Thread: alternative generalisation of binominals law

1. alternative generalisation of binominals law

I have problems with proving some generalisation of binominal law.
It is commonly known that

$\displaystyle x^2-2x+1\geqslant 0$

But, I have problems with proving that

$\displaystyle x^{k+1}-(k+1)x+k >= 0$

is true for $\displaystyle k\geqslant0$.
Can anybody help me?

2. Hi

The inequality is true for $\displaystyle x \geq 0$
For instance for k=2 : $\displaystyle x^3-3x+2 \leq 0$ for x=-3

There are several ways to show the inequality

You can study $\displaystyle f(x) = x^{k+1}-(k+1)x+k$
f is differentiable and $\displaystyle f'(x) = (k+1)(x^k-1)$
The minimum of f is obtained for x=1 and f(1)=0 therefore $\displaystyle f(x) \geq 0$

Second way : since f(1)=0 and f'(1)=0, f(x) can be factorized by (x-1)²
You can show that $\displaystyle f(x) = (x-1) \left(\sum_{j=1}^{k}x^j-k\right)$
and $\displaystyle f(x) = (x-1)^2 \left(\sum_{j=0}^{k-1}(k-j)x^j\right)$
Both factors are positive for x positive

3. Originally Posted by running-gag
Hi

The inequality is true for $\displaystyle x \geq 0$
For instance for k=2 : $\displaystyle x^3-3x+2 \leq 0$ for x=-3

There are several ways to show the inequality

You can study $\displaystyle f(x) = x^{k+1}-(k+1)x+k$
f is differentiable and $\displaystyle f'(x) = (k+1)(x^k-1)$
The minimum of f is obtained for x=1 and f(1)=0 therefore $\displaystyle f(x) \geq 0$

Second way : since f(1)=0 and f'(1)=0, f(x) can be factorized by (x-1)²
You can show that $\displaystyle f(x) = (x-1) \left(\sum_{j=1}^{k}x^j-k\right)$
and $\displaystyle f(x) = (x-1)^2 \left(\sum_{j=0}^{k-1}(k-j)x^j\right)$
Both factors are positive for x positive
Nice.

A third way: induction over k. The case k = 2 has already been established.

Suppose now that
$\displaystyle x^k -kx +k-1 \geq 0$
Then
$\displaystyle x^k \geq kx - k + 1$
Multiplying by x, since $\displaystyle x \geq 0$,
$\displaystyle x^{k+1} \geq kx^2 -kx + x\geq k (2x-1) -kx + x$ ... by the k=2 case
$\displaystyle x^{k+1} \geq (k+1)x -k$
$\displaystyle x^{k+1} -(k+1)x + k \geq 0$
QED