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Math Help - Re-Arranging

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by TheOnlyNameLeft View Post
    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
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  3. #3
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    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.
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  4. #4
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    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.
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  5. #5
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    Quote Originally Posted by TheOnlyNameLeft View Post
    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.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    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
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  7. #7
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    Quote Originally Posted by topsquark View Post
    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})
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  8. #8
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    Quote Originally Posted by CaptainBlank View Post
    or:

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

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