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John Henry 03-30-2004 08:36 AM

Maths Problem
 
I want, if possible, to rearrange this equation in terms of C:

A = B * (1+C)^D - E * ((1+C)^D-1) / C

I've got a feeling it can't be done (I should really be able to tell, with an A-level in maths and an MPhys) so at the moment I'm evaluating it with numerical methods, but that's slow. I was just wondering if there were any mathematical geniuses out there who might know a way to do it, or could at least confirm my fears.

I had wondered about using a substitution of (1+C)^D=F, giving

C=F^(1/D)-1

and

A=BF - EF/C - E

then multiplying through by F and rearranging to get a quadratic, then solving that using what my maths teacher used to call "The Equation", giving:

F=(E/C-A)/(E/C+E)

then rearranging, subbing C back in and multiplying through by C, leaving us with:

(E+BC)(C+C^2)^D-AC+E=0

And that's as far as I can go. Don't know if this is solvable from here. My gf just suggested that integration might help (0 to D dC), but I can't work it out. Also, I don't know if I've worked it out properly so far as my algebra is often dodgy.

Thanks in advance for any help you can provide, people.

gal 03-30-2004 08:55 AM

You can't solve this for C analytically. Go for the numerical approach.

kutulu 03-30-2004 10:23 AM

At first I thought you could go about it using log properties to get rid of the exponents, but that just ends up in a big mess.

Kadath 03-30-2004 10:58 AM

Yeah, I'd say you can't sort that out. But your English-isms are excellent. Maths, dodgy...excellent!

KnifeMissile 03-30-2004 11:50 AM

Through a little bit of algebraic manipulation, you can express this equation like so:

E = (CB - E)(1 + C)^D - CA

As you can see, this is a polynomial of order D + 1, so there exists a general algerbraic solution if and only if D <= 4. Otherwise, you'll have to resort to a numerical approach (which really isn't all that bad, is it?)...

Yakk 03-30-2004 12:56 PM

As an aside
A=BF - EF/C - E
is wrong, it is actually
A=BF - EF/C - E/C
every line after that probably has related errors. Not that important, but still. . .

telekineticfool 03-30-2004 02:42 PM

Quote:

Originally posted by John Henry
I want, if possible, to rearrange this equation in terms of C:

A = B * (1+C)^D - E * ((1+C)^D-1) / C

I've got a feeling it can't be done (I should really be able to tell, with an A-level in maths and an MPhys) so at the moment I'm evaluating it with numerical methods, but that's slow. I was just wondering if there were any mathematical geniuses out there who might know a way to do it, or could at least confirm my fears.

I had wondered about using a substitution of (1+C)^D=F, giving

C=F^(1/D)-1

and

A=BF - EF/C - E

then multiplying through by F and rearranging to get a quadratic, then solving that using what my maths teacher used to call "The Equation", giving:

F=(E/C-A)/(E/C+E)

then rearranging, subbing C back in and multiplying through by C, leaving us with:

(E+BC)(C+C^2)^D-AC+E=0

And that's as far as I can go. Don't know if this is solvable from here. My gf just suggested that integration might help (0 to D dC), but I can't work it out. Also, I don't know if I've worked it out properly so far as my algebra is often dodgy.

Thanks in advance for any help you can provide, people.

I just ran it through TI89's SOLVE operation to solve for C and it couldn't solve for C....it output the following:

(cb-e)*(c+1)^d-ca+e=0

John Henry 03-31-2004 03:08 AM

Thanks for the help, folks. Now that we've pretty comprehensively decided it can't be solved analytically, has anybody got any ideas on the best kind of numerical method for this kind of search-space?

At the moment we're using the bipolar method, simply because it was the easiest thing to code, but this is fairly slow as we need a high degree of accuracy. B and D are fixed for any given search and we are varying values of C to find a set A.

We have tried using linear interpolation, but that seemed to take even longer. Not sure if it might be best to use a combination of two different methods or something.

Yakk 03-31-2004 08:34 AM

John, clarify?

Which values are fixed, and which are you trying to solve for.

John Henry 04-02-2004 01:11 AM

Trying to solve for C. A, B and D are fixed.

Yakk 04-02-2004 07:53 AM

A = B * (1+C)^D - E * ((1+C)^D-1) / C
A B and D are fixed
Solve for C
E is free

A = B * (1+C)^D - E * ((1+C)^D - 1/ C)

Um, mate, given those restrictions, any non-zero value of C is a solution, short of a handful of singularity cases.

Let A B D and C be arbitrary numbers, with C not equal to 0, and (1+C)^D not equal to 1/C

Then:
Let
X := A
Y := B * (1+C)^D
Z := (1+C)^D - 1/ C

Then:
X = Y - E * Z
or
E = (Y-X)/Z

QED

I think I have the wrong question.

John Henry 04-02-2004 03:03 PM

*ahem* it would appear that I've been a bit of a twat. Erm... E is fixed too. Sorry I forgot about E.


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