# Re-Arranging

• Mar 10th 2007, 09:16 AM
TheOnlyNameLeft
Re-Arranging
Hi, I'm stuck on what is probably a very basic re-arranging of an equation but I just can't do it!

Basically, I need to find B in terms of Y:

Y = (B^1.4 - 1) / (B - 1)

Y is actually something a bit more complicated but it's known so it doesn't matter.

Thanks!
• Mar 12th 2007, 10:02 PM
CaptainBlack
Quote:

Originally Posted by TheOnlyNameLeft
Hi, I'm stuck on what is probably a very basic re-arranging of an equation but I just can't do it!

Basically, I need to find B in terms of Y:

Y = (B^1.4 - 1) / (B - 1)

Y is actually something a bit more complicated but it's known so it doesn't matter.

Thanks!

I doubt that there is a smple rearrangement for this

RonL
• Mar 13th 2007, 08:46 AM
Richard Rahl
I plugged it into Maple and got something back that was very confusing, I can post it here if you wish, but I'm not sure I understand it.
• Mar 13th 2007, 04:50 PM
TheOnlyNameLeft
Yeah no harm in putting it up if you can. :)

MATLAB couldn't solve it either. Like the way it's become high school maths now aswell. :D
• Mar 13th 2007, 07:24 PM
ThePerfectHacker
Quote:

Originally Posted by TheOnlyNameLeft
Hi, I'm stuck on what is probably a very basic re-arranging of an equation but I just can't do it!

Basically, I need to find B in terms of Y:

Y = (B^1.4 - 1) / (B - 1)
!

Let, A=B^(.2)

Then we have,
(A^6-1)/(A-1)=Y
Divide through,
1+A+A^2+A^3+A^4+A^5=y

That is a Cycotomic polynomial. It is irreducible.
• Mar 14th 2007, 04:07 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
Let, A=B^(.2)

Then we have,
(A^6-1)/(A-1)=Y
Divide through,
1+A+A^2+A^3+A^4+A^5=y

That is a Cycotomic polynomial. It is irreducible.

This is a clever substitution. But how on Earth did you get that form in terms of A?

Y = (A^7 - 1)/(A^{1/0.2} - 1) = (A^7 - 1)/(A^5 - 1)

Y = (A^6 + A^5 + A^4 + A^3 + A^2 + A + 1)/(A^4 + A^3 + A^2 + A + 1)
if I did that right.

-Dan
• Mar 14th 2007, 06:57 AM
ThePerfectHacker
Quote:

Originally Posted by topsquark
This is a clever substitution. But how on Earth did you get that form in terms of A?

I used the identity,

x^n-y^n=(x-y)(x^{n-1}y+...+xy^{n-1})
• Mar 14th 2007, 07:39 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlank
or:

(A^7-1)/(A^5-1)=Y

RonL

It seems I made another mistake (that is typical of me).