Remainder of dividing polynomial of the $n$th degree

Question

What is the remainder when dividing the polynomial $$P(x)=x^n+x^{n-1}+\cdots+x+1$$ with the polynomial $$x^3-x$$ if $n$ is a natural odd number?

So, what I know so far is:

$$P(x)=Q(x)D(x)+R(x)$$

In this case I'll call $Q(x) = x^3-x$

$$Q(x) = 0 \iff x=\pm1$$

So from here:

$\begin{array} {l}P(1):\qquad&1^n+1^{n-1}+\cdots+1^1+1=R(1)\\P(-1):&(-1)^n+(-1)^{n-1}+\cdots+(-1)^1+1=R(-1)\end{array}$

Here is where I assumed that $R(x)=ax^2+b$ since $Q(x)=x^3-x$

And from the equations (assuming $n = \{2k+1 \mid k\in\mathbb{N}\}$):

$\begin{array}{l}P(1):\\&R(1)=n\\P(-1):\\&R(-1)=0\end{array}$

And here I get to:

$$\left\{ \begin{array}{ll} ax^2+by &=n \\ ax^2+by &=0 \\ \end{array} \right.$$

I would like to know where did I go wrong? Is my assumption for $R(x)$ incorrect, my calculation of the $P(1)$ and $P(-1)$ wrong or is there something else I didn't think about?

$$Q(x) = 0 \iff x\in{0,\pm1}$$ – Michael Hardy Oct 02 '20 at 21:50 — Michael Hardy, Oct 02 '20 at 21:50

Bernard · Accepted Answer · 2020-10-05T15:07:49.603

Hint:

Experimenting the first few values of $n$: $\:n= 3,5,7,9,11$, you may conjecture the remainder for the general polynomial $P_n(x)=x^n+x^{n-1}+\dots+x^3+x^2+x+1\:$ is $$R_n(x)=kx^2+(k+1)x+1,\quad\text{ where } k=\left\lfloor \frac n2 \right\rfloor.$$ and try to prove it by induction on $n$ (odd).

You can prove the inductive step (case $n \implies$ case $n+2$) you can try rewriting $$ P_{n+2}(x)= x^2 P_n(x)+ (x+1)=x^2\bigl[ Q_n(x)(x^3-x)+R_n(x)\bigr]+(x+1), $$ so the remainder for $P_{n+2}$ will be as well the remainder for $$x^2 R_n(x)+(x+1)=x^2(kx^2+(k+1)x +1)+(x+1)=kx^4+(k+1)x^3+x^2+x+1.$$ You'll only have to prove that the coefficient of $x^2$ thus obtained is $\left\lfloor\dfrac{n+2}2\right\rfloor$ and the coefficient of $x$ is $1$ more.

R. J. Mathar · Answer 2 · 2020-10-11T13:27:47.967

0

The remainder is $$\left(\frac{n+1}{2}\frac{1}{x-1}-\frac{1}{x}\right)(x^3-x) = \frac{(x+1)(2+(n-1)x)}{2}$$

edited Oct 11 '20 at 13:27

answered Oct 02 '20 at 22:32

R. J. Mathar

2,324

2

The remainder is a polynomial, not a rational fraction. – Bernard Oct 02 '20 at 22:50
@Bernard, this rational function simplifies to the correct polynomial. – lhf Oct 05 '20 at 18:35
But you don't obtain $\Bigl\lfloor\frac n2\Bigr\rfloor$ as the leading coefficient. Or do I misunderstand your notations? – Bernard Oct 05 '20 at 18:48
@Bernard, I think $[\cdots]$ here are just parentheses. – lhf Oct 06 '20 at 09:48
@lhf: That's the way I understood them finally, and I had to convince myself it was the same result as mine. – Bernard Oct 06 '20 at 10:38

lhf · Answer 3 · 2020-10-06T13:45:20.257

Your approach is fine but you've missed the root $x=0$ of $x^3-x$.

We get $R(0)=P(0)=1$, $R(1)=P(1)=n+1$, $R(-1)=P(-1)=0$.

The quadratic polynomial that interpolates this data is $\frac{n-1}{2} x^2+\frac{n+1}{2}x+1$.

The same approach works when $n$ is even (the interpolation data is different because $P(-1)=1$) and we get $\frac{n}{2} x^2+\frac{n}{2}x+1$.

In both cases, we get the nice recurrence: $R_{n+2}=R_n+x^2+x$, with $R_0=1$ and $R_1=x+1$.

Remainder of dividing polynomial of the $n$th degree

3 Answers3