# Thread: can anyone think of more downward sloping convex functions?

1. ## can anyone think of more downward sloping convex functions?

Are all negative sloped convex functions of this basic form?

$y=\frac{k}{x}$ or $y=\frac{k}{\ln x}$

(k=any positive real number)

Is there some theorem out there defining the general form of negative sloped convex functions?

Thanks

2. There's loads more. How about $1/x^{n},$ for $n>0?$ How about $e^{-x}?$ Don't forget linear combinations of functions such as these, although I'm not sure that linear combinations of downward-sloping convex functions are necessarily downward-sloping and convex. Still, I imagine you could find a number of examples.

I don't think you're going to be able to come up with a general formula for all downward-sloping convex functions. The issue here is that the most information you have is that

$f'(x)<0$ and $f''(x)>0.$

That is not enough information to determine $f.$

[EDIT] Absolutely LOVE your signature. That is a pet peeve of mine as well, which is one reason I detest research papers. One book you might find interesting is N. David Mermin's book Boojums All the Way Through.

3. Ackbeet-

Thanks for this response. Yes I see it was quite silly of me to think there were only those forms. Duh!

But it does seem that in every case x or some function of x is in the denominator of
f(x). Would it be rash to conclude that all downward sloping convex curves are of the form:

$y=\frac{k}{f(x)}$ ?

Glad you appreciate the signature! It's refreshing to see a mathematician say things like that from time to time.

4. Yes, it would be rash. What about functions that don't converge to y = 0, like

$y=\dfrac{1}{x}+1?$

$y=\dfrac{1+x}{x},$

but then you don't have a constant in the numerator as per your suggested form.

May I ask why you are trying to characterize all functions of these types? Maybe there's another way to approach your original problem (which you haven't stated, by the way. Maybe you could do that, please?)

Cheers.

5. Originally Posted by Ackbeet
May I ask why you are trying to characterize all functions of these types? Maybe there's another way to approach your original problem (which you haven't stated, by the way. Maybe you could do that, please?)

Cheers.
I am interested in the properties of the dark blue curve in the attached graph.

That curve is the locus of all points of tangency between a downward sloping convex function and a tangent line whose x-axis intercept is increasing.

The line is $y=-\frac{k}{m}x+k$

k, the y-axis intercept, is kept constant while m is increased, moving the x-axis intercept out.

The downward sloping convex function in this case is $y=\frac{a}{\frac{x}{a}-1}$

(where a is the scalar by which the function must be adjusted to remain tangent to the shifting line)

In this case the dark blue curve clearly has the following properties:

$f'(x)>0$ and $f''(x)>0$

It seems that in the case of functions of the form $y=\frac{a}{\left(\frac{x}{a}\right)^n}$ the dark blue curve is always a horizontal line.

Are there any downward sloping convex functions that would produce a negatively sloped dark blue curve? Or perhaps one that alternates between negative and positive slope?

These are the real questions I am interested in. Appreciate the help.

6. I think you meant that for the dark blue curve,

$f'(x)>0, f''(x)<0.$ Right?

I wonder if we can find the equation of the dark blue line. The tangent lines are, as you've said,

$y_{\tan}=-\dfrac{k}{m}\,x+k,$ and

$y_{\text{curve}}=\dfrac{a^{2}}{x-a}.$

We would like to determine $a=a(k,m).$ To do this, we equate the functions:

$-\dfrac{k}{m}\,x+k=\dfrac{a^{2}}{x-a}.$

We solve this for $x$ to obtain

$x=\dfrac{1}{2}\left(a+m\pm\dfrac{\sqrt{a^{2}(k-4m)-2akm+km^{2}}}{\sqrt{k}}\right).$

However, we want there to be only one solution, arguing from a graphical standpoint. Therefore, the square roots in the numerator must vanish, which implies that

$a^{2}(k-4m)-2akm+km^{2}=0,$ with solution for $a$ of

$a=\dfrac{\sqrt{k}\,m}{\sqrt{k}\pm 2\sqrt{m}}.$

There's a lot more to this problem. It's a difficult one, and I'm not seeing something here.

7. Originally Posted by Ackbeet
I think you meant that for the dark blue curve,

$f'(x)>0, f''(x)<0.$ Right?
Yes, forgive me.

Originally Posted by Ackbeet
I wonder if we can find the equation of the dark blue line.
I plotted the blue curve parametrically, as the explicit solution is way too long to fit into my grapher.

$Y(t)=\frac{2kt-k\sqrt{kt}}{\abs{4t-k}}$

$X(t)=\frac{t\sqrt{kt}+2t^2-kt}{\abs{4t-k}}$

where t=m
This plot is good only up to where the denominator is 0.

In solving for "a", the one thing you are missing is that you have $\dfrac{a^{2}}{x-a}.$ where it should be $\dfrac{a}{\frac{x}{a}-1}$. Following the same steps, then, you arrive at the positive root:

$a=\frac{2m\sqrt{km}-km}{4m-k}$

Originally Posted by Ackbeet
There's a lot more to this problem. It's a difficult one, and I'm not seeing something here.
I have gotten as far as I have only because I am using a CAS to solve my equations. Making the downward sloping curve only slightly more complicated, such as $\dfrac{a}{(\frac{x}{a})^2-1}$, the solution for "a" and for the parametric equations becomes MUCH more intractible, even with a CAS.

Not wanting to abuse your good will, I would be content if, just intuitively, you would offer your best guess as to:

1) If, for any downward sloping convex curve with positive asymptotes, the slope of the dark blue curve will always be 0 or positive.
2) In the event that it is possible for the dark blue slope to be negative, does it seem to you there might be cases where it will also alternate between negative and positive? Or can we rule this possibility out completely?

Thank you

8. Well, isn't it true that

$\dfrac{a}{\dfrac{x}{a}-1}=\dfrac{a}{\dfrac{x}{a}-\dfrac{a}{a}}=\dfrac{a}{\dfrac{x-a}{a}}=a\,\dfrac{a}{x-a}=\dfrac{a^{2}}{x-a}?$

But, more importantly, I think you can get a negatively-sloped dark blue curve. Try a function like this on for size:

$y=\frac{a^{2}}{x-a}-5a.$

Here, the vertical asymptotes are moving to the right with increasing $a,$ and the horizontal asymptotes are moving down with increasing $a.$ But you want the downward movement of the horizontal asymptotes to be faster than the right-ward movement of the vertical asymptote. Hence the 5. I haven't worked this one through, but let me know what you get.

9. Originally Posted by Ackbeet
Well, isn't it true that

$\dfrac{a}{\dfrac{x}{a}-1}=\dfrac{a}{\dfrac{x}{a}-\dfrac{a}{a}}=\dfrac{a}{\dfrac{x-a}{a}}=a\,\dfrac{a}{x-a}=\dfrac{a^{2}}{x-a}?$

Originally Posted by Ackbeet
But, more importantly, I think you can get a negatively-sloped dark blue curve. Try a function like this on for size:

$y=\frac{a^{2}}{x-a}-5a.$

Here, the vertical asymptotes are moving to the right with increasing $a,$ and the horizontal asymptotes are moving down with increasing $a.$ But you want the downward movement of the horizontal asymptotes to be faster than the right-ward movement of the vertical asymptote. Hence the 5. I haven't worked this one through, but let me know what you get.
I find what you say about asymptotes interesting and will give it further thought. However, I did the calculations using the function you suggested and the locus of the points of tangency for this function does not yield a downward sloping dark blue curve (as shown in attachment).

10. That's weird. It looks like the vertical asymptotes are increasing, as well as the horizontal! Are you sure you've used a (-5a) at the end there? If so, try a positive 5a instead.

11. Originally Posted by Ackbeet
That's weird. It looks like the vertical asymptotes are increasing, as well as the horizontal! Are you sure you've used a (-5a) at the end there? If so, try a positive 5a instead.
Aha! Yes indeed. I had graphed with a +5 instead of -5.

When I do the graph with a -5, as you originally suggested, the slope is negative. Not only that, but then it also hits an inflection point and becomes positive. (Graph attached)

But can this result be achieved for curves that are completely within the first quadrant? The curve you suggested strays over into negative territory.

In any case, thanks very much!