Polynomial?
https://www.theanswerbank.co.uk/Science/Question1771392.html
Question posted by Rev. Green on 15:18 Sun 31st Oct 2021 in The AnswerBank.enThe AnswerBankMon, 20 May 2024 21:18:58 +0100Science1800... is less than b. I'm using a formula involving logarithms. Is there a quotient of polynomials?
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LALASciencehttps://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13055333Sun, 31 Oct 2021 14:34:16 +0000f(0) can't "equal" infinity, and f("infinity") can't "equal" zero, so I assume you mean "tends to" in both cases.
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?
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LALASciencehttps://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13057610Tue, 02 Nov 2021 20:28:45 +0000A, B, and X are given. F(A) = 2, F(B) = 1. I'd like to know F(X). The formula I have is something like (B/X) to the power of ((log2)/log(A/B)). You are correct, if A happens to be half B the answer is trivial.
https://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13057695
LALASciencehttps://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13057695Tue, 02 Nov 2021 21:50:34 +0000The function f(x) = 1/g(x) for g(x) = x[(b-2a)x + (2a^2 - b^2)] does the job as long as sqrt(2)a < b < 2a. That's the best I can come up with for now.
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.
https://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13057999
LALASciencehttps://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13057999Wed, 03 Nov 2021 10:18:46 +0000Actually, 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 < (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.
https://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13058046
LALASciencehttps://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13058046Wed, 03 Nov 2021 11:22:07 +0000Thanks for all your work Jim. Sixty years ago I'd have been able to find a good approximation before you could say "Pafnuty Chebyshev", but not now. But you are correct, it is easier to work on the inverse: f(0)=0; f(A)=0.5; f(B)=1 and all can be scaled to let B=1. At present, all of the values of X will be close to B, so a straight line fit through (A/B, 0.5) and (1,1) will be good enough.
https://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13058230
LALASciencehttps://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13058230Wed, 03 Nov 2021 15:12:44 +0000If you're going to use a linear fit after all that I'm going to get peeved :P
What's this even for, anyway?
https://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13058347
LALASciencehttps://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13058347Wed, 03 Nov 2021 18:12:13 +0000Sorry Jim. I hoped someone with a more nimble brain than mine would give a simple formula or say that one was impossible. However, you'll be happy to know how important it is: each day a sudoku is published and two target times for solving it are given, one (A) for two points and one (B) for one point. The times vary greatly from day to day, as does their ratio. X is then a measure of (inverse) senility. I'm down to about 0.9 now.
https://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13058427
LALASciencehttps://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13058427Wed, 03 Nov 2021 19:38:08 +0000Ah, that's a relief, I thought for a second it was going to be something so unimportant it would make me question my decision to spend an hour or more on it :/
https://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13058490
LALASciencehttps://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13058490Wed, 03 Nov 2021 21:10:25 +0000I read your answers Jim
How is Croatia?
https://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13058496
LALASciencehttps://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13058496Wed, 03 Nov 2021 21:22:11 +0000Fine, but I'll address it in its separate post at some point. Until then, stay tuned...
https://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13058498
LALASciencehttps://www.theanswerbank.co.uk/Science/Question1771392.html?answer=13058498Wed, 03 Nov 2021 21:27:38 +0000