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# Polynomial?

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I need a formula that monotonically decreases from f(0)=infinity to f(infinity)=0 and passes through the points f(a)=2 and f(b)=1, where a

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Actually, I tell a lie. A cubic would do. I'd realised this earlier but hadn't followed through properly. So, define f(x) = 1/g(x), g(x) = α x^3 + β x^2 + γ x, and impose f(a) = 2, f(b) = 1, with 0 < a < b. Then β = (b - 2a)/(2ab(a-b)) - (a+b)α ; γ = (2a^2-b^2)/(2ab(a-b))+ abα ; And α is chosen such that β^2 - 4 αγ < 0, which gives firstly (2a + b - 2 sqrt(2ab))/(2ab(a-b)^2) < α <...

12:22 Wed 03rd Nov 2021

In that case, any variation of f(x) = 1/x^n does the job. Then, e.g for n=1, f(1/2) = 2, f(1) = 1 satisfies your condition.

Is there another constraint you need? eg, specific values for a and b?

The "monotonically decreases" part makes finding a polynomial only solution at least trickier, because you have to avoid zeroes in the denominator, or at least push them over to the negative x axis. And with a, b, free then there's not a lot you can do, I don't think, to avoid at least *some* parameter space that has a real, positive solution to g(x) = 0.

-- answer removed --

So, define f(x) = 1/g(x), g(x) = α x^3 + β x^2 + γ x, and impose f(a) = 2, f(b) = 1, with 0 < a < b. Then

β = (b - 2a)/(2ab(a-b)) - (a+b)α ;

γ = (2a^2-b^2)/(2ab(a-b))+ abα ;

And α is chosen such that β^2 - 4 αγ < 0, which gives firstly

(2a + b - 2 sqrt(2ab))/(2ab(a-b)^2) < α < (2a + b + 2 sqrt(2ab))/(2ab(a-b)^2)

Not every choice of α in this region satisfies the "monotonic increasing" definition, but I think that there will usually exist at least one choice of α that works.

Still, we have a problem with *extremely* nontrivial constraints, and three free parameters, so I would have to study further and I've already spent more time than I should have when I have work to do. But hope this helps!

If you wanted to study it from here yourself, I'd recommend setting b = 1, say, so that 0 < a < 1, and α is still free, but the problem will be easier to study in this case as it's now a two-parameter problem.