Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

... Find all the real roots of $P_n (X)$, for each $n$.

Help! I'm completely stuck on this question. I started out by finding $P_n (X)$ for various $n$ up to $n=5$, and then I found that the only real solution for each $n$ is $x=0$. Here is a picture of my polynomial work, keeping in mind there is a mistake in the picture for $n=5$, which should be $x^5-3x^4+4x^3-2x^2+x$

I know I can use induction to show that $0$ is a solution for all $n$, but how can I show it is the unique real solution for all $n$, too?

I asked elsewhere and was provided with an equation that can help, only I have no idea how they derived it, and that thread got ignored and eventually lost, on another forum.

Please help!

share|improve this question
What do you mean, this is not simply homework anymore? Note my previous answer had a mistake which I have fixed. There is only one real root: $0$. –  1015 Mar 14 '13 at 2:15
add comment

3 Answers

Here is how you can derive the formula if you don't know it a priori.

Fix $t\neq 2$ and consider the sequence $a_n=P_n(t)$: $$ a_0=0\quad a_1=t\quad a_{n+2}=ta_{n+1}+(1-t)a_n \quad\forall n\geq 0. $$ This is homogeneous and linear, so we know how to find a closed form. You can find the general method here.

First consider the characteristic equation $$ r^2-tr+(t-1)=0. $$ The discriminant is $(t-2)^2>0$ and the quadratic formula yields two distinct roots $$ r_1=1\qquad r_2=t-1. $$ Now we know that there exist two constants $\lambda,\mu$ such that $$ a_n=\lambda \;r_1^n+\mu \;r_2^n=\lambda+\mu (t-1)^n. $$ Considering the intial conditions, we can compute $\lambda$ and $\mu$ and we find $$ a_n=P_n(t)=\frac{t((t-1)^n-1)}{t-2}. $$

In the case $t=2$, $$ a_{n+2}=2a_{n+1}-a_n\quad\Leftrightarrow\quad a_{n+2}-a_{n+1}=a_{n+1}-a_n. $$ so $a_{n+1}-a_n$ is constant equal to $a_1-a_0=2$. Hence, an easy induction and the fact that $a_0=0$ yield $$ a_n=P_n(2)=2n. $$

Note: of course, you could directly deduce that $P_n(2)=2n$ by taking the limit as $t$ tends to $2$ in the formula for $P_n(t)$ when $t\neq 2$.

Now the real roots: $$ P_n(t)=0\qquad\Leftrightarrow \qquad t((t-1)^n-1)\quad\mbox{and}\quad t\neq 2 $$ $$ \quad\Leftrightarrow\quad \{t=0\quad\mbox{or}\quad (t-1)^n=1\}\quad\mbox{and}\quad t\neq 2. $$ Now observe that in $\mathbb{R}$, we have $(t-1)^n=1$ implies $t-1=\pm 1$, ie $t=0$ or $2$. Since $P_n(2)=2n$ this does not add any real root to $0$.

So there is only one real root for $P_n$, for all $n\geq 1$: that's $0$.

share|improve this answer
Sorry, what's linear exactly? In your 4th line, that is. –  user65132 Mar 5 '13 at 14:42
Recursive homogeneous linear of order $2$ means there exist constants $u,v$ such that $a_{k+2}=ua_{k+1}+va_k$. See here en.wikipedia.org/wiki/Recurrence_relation for the general theory. –  1015 Mar 5 '13 at 14:44
Ah, I get it now. But your characteristic equation isn't like the one from the wiki page, what's up with that? –  user65132 Mar 5 '13 at 15:19
I can't respond to Julien for some reason. Thanks a lot, and I finally get how to solve this recurrence relation after a few days of banging my head at it, but I have one qualm. You say 2 is a real root, but if you substitute $P_2 (X)=X^2$ then the answer you get for that is 4, not 0. So certainly this means that two cannot be a root? –  user65132 Mar 7 '13 at 22:36
@user65132: You have accidentally created two accounts, which is why you were not able to edit your post directly. Here is the process to merge your accounts: From any page footer -> 'contact us' >> 'Merge user profiles' –  Zev Chonoles Mar 7 '13 at 22:43
add comment

Suppose $X \in (0, 1)$. Then, an easy inductive argument shows $P_n(X) > 0$ for all $n > 0$.

Suppose $X \in (1, \infty)$. Then

$$P_n(X) = X (P_{n-1}(X) - P_{n-2}(X)) + P_{n-2}(X) $$

Can we show $P_n(X) > P_{n-1}(X)$ for all $n > 0$? If so, we have another easy inductive argument that $P_n(X) > 0$ for all $n > 0$. Simplifying,

$$P_n(X) - P_{n-1}(X) = (X-1) (P_{n-1}(X) - P_{n-2}(X))$$

Oh wait, this immediately suggests a drastic simplification to the problem, since we can easy solve this recurrence for $P_n(X) - P_{n-1}(X)$:

$$P_n(X) - P_{n-1}(X) = (X-1)^{n-1} (P_1 (X) - P_0(X)) = (X-1)^{n-1} X $$ and this one is easy to solve for $P_n(X)$:

$$P_n(X) = P_n(X) - P_0(X) = \sum_{i=0}^{n-1} (X-1)^i X = X \frac{(X-1)^n - 1}{(X-1) - 1}$$

So the roots of $P_n(X)$ are $0$, the roots of $(X-1)^n - 1$, and possibly roots of $(X-1) - 1$.

share|improve this answer
It may seem odd that I switched my solution method in the middle; what I've written is a sketch of my actual train of thought when solving the problem. Rather than post a canned solution, I'm hoping to demonstrate the actual process of exploring a problem, spotting a useful fact, then using it. –  Hurkyl Mar 5 '13 at 14:17
add comment

To verify that $P_n(x)=x\left(\left(x-1\right)^{n-1}+\left(x-1\right)^{n-2}+\ldots+\left(x-1\right)+1\right)$ for $n\geq 1$ you could use the recurrence from which $P_n$ is defined.

Let's denote by $Q_n(x)$ the polynomial $x\left(\left(x-1\right)^{n-1}+\left(x-1\right)^{n-2}+\ldots+\left(x-1\right)+1\right)$.
By noting that $Q_1(x)=x=P_1(x)$ and $Q_2(x)=x^2=P_2(x)$, is enough to show that $$Q_n(x)=xQ_{n-1}(x)-(x-1)Q_{n-2}(x), \text{ for all } n\geq 3.$$

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.