3
$\begingroup$

I'm having a bit trouble with this excercise:

The problem:
Let there be a polynomial $f(x)=a_1x^{t_1} + a_2x^{t_2} + ... + a_nx^{t_n}$ Where $t_1, t_2, ..., t_n$ are not-negative integers. The polynomial has a root $b$ which occurs $n$ times. Prove that $b = 0$.

What I have so far:
I can presume that $a_1, a_2, ..., a_n \neq 0$
If $n = 1$, then it's obviously true.
If $n = 2$, I tried using this:
$f^{(n-1)}(b) = 0 \\ f^{(n)}(b) = 0$
So:
$f(b) = a_1b^{t_1} + a_2b^{t_2} = 0 \\ f'(b) = t_1*a_1b^{(t_1-1)} + t_2*a_2b^{(t_2-1)} = 0 \\ f''(b) = (t_1-1)t_1*a_1b^{(t_1-2)} + (t_2-1)t_2*a_2b^{(t_2-2)} \neq 0$
Why can't $b \neq 0$ be true?

  • 0
    @user30620: $(x-b)^n$ has $n+1$ terms if and only if $b\neq0$...2012-05-06

1 Answers 1

2

You have $b^kf^{(k)}(b)=0$ for $k=0,\dots,n-1$, i.e. $\sum_{j}p_i(t_j)b^{t_j}a_j=0$, where $p_i(x)=x(x-1)\dots(x-i+1)$. If $b^{t_j}a_j\neq 0$ for at least one $j$ then there are some $c_0,\dots,c_{n-1}$, not all zero, such that $\sum_i c_ip_i(t_j)=0$ for all $j$'s (that's where linear algebra is used). But the polynomial $\sum_i c_ip_i(x)$ has degree at most $n-1$, so it can't have $n$ different roots.

  • 0
    Are $u$ and $v$ taken from the matrix $M$? Also are you implying that $\sum_i c_ip_i(x)$ is equal to $f(x)$?2012-05-04