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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Fri, 28 Mar 2003 17:29:27 GMT
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Viewed:
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497 times
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> Whew.
>
> Of course, now here's a totally different question. In order to get that
> point, I cheated. I couldn't solve:
>
> e^(B/4) + e^(-B) = 2
>
> using algebra, but using other means, I figured out that B has 2 roots: 0
> and 2.62502392. How would your average math wiz solve for B above?
>
> DaveE
I don't see an easy way to do this. I tried substituting
x=e^B
Then I get a polynomial of degree 5. Only the positive roots lead to real
values of B. (since e^any real number > 0)
This is solvable, but very difficult to do by hand, especially since all the
roots are positive (I didn't solve, but I used Descartes' rule of signs to
figure out how positive & negative roots there are. There are no negative
roots, so they all must be positive, although some could be repeats).
I also went back to the original equation and points, and tried to solve
that by eliminating B first and then solving for A and C. This time, I get
a polynomial of degree 4. Again, not easy to solve.
Iteration might be best, i.e. try a set of A, B, & C, see if they're close
in point one. Plug that into point 2 and adjust B only. Plug the new A, B,
& C into point 3 and adjust C only. Plug those new A,B,&C into point 1 and
adjust A only. Keep going until you reach the tolerances that you want (2
digits, 3 digits, etc).
I tried this and came up with similar values of A, B, & C as you did. If
you're using 3 significant digits, your values are quite good. This is
pretty easy to do using a spreadsheet program, such as MS Excel or Works.
It's even easier if you can write a short program in C++, Java, or just
about any other programming language. I didn't bother writing a program,
but I've done it in the past for similar problems.
Even math wizs would have trouble with this equation. This is because the
conditions (points) that you chose are not the traditional conditions for
solving this expression. The conditions are usually initial (q=0), final
(q=infinity) and the original differential equation. The initial and final
points are independant of B, thus giving quick solutions ot A & C. The
diffeq provides an easy way to solve for B, since this is the equation that
relates the rate/slope to position. 1/B is usually referred to as a time
constant.
It's still solvable, but as you saw, without the traditional conditions,
it's a lot more work.
Anyway, you have a solution now. Hope your rating system works as planned.
John
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Message is in Reply To:
 | | Re: On a scale of 0 to 10?
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| (...) aHA! Much oblige! I now get (approx): s = 5.39 * e^(2.625q) - 0.39 Whew. Of course, now here's a totally different question. In order to get that point, I cheated. I couldn't solve: e^(B/4) + e^(-B) = 2 using algebra, but using other means, I (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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