MHB CaptainBlacks Occasional Problem #3

[CaptainBlack]

Prove that the polynomials:

\[P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1),\ \ \ (n=1,2...)\]

have no real roots.

CB

 

[Opalg]

Still no replies to this one? I started by graphing $P_1(x)$ and $P_2(x)$, both of which have a positive global minimum at $x=1$ (and hence no real roots). That made me look for a factor of $(x-1)^2$, and I found that

$P_1(x) = x^2-2x+3 = (x-1)^2+2,$

$P_2(x) = x^4-2x^3+3x^2-4x+5 = (x^2+2)(x-1)^2+3,$

$P_3(x) = x^6-2x^5+3x^4-4x^3+5x^2-6x+7 = (x^4+2x^2+3)(x-1)^2+4.$

That should be enough of a pattern to suggest a solution. (Wink)

 

[alexmahone]

Still no replies to this one? I started by graphing $P_1(x)$ and $P_2(x)$, both of which have a positive global minimum at $x=1$ (and hence no real roots). That made me look for a factor of $(x-1)^2$, and I found that

$P_1(x) = x^2-2x+3 = (x-1)^2+2,$

$P_2(x) = x^4-2x^3+3x^2-4x+5 = (x^2+2)(x-1)^2+3,$

$P_3(x) = x^6-2x^5+3x^4-4x^3+5x^2-6x+7 = (x^4+2x^2+3)(x-1)^2+4.$

That should be enough of a pattern to suggest a solution. (Wink)

Claim: $P_n(x)=(x-1)^2R_n(x)+n+1$

Let $Q_n(x)=P_n(x)-n-1=x^{2n}-2x^{2n-1}+3x^{2n-2}-\ldots-2nx+n$

$Q_n(1)=1-2+3-\ldots-2n+n=0$

$Q_n'(x)=2nx^{2n-1}-2(2n-1)x^{2n-2}+3(2n-2)x^{2n-3}-\ldots-2n$

$Q_n'(1)=2n-2(2n-1)+3(2n-2)-\ldots-2n=0$ (1st term cancels the last term, 2nd term cancels the 2nd last term etc.)

$Q_n(1)=Q_n'(1)=0\implies Q_n(x)=(x-1)^2R_n(x)$

So, $P_n(x)=(x-1)^2R_n(x)+n+1$

Now we just need to prove that $R_n(x)$ is always positive. It looks like $R_n(x)=x^{2n-2}+2x^{2n-4}+3x^{2n-6}+\ldots+n$. This needs a proof.

 

[CaptainBlack]

Prove that the polynomials:

\[P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1),\ \ \ (n=1,2...)\]

have no real roots.

CB

A slight hint: It is obvious that these polynomials have no negative roots, so we need only consider the posibility (or not) of positive roots.

CB

 

[anemone]

Still no replies to this one? I started by graphing $P_1(x)$ and $P_2(x)$, both of which have a positive global minimum at $x=1$ (and hence no real roots). That made me look for a factor of $(x-1)^2$, and I found that

$P_1(x) = x^2-2x+3 = (x-1)^2+2,$

$P_2(x) = x^4-2x^3+3x^2-4x+5 = (x^2+2)(x-1)^2+3,$

$P_3(x) = x^6-2x^5+3x^4-4x^3+5x^2-6x+7 = (x^4+2x^2+3)(x-1)^2+4.$

That should be enough of a pattern to suggest a solution. (Wink)

By following the hints given by Opalg, I see that

$P_n(x) = x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1) = (x^{2n-2}+2x^{2n-4}+3x^{2n-6}+...+n)(x-1)^2+(n+1),\ \ \ (n=1,2...)$

It's obvious that $P_n(x)\ne 0$ for all $ x\in R$.

Thus it suggests that \[P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1),\ \ \ (n=1,2...)\] has no real roots must be true.

Am I on the right track?

 

[alexmahone]

Am I on the right track?

Yes. You have to prove this, though:

$P_n(x) = x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1) = (x^{2n-2}+2x^{2n-4}+3x^{2n-6}+...+n)(x-1)^2+(n+1),\ \ \ (n=1,2...)$

 

[anemone]

A slight hint: It is obvious that these polynomials have no negative roots, so we need only consider the posibility (or not) of positive roots.

CB

If I choose to use the hint provided by CB, I'll start to approach the problem by using the Descartes' rule of signs.

First, I assume the $ P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-...-2nx+(2n+1)$ has real roots.

From $ P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-...-2nx+(2n+1)$, I get
$ P_n(-x)=(-x)^{2n}-2(-x)^{2n-1}+3(-x)^{2n-2}-...-2n(-x)+(2n+1)$
$ P_n(-x)=x^{2n}+2x^{2n-1}+3x^{2n-2}-...+2nx+(2n+1)$
No sign changes between the coefficients of x's, that means if the polynomial $P_n(x)$ has any real roots, then it certainly has no real negative roots.

Now, I consider the case where $ P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-...-2nx+(2n+1)$.
I see that the number of sign changes equals to 2n, so, based on the Descartes' rule of signs I can conclude that the number of positive roots of the polynomial $P_n(x)$is either equal to the number of sign differences between consecutive nonzero coefficients, i.e. 2n or is less than it by a multiple of 2.

That is to say, we're now need to consider two cases:
I: the number of positive real roots = 2n or
II: the number of positive real roots = 2n-2k, $k=1, 2, ..., n$ and the number of positive complex roots in the form $a\pm b\sqrt {c}$ where $a>b\sqrt {c}$ =2k, $k=1, 2, ..., n$But we can also tell right from the start that the product of all roots of $ P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-...-2nx+(2n+1)$ is -(2n+1) for n=1, 2, ..., that is, the product of all the real positive roots (be it two integers/fractions or two pair of positive complex roots) = -ve.

Obviously this leads to contradiction and we can conclude at this point that the polynomial $ P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-...-2nx+(2n+1)$ has no real roots.

Edit: I made a horrible mistake by saying that the product of all roots = - (2n+1) which is wrong. It should be (2n+1). So, my effort in proving in this post have been amounted to zero, zip, nothing. Please kindly dismiss it.

---------- Post added at 07:02 ---------- Previous post was at 07:00 ----------

Yes. You have to prove this, though:

I think the pattern could be generalized from the hint given by Opalg.

 

[CaptainBlack]

If I choose to use the hint provided by CB, I'll start to approach the problem by using the Descartes' rule of signs.

First, I assume the $ P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-...-2nx+(2n+1)$ has real roots.

From $ P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-...-2nx+(2n+1)$, I get
$ P_n(-x)=(-x)^{2n}-2(-x)^{2n-1}+3(-x)^{2n-2}-...-2n(-x)+(2n+1)$
$ P_n(-x)=x^{2n}+2x^{2n-1}+3x^{2n-2}-...+2nx+(2n+1)$
No sign changes between the coefficients of x's, that means if the polynomial $P_n(x)$ has any real roots, then it certainly has no real negative roots.

Now, I consider the case where $ P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}-...-2nx+(2n+1)$.
I see that the number of sign changes equals to 2n, so, based on the Descartes' rule of signs I can conclude that the number of positive roots of the polynomial $P_n(x)$is either equal to the number of sign differences between consecutive nonzero coefficients, i.e. 2n or is less than it by a multiple of 2.

That is to say, we're now need to consider two cases:
I: the number of positive real roots = 2n or
II: the number of positive real roots = 2n-2k, $k=1, 2, ..., n$ and the number of positive complex roots in the form $a\pm b\sqrt {c}$ where $a>b\sqrt {c}$ =2k, $k=1, 2, ..., n$

What are positive complex roots, and where will you find a reference to what Descartes rule of signs says about this sort of sign of the complex roots?

CB

 

[anemone]

My mistakes.
I want to clarify that the following observations were not right.
"That is to say, we're now need to consider two cases:
I: the number of positive real roots = 2n or
II: the number of positive real roots = 2n-2k, $k=1, 2, ..., n$ and the number of positive complex roots in the form $a\pm b\sqrt {c}$ where $a>b\sqrt {c}$ =2k, $k=1, 2, ..., n$"
Reason:
1. I didn't know why on Earth I called $a\pm b\sqrt {c}$ as complex roots. I'm sorry.
2. This "...based on the Descartes' rule of signs I can conclude that the number of positive roots of the polynomial $P_n(x)$is either equal to the number of sign differences between consecutive nonzero coefficients, i.e. 2n or is less than it by a multiple of 2." was taken at http://en.wikipedia.org/wiki/Descartes'_rule_of_signs whereas the two generated cases were my own conclusions.I'm sorry for confusing you, CB.
My apologies.

 

[Opalg]

Yes. You have to prove this, though:

$P_n(x) = x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1) = (x^{2n-2}+2x^{2n-4}+3x^{2n-6}+...+n)(x-1)^2+(n+1),\ \ \ (n=1,2...)$

My proof of that was just to multiply out the product on the right hand side:

$$\begin{aligned}(x^{2n-2}+2x^{2n-4}+3x^{2n-6}+...+n)(x^2-2x+1) \\ =& x^{2n}\phantom{{}-2x^{2n-1}}+2x^{2n-2}\phantom{{}-4x^{2n-3}}+3x^{2n-4} \phantom{{}-2x^{2n-1}}+ \ldots \\ &\phantom{x^{2n}} - 2x^{2n-1} \phantom{{}+2x^{2n-2}} - 4x^{2n-3} \phantom{{}+2x^{2n-2}} -6x^{2n-5} -\ldots \\ &\phantom{x^{2n} - 2x^{2n-1}} +\phantom{1}x^{2n-2}\phantom{{}-2x^{2n-1}} + 2x^{2n-4} \phantom{{}-2x^{2n-1}} +\ldots \\ &\phantom{x}\\ =& x^{2n}-2x^{2n-1}+3x^{2n-2}-4x^{2n-3} + 5x^{2n-4}-6x^{2n-5}+\ldots \end{aligned}$$

For $0\leqslant k\leqslant n-1$ the coefficient of $x^{2n-2k}$ is $k+(k+1)=2k+1$ and the coefficient of $x^{2n-2k-1}$ is $-2(k+1)$. The constant term is $n$. So the resulting polynomial is $P_n(x) - (n+1)$.

 

[CaptainBlack]

Solution: CaptainBlacks Occasional Problem #3

This again is from the Purdue PotW


Prove that the polynomials:



\[P_n(x)=x^{2n}-2x^{2n-1}+3x^{2n-2}- ... -2nx + (2n+1),\ \ \ (n=1,2...)\]

have no real roots.


================================================

Solution:

Spoiler

These obviously have no negative roots (either from Descartes rule of signs or observing that the odd powers of \(x \) all have -ve signs)


So for \(x>0\) consider:
\[P_n(x)+xP_n(x)=x(x^{2n}-x^{2n-1}+x^{2n-2}- ... - x+1) + (2n+1) \]
Now the bracketed term is a finite geometric series and so we may replace it with its sum to get:
\[P_n(x)(1+x)=x \left[ \frac{1+x^{2n+1}}{1+x}\right] + (2n+1) \]
or:
\[P_n(x)=x \left[ \frac{1+x^{2n+1}}{(1+x)^2}\right] + \frac{2n+1}{1+x}>0 \]
and so as \(P_n(0) \ne 0\) we have \(P_n(x)>0\) for all real \(x\)

The above is somewhat different from the solution given on the Purdue PotW site, mainly in that it avoids an unexplained step, which while probably valid I find opaque.

CB