1. ## 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!

2. 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

3. 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.

4. 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.

5. 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.

6. 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

7. 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})

8. Originally Posted by CaptainBlank
or:

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

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