Subject:
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Re: On a scale of 0 to 10?
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Wed, 26 Mar 2003 18:18:55 GMT
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Viewed:
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340 times
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In lugnet.off-topic.geek, David Eaton writes:
> In lugnet.off-topic.geek, Adrian Drake writes:
> > In lugnet.off-topic.geek, David Eaton writes:
> > > I've got a function that represents a quality rating, based on 2 parameters,
> > > which are on a scale of 0 to 1:
> > >
> > > (d-k)*k => quality (q)
*snip*
>
> Thought about that... If necessary I guess, but I'm pretty sure there's a
> way to do it. I think what I need is:
>
> s = Ae^(q + B) + C
>
> And at the moment, I've solved:
>
> C = -Ae^(B - 1)
> A = 10 / (e^(B + .25) - e^(B - 1))
>
> And I'm scratching my head like crazy to solve for B:
>
> e^(B + .25) + e^(B - 1) = 2e^B
>
> It's been too long since algebra! Anyone know how to solve for B here?
>
> DaveE
summed exponents like that can be re-written in this way:
e^B * e^.25 + e^B * e^-1 = 2e^B
(e^.25 + e^-1) * e^B = 2e^B
Uh oh. Divide both sides by e^B and we've got
(e^.25 + e^-1) = 2
1.652 = 2
Which ain't right. I think your math is wrong somewhere.
Adrian
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Message has 1 Reply:
 | | Re: On a scale of 0 to 10?
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| (...) Oh yeah! (...) Oh... yeah. (...) Hm. Double checked the math-- it appears solid [1], which would mean that there's no viable solution for: s = Ae^(q + B) + C for coordinates (-1,0), (0,5), (0.25,10). Darn. Hm. I guess that also means that any (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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Message is in Reply To:
| | Re: On a scale of 0 to 10?
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| (...) Thought about that... If necessary I guess, but I'm pretty sure there's a way to do it. I think what I need is: s = Ae^(q + B) + C And at the moment, I've solved: C = -Ae^(B - 1) A = 10 / (e^(B + .25) - e^(B - 1)) And I'm scratching my head (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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