Jon P. answered • 02/09/15
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Are you asking what the identity is?
xp - 1 = (x - 1) (xp-1 + xp-2 + ... + 1)
You can see this in the more commonly seen identities:
x2 - 1 = (x - 1) (x + 1)
x3 - 1 = (x - 1) (x2 + x + 1)
The proof can be done by mathematical induction, but if you multiply out the first few terms of the expression on the right and gather like terms, you'll see that every power of x other than p and 0 is canceled out.
Is that what you wanted to know?
Jon P.
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Oh, I see. You want to know how to get from
(x - 1){q(x^(p-1) + x^(p-2) + ... + 1) - p(x^(q-1) + x^(q-2) + ... + 1)} > 0
to
(x - 1){q(x^(p-1) + x^(p-2) + ... + x^q) - (p - q)(x^(q-1) + x^(q-2) + ... + 1)} > 0
Is that it?
It's like this:
(x - 1){q(x^(p-1) + x^(p-2) + ... + 1) - p(x^(q-1) + x^(q-2) + ... + 1)} > 0
Split the first main term into two parts:
(x - 1){q(x^(p-1) + x^(p-2) + ... + x^q) + q(x^(q-1) + x^(q-2) + ... + 1) - p(x^(q-1) + x^(q-2) + ... + 1)} > 0
Move the second part of the split first main term to be the last term:
(x - 1){q(x^(p-1) + x^(p-2) + ... + x^q) - p(x^(q-1) + x^(q-2) + ... + 1) + q(x^(q-1) + x^(q-2) + ... + 1) } > 0
Now there are THREE main terms that are multiples of polynomials in x. However, notice that the polynomials are the same for the second and third terms. So factor it out:
(x - 1){q(x^(p-1) + x^(p-2) + ... + x^q) - (p - q)(x^(q-1) + x^(q-2) + ... + 1) } > 0
That's the line you were attempting to get to.
The reason this depends on p being > q is that you can only split the original first main term the way I described if there are more terms in the p-1 degree polynomial in x than in the q-1 degree polynomial in x, in the inequality you are starting from.
I hope I'm answering the right question this time!
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02/10/15
Gerry H.
Yep, that's it. Perfect. Many thanks for your help Jon.
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02/10/15
Gerry H.
Hello Jon,
Many thanks for your reply. No I'm afraid it doesn't answer my question. I refer to the top of page 43 of Chrystal's Algebra vol 2. It is a justification for an inequality and I want to know exactly how he gets the last line shown in my question.
02/09/15