Prove a Polynomial has no real roots

[lfdahl]

Prove that polynomials of the form:\[P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-...-2nx+2n+1, \: \: n = 1,2,...\]- have no real roots.

 

[Opalg]

Prove that polynomials of the form:\[P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-...-2nx+2n+1, \: \: n = 1,2,...\]- have no real roots.

[sp]Let $Q_n(x) = (x-1)^2(x^{2n-2} + 2x^{2n-4} + 3x^{2n-6} + \ldots + n) + (n+1)$. Then $Q_n(x)>0$ for all $x$. (In fact, $Q_n(x)$ has minimum value $n+1$, when $x=1$, because each term in the expression for $Q_n(x)$ is an even power of either $x$ or $x-1$.) So $Q_n(x)$ has no real roots. It will therefore be sufficient to show that $P_n(x) = Q_n(x).$

To prove that by induction, the base case $P_1(x) = Q_1(x) = x^2 - 2x + 3$ is easy to check.

Suppose that $P_n(x) = Q_n(x)$ for some $n$. Then $$\begin{aligned} P_{n+1}(x) &= x^{2n+2}-2x^{2n+1}+3x^{2n}- \ldots -(2n+2)x + (2n+3) \\ &= x^2\bigl(x^{2n}-2x^{2n-1}+3x^{2n-2}- \ldots + (2n+1)\bigr) -(2n+2)x + (2n+3) \\ &= x^2P_n(x) -(2n+2)x + (2n+3) \\ &= x^2Q_n(x) -(2n+2)x + (2n+3) \\ &= x^2(x-1)^2(x^{2n-2} + 2x^{2n-4} + 3x^{2n-6} + \ldots + n) + (n+1)x^2 -(2n+2)x + (2n+3) \\ &= (x-1)^2(x^{2n} + 2x^{2n-2} + 3x^{2n-4} + \ldots + nx^2) + (n+1)(x-1)^2 + (n+2) \\ &= (x-1)^2\bigl(x^{2n} + 2x^{2n-2} + 3x^{2n-4} + \ldots + nx^2 + (n+1)\bigr) + (n+2) \\ &= Q_{n+1}(x) .\end{aligned}$$ That completes the inductive step.[/sp]

 

[lfdahl]

[sp]Let $Q_n(x) = (x-1)^2(x^{2n-2} + 2x^{2n-4} + 3x^{2n-6} + \ldots + n) + (n+1)$. Then $Q_n(x)>0$ for all $x$. (In fact, $Q_n(x)$ has minimum value $n+1$, when $x=1$, because each term in the expression for $Q_n(x)$ is an even power of either $x$ or $x-1$.) So $Q_n(x)$ has no real roots. It will therefore be sufficient to show that $P_n(x) = Q_n(x).$

To prove that by induction, the base case $P_1(x) = Q_1(x) = x^2 - 2x + 3$ is easy to check.

Suppose that $P_n(x) = Q_n(x)$ for some $n$. Then $$\begin{aligned} P_{n+1}(x) &= x^{2n+2}-2x^{2n+1}+3x^{2n}- \ldots -(2n+2)x + (2n+3) \\ &= x^2\bigl(x^{2n}-2x^{2n-1}+3x^{2n-2}- \ldots + (2n+1)\bigr) -(2n+2)x + (2n+3) \\ &= x^2P_n(x) -(2n+2)x + (2n+3) \\ &= x^2Q_n(x) -(2n+2)x + (2n+3) \\ &= x^2(x-1)^2(x^{2n-2} + 2x^{2n-4} + 3x^{2n-6} + \ldots + n) + (n+1)x^2 -(2n+2)x + (2n+3) \\ &= (x-1)^2(x^{2n} + 2x^{2n-2} + 3x^{2n-4} + \ldots + nx^2) + (n+1)(x-1)^2 + (n+2) \\ &= (x-1)^2\bigl(x^{2n} + 2x^{2n-2} + 3x^{2n-4} + \ldots + nx^2 + (n+1)\bigr) + (n+2) \\ &= Q_{n+1}(x) .\end{aligned}$$ That completes the inductive step.[/sp]

Thankyou very much, Opalg for another brilliant answer!

Spoiler

I am curious . Please explain how on Earth you came up with the expression for $Q_n(x)$.
Thankyou in advance.

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An alternative approach:

Spoiler

Note, that $P_n(x) > 0$ for $x \le 0$, since the terms with negative coefficients are multiplied by odd powers of $x$.

Now,
\[P_n(x)+xP_n(x) = x\left ( x^{2n}-x^{2n-1}+x^{2n-2}-...-x+1 \right )+2n+1\]
- so
$$P_n(x) = x\frac{x^{2n+1}+1}{(x+1)^2}+\frac{2n+1}{x+1} \Rightarrow P_n(x) > 0,\: \: x>0.\]
Done.

 

[Opalg]

Spoiler

I am curious . Please explain how on Earth you came up with the expression for $Q_n(x)$.

[sp]That just came from looking at what happens for small values of $n$. You soon find that each $P_n(x)$ (for $n=1,2,3$) has its minimum value when $x=1$, and that this minimum value is $n+1$. That leads to the discovery that $P_n(x) - (n+1)$ has a factor $(x-1)^2$. The quotient of $P_n(x) - (n+1)$ by that factor is $Q_n(x).$[/sp]

Spoiler

$$P_n(x) = x\frac{x^{2n+1}+1}{x+1}+2n+1\Rightarrow P_n(x) > 2n+1,\: \: x>0.$$

[sp]That should be $$(1+x)P_n(x) = x\frac{x^{2n+1}+1}{x+1}+2n+1$$, which certainly shows that $P_n(x)>0$ for $x>0$. But it is not so clear from that expression what the minimum value of $P_n(x)$ should be (in fact, it is $n+1$, not $2n+1$).[/sp]

 

[lfdahl]

[sp]That should be $$(1+x)P_n(x) = x\frac{x^{2n+1}+1}{x+1}+2n+1$$, which certainly shows that $P_n(x)>0$ for $x>0$. But it is not so clear from that expression what the minimum value of $P_n(x)$ should be (in fact, it is $n+1$, not $2n+1$).[/sp]

Spoiler

Oh, yes. I´m sorry for my typo ... I´ll edit my solution right away.
Thankyou, also for explaining the idea that led to the expression of $Q_n(x)$.