneath the segment cut by the curve, then being prised open to form
an arc, thence finally to be pulled into the shape of the curve for f(x).

Any way to do that without a pair of pliers, mathematicians?

With thanks in advance.

[Frederick Williams]

Jack wrote:
>
> I think this can be proven through the fact that the curve for g(x) will cut
> through a circle at most twice. I am imagining the circle being broken at a
> point beneath the segment cut by the curve, then being prised open to form
> an arc, thence finally to be pulled into the shape of the curve for f(x).

I don't get that. Can you post a link to a drawing?