Proving a polynomial to be positive for all real values

Question

I'm trying to prove the following theorem: $$\sum_{j=0}^{2n} (-x)^j >0\ \forall x \in \mathbb{R}, n \in \mathbb{N} $$ I have verified this theorem for $n=1$ (just a quadratic) and for $n=2$ (by simple factoring). However I'm stuck for higher values of $n$ and generalizing it. I'm thinking of proving this by proving it to be positive for one value of $x$ ($x=0$ should suffice) and then proving it has no real roots. I'm not sure how to do this though.

Answer 1

If $x=-1$, it is trivial. Otherwise, $$ \sum_{j=0}^{2n}(-x)^j=\frac{x^{2n+1}+1}{x+1}, $$ and:

• if $x>-1$, the numerator and the denominator of that fraction are both greater than $0$;
• if $x<-1$, the numerator and the denominator of that fraction are both smaller than $0$.

Answer 2

Perhaps the most straightforward way is to compare individual terms.

First, the trivial cases. If $x<0$ all terms are positive, and if $x=0$ the first term is positive and the rest are $0$. So we just need to deal with $x>0$.

Now the terms are alternating positive and negative. Also the absolute values are either increasing ($x>1$), decreasing ($x<1$) or all equal. If they are increasing then

$$\sum_{j=0}^{2n}(-x)^j=\left(x^{2n}-x^{2n-1}\right)+\cdots+\left(x^{2}-x^{1}\right)+x^0,$$ which is positive because the difference of each pair is positive, as is the unpaired term. If they are decreasing then similarly you can write $$\sum_{j=0}^{2n}(-x)^j=\left(x^{0}-x^{1}\right)+\cdots+\left(x^{2n-2}-x^{2n-1}\right)+x^{2n},$$ with the same conclusion. If $x=1$ then either of these works (the pairs all contribute $0$ but the remaining term is $1$).

Answer 3

Using the principle of mathematical induction, as suggested by @Digitallis in the comments, we prove the stronger statement that, for real $x$ and natural $n$, $$\sum_{j=0}^{2n} (-x)^j \ge \frac{1}{2}.$$

(1) Base step: We have \begin{align*} 1-x+x^2 &= (x-1/2)^2+3/4 \ge 3/4 \ge 1/2. \end{align*}

(2) Inductive step: We assume that $\sum_{j=0}^{2k}(-x)^j \ge 1/2$ for some $k\ge 1$. We must show that $\sum_{j=0}^{2(k+1)}(-x)^j \ge 1/2$. But
\begin{align*} \sum_{j=0}^{2(k+1)}(-x)^j &= 1-x+\sum_{j=2}^{2n+2}(-x)^j \\ &= 1-x+x^2\sum_{j=0}^{2n}(-x)^j \\ &\ge 1-x+\frac12 x^2 \\ &= \frac12(x-1)^2+\frac12 \\ &\ge \frac12. \end{align*}