Prove $f=1+x+x^2+x^3+\cdots+x^n$ has no multiple roots.

Question

Prove $f=1+x+x^2+x^3+\cdots+x^n$ has no multiple roots.

My attempt:

Consider the polynomial $g=(x-1)(1+x+x^2+x^3+\cdots+x^n)$

As $f\mid g, g$ all the roots of $f$ are roots of $g$.

This means I have to prove the statement:

If $c$ is a multiple root of $f \Longrightarrow c$ is multiple a root of $g$.

Equivalently:

If $c$ is not a multiple root of $g \Longrightarrow c$ is not a multiple root of $f$.

I'll try to prove this last statement:

Suppose $\exists c$ such that $c$ is a multiple root of $g$, then

$$g(c)=0 \implies g'(c)=0$$

$$ g=(x-1)(1+x+\cdots+x^n) \implies g'=nx^n $$ $$ g'(c)=nc^n=0 \implies c=0 $$ But $$ g(0)\neq 0 $$ And we've found the contradiction: $\not \exists c: g(c)=g'(c)=0$. Then $f$ has no multiple roots.

Is this correct?

Also: Assuming this is correct, is it possible to prove $\sum_{k=0}^n \frac {x^n}{n!} $ has no multiple roots in a similar manner?

Answer 1

Your proof is correct. Here is an alternative.

We have $f(1) = n+1$, so $1$ is not a root, and similarly, $f(0) = 1$, so $0$ is not a root.

Since $f(x) = {1 -x^{n+1} \over 1-x}$, if $x$ is a root of $f$, then $x^{n+1} = 1$, so just need to show that $f'(x) \neq 0$ whenever $x^{n+1} = 1$.

Since $f'(x) = {1 -x^{n+1} \over (1-x)^2} - (n+1) {x^n \over 1-x }$, we see that $f'(x) = - (n+1) {x^n \over 1-x } \neq 0$ whenever $f(x) = 0$.

For your other question, it is not clear what you mean by a similar manner.

Let $f_n(x) = 1+x+ {x^2 \over 2!}+\cdots+{x^n \over n!}$. We note that $f_n(0) = 1$, so $0$ is not a root. Then we have $f_n(x) = f_n'(x)+ {x^n \over n!}$, and so if $f_n(x) = 0$, we must have $f_n'(x) \neq 0$, hence any root is simple.

Answer 2

All roots of $f(x)$ also are roots of $(x-1)f(x)= x^{n+1}-1.$ Since $x^{n+1}-1$ is coprime with its derivative $(n+1)x^n$ then $x^{n+1}-1$ has no multiple roots. Thus and $f(x)$ has no multiple roots.

Answer 3

If you are just talking about real roots, it depends on the value of n: If n is odd, the root is -1 but if n is even, no real roots exist.

However if you are talking about all possible roots, then there exist exactly n roots, no matter the value of n. The roots of these particular equation are commonly referred to as the nth roots of unity