proof of inequalities for difference of powers
1 First Inequality
We have the factorization
un-vn=(u-v)n-1∑k=0ukvn-k-1. |
Since the largest term in the sum is is un-1 and the smallest is
vn-1, and there are n terms in the sum, we deduce the following
inequalities:
n(u-v)vn-1<un-vn<n(u-v)un-1 |
2 Second Inequality
This inequality is trivial when x=0. We split the rest of the proof into two cases.
2.1 -1<x<0
In this case, we set u=1 and v=1+x in the second inequality above:
1-(1+x)n<n(-x) |
Reversing the signs of both sides yields
nx<(1+x)n-1 |
2.2 0<x
In this case, we set u=1+x and v=1 in the first inequality above:
nx<(1+x)n-1 |
3 Third Inequality
This inequality is trivial when x=0. We split the rest of the proof into two cases.
3.1 -1<x<0
Start with the first inequality for differences of powers, expand the left-hand side,
nuvn-1-nvn<un-vn, |
move the vn to the other side of the inequality,
nuvn-1-(n-1)vn<un, |
and divide by vn to obtain
nuv-n+1<(uv)n. |
Taking the reciprocal, we obtain
(vu)n<vv+n(u-v)=1-n(u-v)v+n(u-v) |
Setting u=1 and v=1+x, and moving a term from one side to the other, this becomes
(1+x)n-1<nx1-(n-1)x. |
3.2 0<x<1/(n-1)
Start with the second inequality for differences of powers, expand the right-hand side,
un-vn<nun-nun-1v |
move terms from one side of the inequality to the other,
nun-1v-(n-1)un<vn |
and divide by un to obtain
nvu-n+1<(vu)n |
When the left-hand side is positive, (i.e. nv>(n-1)u) we can take the reciprocal:
(uv)n<uu-n(u-v)=1+n(u-v)u-n(u-v) |
Setting u=1+x and v=1, and moving a term from one side to the other, this becomes
(1+x)n-1<nx1-(n-1)x |
and the positivity condition mentioned above becomes (n-1)x<1.
Title | proof of inequalities for difference of powers |
---|---|
Canonical name | ProofOfInequalitiesForDifferenceOfPowers |
Date of creation | 2013-03-22 15:48:45 |
Last modified on | 2013-03-22 15:48:45 |
Owner | Mathprof (13753) |
Last modified by | Mathprof (13753) |
Numerical id | 14 |
Author | Mathprof (13753) |
Entry type | Proof |
Classification | msc 26D99 |