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CaptainBlacks Occasional Problem #3

CaptainBlack

Well-known member
Jan 26, 2012
890
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
 
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Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
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

Active member
Jan 26, 2012
268
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.
 
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CaptainBlack

Well-known member
Jan 26, 2012
890
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

MHB POTW Director
Staff member
Feb 14, 2012
3,753
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

Active member
Jan 26, 2012
268

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,753
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.
 
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CaptainBlack

Well-known member
Jan 26, 2012
890
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

MHB POTW Director
Staff member
Feb 14, 2012
3,753
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

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
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

Well-known member
Jan 26, 2012
890
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:
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
 
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