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I'm studying for an exam and trying to prove whether 1 is a triple root for the polynomial: $$x^{2n+1}-(2n+1)x^{n+1}+(2n+1)x^n-1$$ for every $n\geq1$.

In our math class we never solved such a problem. So far we only used horner's scheme to prove that someone is a root, double root or triple root.

Can you please help me solve this problem?

Bumblebee
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Jane
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    Are you familiar with the fact that $a$ is a root of multiplicity $k\gt 1$ if and only if $(x-a)^{k-1}$ divides both $f(x)$ and $f'(x)$? – Arturo Magidin Nov 10 '21 at 18:36
  • Similarly, $a$ is a root of multiplicity $k$ if and only if $f(a)=f'(a)=\cdots=f^{(k-1)}(a)=0$, where $f^{(n)}(x)$ is the $n$th derivative of $f(x)$. – Greg Martin Nov 10 '21 at 18:48
  • @ArturoMagidin I'm not sure how I would apply that theorem. – Jane Nov 10 '21 at 18:51
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    By checking if $1$ is a root of $f$ and of $f'$, and then repeating the process to see if it is a double root of $f'$, etc. – Arturo Magidin Nov 10 '21 at 18:53
  • @ArturoMagidin And $f'$ is the derivation of the polynome? – Jane Nov 10 '21 at 18:57
  • (Without using derivatives,) You can use synthetic division to quickly divide out by $ x-1$ 3 times. Yes, there is a bunch of algebra involved. Can you show what you've tried? – Calvin Lin Nov 10 '21 at 21:02

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Assume that $a$ is a $k$-ary root of some polynomial $p \in \Bbb R[x]$. Then you might write $p$ as $p=(x-a)^kq$ for some polynomial $q$ where $q(a) \neq 0$. Since $p$ (as a function on $\Bbb R$) is differentiable we may differentiate it and find $$ \mathrm{D}p(x) = k(x-a)^{k-1}q(x) + (x-a)^k \mathrm{D}q(x) = (x-a)^{k-1}(kq(x) + (x-a)\mathrm{D}q(x)). $$ Note that this means that $a$ is a $(k-1)$-ary root of $\mathrm{D}p$. If we on the other hand have $\mathrm{D}^k p(a) = 0$ and $\mathrm{D}^{k+1} p(a) \neq 0$, then $\mathrm{D}^kp(x)=(x-a)q(x)$ is integrable and thus $$\begin{align*} \mathrm{D}^{k-1} p(x) &= \int \mathrm{D}^k p(x) \mathrm{d}x \\ &= \int (x-a)q(x) \mathrm{d}x \\ &= (x-a)^2 \frac{q(x)}{2} - \int (x-a)^2 \frac{\mathrm{D}q(x)}{2} \mathrm{d}x \\ &= (x-a)^2 \frac{q(x)}{2} - (x-a)^3 \frac{\mathrm{D}q(x)}{3!} + \int (x-a)^3 \frac{\mathrm{D^2}q(x)}{3!} \mathrm{d}x \\ &\vdots\\ &=(x-a)^2 \sum_{k=0}^{\mathrm{deg}(q)} (-1)^k(x-a)^k \frac{\mathrm{D}^kq(x)}{k!} \quad (+C) \end{align*} $$ so $\mathrm{D}^{k-1}p$ has a double root at $a$ if $\mathrm{D}^k p$ has a singular one.

So to show that 1 is a triple root of your polynomial just show that the polynomial and it's first few derivatives are $0$.

Arturo Magidin
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SV-97
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$$\begin{align*} x^{2n+1}&-(2n+1)x^{n+1}+(2n+1)x^n-1\\ &=x^{2n+1}-1-(2n+1)[ x^{n+1}- x^n]\\ &=(x-1)(x^{2n}+x^{2n-1}+\cdots+1)-(2n+1)(x-1)x^n\\ &=(x-1)(x^{2n}+x^{2n-1}+\cdots+x^{n+1}+x^{n-1}+\cdots+1-2nx^n)\\ &=(x-1)[x^n(x^n-1)+x^{n-1}(x^n-x)+\cdots+x^n(x-1)\\ &\qquad\qquad+x^{n-1}(1-x)+x^{n-2}(1-x^2)+\cdots+x(1-x^{n-1})+1-x^n] \end{align*}$$ Combining corresponding terms from opposite ends of the expression inside the square bracket we have: $$ =(x-1)[(x^n-1)^2+x(x^{n-1}-1)^2+...+x^{n-1}(x-1)^2] $$ It is clear that each term inside the square bracket is divisible by $(x-1)^2$ so the whole expression is divisible by $(x-1)^3$

am301
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  • Side note: It might be easier to see the final line via $ \sum (x^{2n-k} - 2x^n + x^k) = \sum x^k ( x^{n-k} - 1 ) ^2 $, which essentially combines both of your steps. – Calvin Lin Nov 10 '21 at 21:06
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Let $f_n(x)=x^{2n+1}-(2n+1)x^{n+1}+(2n+1)x^n-1$. We will prove the statement by induction on $n$.

For $n=1$ we get $f_1(x)=x^3-3x^2+3x-1=(x-1)^3$ and the statement holds.

Now assume $f_n(x)$ has a factor $(x-1)^3$, then take \begin{align} f_{n+1}(x)-xf_n(x) &=x^{2n+3}-x^{2n+2}-2x^{n+2}+2x^{n+1}-1+x\\ &=x^{2n+2}(x-1)-2x^{n+1}(x-1)+(x-1)\\ &=(x-1)(x^{2n+2}-2x^{n+1}+1)\\ &=(x-1)(x^{n+1}-1)^2\\ &=(x-1)^3\left(\frac{x^{n+1}-1}{x-1}\right)^2\\ &=(x-1)^3\left(1+x+x^2+\dots+x^n\right)^2. \end{align} The RHS has a factor $(x-1)^3$, so does $f_n(x)$ (by the induction hypothesis), hence $f_{n+1}(x)$ has a factor $(x-1)^3$.

Sil
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