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 19:06:10 GMT
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Viewed:
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352 times
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In lugnet.off-topic.geek, Adrian Drake writes:
> 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
Oh yeah!
> (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.
Oh... yeah.
> I think your math is wrong somewhere.
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 similar equation is out too, in the format:
s = AD^(q + B) + C
Hmm. Is there really no mathematical way to do this? That just goes against
every math-fiber in my body! There's *got* to be a way! Should I instead be
solving for:
s = AB^x + C ?
or some other wacky formula? What gives?
Still pondering,
DaveE
[1] for the truly bored at heart, my math:
s = Ae^(q + B) + C
Given:
10 = Ae^(.25 + B) + C
5 = Ae^(0 + B) + C
0 = Ae^(-1 + B) + C
Solve for C:
0 = Ae^(-1 + B) + C
C = -Ae^(B - 1)
Solve for A:
10 = Ae^(.25 + B) + C
10 = Ae^(B + .25) - Ae^(B - 1)
A = 10 / (e^(B + .25) - e^(B - 1))
Solve for B:
5 = Ae^(0 + B) + C
5 = (10e^B) / (e^(B + .25) - e^(B - 1)) - Ae^(B - 1)
5 = (10e^B) / (e^(B + .25) - e^(B - 1)) - (10e^(B - 1)) / (e^(B + .25) -
e^(B - 1))
e^(B + .25) - e^(B - 1) = 2e^B - 2e^(B - 1)
e^(B + .25) + e^(B - 1) = 2e^B
(e^B)*(e^.25) + (e^B)*(e^-1) = (e^B)*2
e^.25 + e^-1 = 2
1.284 + .368 = 2
WRONG!
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Message has 1 Reply:
 | | Re: On a scale of 0 to 10?
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| Hi all. The reason the equation is unsolvable is because B is an unneeded constant. Why? A*e^(q+B)=A*(e^q)*(e^B) It is impossible to distinquish between A & e^B. What you need to solve is an equation of the form s=A*exp(B*q) + C This is the form of (...) (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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| (...) *snip* (...) 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 (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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