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Homework Help: Proof of x^n - y^n =

  1. Oct 26, 2013 #1

    The problem to be solved is [itex]x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y +...+xy^{n-2} + y^{n-1})[/itex]. The first thing I noticed was that the term on the RHS side of the equation could possibly be expressed as an infinite series. And so, I undertook to find the general term:

    I noticed that what was being summed was a product, the first term being [itex]x^{n-1}y^0[/itex], and the last [itex]x^0y^{n-1}[/itex], thus my general term needed to at least capture this.

    Finally, [itex](x^{n-2} + x^{n-1}y +...+xy^{n-2} + y^{n-1}) = \sum_{i=0}^{n-1} x^{n-1-i}y^i[/itex]

    [itex] \sum_{i=0}^{n-1} x^{n-1-i}y^i = x^{n-1} \sum_{i=0}^{n-1} x^{-i}y^i = x_{n-1} \sum_{i=0}^{n-1} (\frac{y}{x})^i[/itex]

    This I knew to be the geometric series, the point at which I went wrong:

    [itex]x_{n-1} \sum_{i=0}^{n-1} (\frac{y}{x})^i = x^{n-1} (\frac{1-(\frac{y}{x})^n}{1-\frac{y}{x}})[/itex]

    I substituted in n rather than n-1. This, however, wasn't the most disconcerting thing, as I am sure you found it disconcerting: even with the incorrect substitution, I was able to properly solve the problem.

    Why is that so? How can I fix this?
    Last edited: Oct 26, 2013
  2. jcsd
  3. Oct 26, 2013 #2


    Staff: Mentor

    The first and second terms in the second factor have incorrect exponents (and you're missing a right parenthesis). These terms should be xn -1 and xn - 2y.

    The terms in the second factor are NOT an infinite series. This is a finite sum with n terms, with the same n that appears on the left side.
  4. Oct 26, 2013 #3
    If the purpose of the problem was to show that this is an identity you first have to have a correct identity and I think you have a typo: you wrote ##x^{n-2} + x^{n-1}y ...##. Hopefully you meant ##x^{n-1} + x^{n-2}y ...##

    To show it is true you have 2 choices:

    1. You can divide ##(x^n -y^n)## by (x-y) and show you get the sum of the terms

    2. You can multiply out the right hand side and combine like terms to show that you get the left hand side.

    Both these choices involve ordinary algebra, and your fancier computations are not required.

    If the problem is something else, you were not clear about that.

    A note: When you are using summation signs you must be very careful to keep your indices straight. Even your first sum is not written correctly.
  5. Oct 26, 2013 #4

    Ray Vickson

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    [tex]t_n = x^{n-1} + x^{n-2} y + \cdots + x y^{n-2} + y^{n-1},[/tex]
    so you want to prove that ##x^n - y^n = (x-y) t_n.## This almost cries out for the use of induction on n.
    Last edited: Oct 26, 2013
  6. Oct 27, 2013 #5
    One could certainly use induction, but division is easier.
  7. Oct 27, 2013 #6

    Ray Vickson

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    Yes, you already mentioned division. I am mentioning another method that the OP may, or may not prefer. I do not know what the OP knows or does not know. If he does not know about induction my suggestion is not helpful; if he does not know how to do (algebraic) division, your suggestion will not be helpful to him.
  8. Oct 27, 2013 #7
    Yes, Ray, all true.
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