Maths Problem - Tilted Forum Project Discussion Community

John Henry · 03-30-2004, 08:36 AM

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.

03-30-2004, 08:36 AM	#1 (permalink)
John Henry Addict Location: Grey Britain	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. __________________ "No one was behaving from very Buddhist motives. Then, thought Pigsy, he was hardly a Buddha, nor was he a monkey. Presently, he was a pig spirit changed into a little girl pretending to be a little boy to be offered to a water monster. It was all very simple to a pig spirit." Last edited by John Henry; 03-30-2004 at 08:57 AM..

03-30-2004, 10:58 AM	#4 (permalink)
Kadath Muffled Location: Camazotz	Yeah, I'd say you can't sort that out. But your English-isms are excellent. Maths, dodgy...excellent! __________________ it's quiet in here

03-30-2004, 12:56 PM	#6 (permalink)
Yakk Wehret Den Anfängen! Location: Ontario, Canada	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. . . __________________ Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.

03-31-2004, 03:08 AM	#8 (permalink)
John Henry Addict Location: Grey Britain	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. __________________ "No one was behaving from very Buddhist motives. Then, thought Pigsy, he was hardly a Buddha, nor was he a monkey. Presently, he was a pig spirit changed into a little girl pretending to be a little boy to be offered to a water monster. It was all very simple to a pig spirit."

03-31-2004, 08:34 AM	#9 (permalink)
Yakk Wehret Den Anfängen! Location: Ontario, Canada	John, clarify? Which values are fixed, and which are you trying to solve for. __________________ Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.

03-30-2004, 08:55 AM	#2 (permalink)
gal Insane	You can't solve this for C analytically. Go for the numerical approach.

03-30-2004, 10:23 AM	#3 (permalink)
kutulu Junkie	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.

03-30-2004, 11:50 AM	#5 (permalink)
KnifeMissile Location: Waterloo, Ontario	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?)...

04-02-2004, 01:11 AM	#10 (permalink)
John Henry Addict Location: Grey Britain	Trying to solve for C. A, B and D are fixed. __________________ "No one was behaving from very Buddhist motives. Then, thought Pigsy, he was hardly a Buddha, nor was he a monkey. Presently, he was a pig spirit changed into a little girl pretending to be a little boy to be offered to a water monster. It was all very simple to a pig spirit."

04-02-2004, 07:53 AM	#11 (permalink)
Yakk Wehret Den Anfängen! Location: Ontario, Canada	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. __________________ Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.

04-02-2004, 03:03 PM	#12 (permalink)
John Henry Addict Location: Grey Britain	ahem it would appear that I've been a bit of a twat. Erm... E is fixed too. Sorry I forgot about E. __________________ "No one was behaving from very Buddhist motives. Then, thought Pigsy, he was hardly a Buddha, nor was he a monkey. Presently, he was a pig spirit changed into a little girl pretending to be a little boy to be offered to a water monster. It was all very simple to a pig spirit."