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Question

I would like to show for arbitrary $n \in \mathbb{N}$, that the polynomial: $$ p(x) := \sum_{k=1}^{n-1} (n-k) x^k $$ is non-decreasing.

Start Of Solution

We can write its derivative by: $$ p'(x) = \frac{(-(x + 1) (x^n-1) + n (x - 1) (x^n + 1))}{(x - 1)^3}, $$ as $(-1 + x)^3 < 0$ for $x \in ]-1,1[$ it suffices to show for arbitrary $n$ that: $$ n(x-1) (x^n + 1) \leq (x+1)(x^n - 1), $$ for $n = 1$ this is trivial (as both sides become equal). Thus we assume the inequality to hold for $n$ and show it for $n+1$, here I use $x^{n+1} + 1 = x(x^n+1) + (1-x)$ to use the induction hypothesis but I don't achieve the inequality through this method.

No positive roots

By Descartes' Sign rule we see that $p'(x) = \sum\limits_{k=1}^{n-1}(n-k)k x^{k-1}$ has no positive roots, thus it suffices to show

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    How do you get the formula for $p'$? – 2017-01-03
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    Mathematica to be honest. – 2017-01-03
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    Using your previous (now deleted) comment: $(x^n - 1) = (x-1) (1+\dots+x^{n-1})$, I have narrowed it down to showing: $ \frac{\frac{1}{2} + x + \dots + x^{n-1} + \frac{x^n}{2}}{n} \leq \frac{x^n+1}{2} $. Trying to show this by induction now. – 2017-01-03
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    Okay now I just need to show that $2x^n \leq x^{n+1} + 1$. – 2017-01-03
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    But this last inequality doesn't hold for $n$ even – 2017-01-03

1 Answers 1

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I found it I think. Let's replace $n$ by $n+1$ to obtain: $$ p(x) = \sum_{k=1}^n(n+1-k)x^k, $$ now we let the sum go from $0$ to $n-1$: $$ p(x) = \sum_{k=0}^{n-1} (n-k)x^{k+1}, $$ taking the derivative yields: $$ p'(x) = \sum_{k=0}^{n-1}(n-k)(k+1)x^k $$ it now suffices to show that $p'(x) \geq 0$ on $]-1,1[$.

One can check (for example with mathematica) that we have: $$ p'(x) = \frac{n(x^{n+1} + 1) - 2 (x^n + \dots + x))}{(x-1)^2}, $$

thus we just need to show that $$\frac{(x^{n+1} + 1)}{2} \leq \frac{x^n+\dots+x}{n}.$$ For $x \in ]-1,0]$ we have $\frac{(x^{n+1} + 1)}{2} \geq 0$ while $ \frac{x^n+\dots+x}{n} = x(1-x^n)/(1-x) \leq 0$, so we may assume $x \in [0,1[$. We do this part by induction skipping the case $n = 1$. We have: $$ \frac{x^{n+1} + \dots + x}{n+1} = \frac{nx \frac{x^n + \dots + x}{n} + x}{n+1} \leq \frac{nx \frac{x^{n+1} + 1}{2}+x}{n+1} \overset{?}{\leq} \frac{x^{n+2} + 1}{2}. $$ It remains to show that the last inequality holds. Subtracting the left hand side from the right hand side we obtain: $$ \frac{(x^{n+2} - ((n+2)x - n-1)}{2(m+1)}, $$ which is positive as $x^{n+2}$ is convex and the tangent line at $1$ is given by $y = ((n+2)x - n -1)$. This completes the proof.