OK, I have no idea what this is asking... I don't know what to do. Can someone help please?? thanks.
Hello qtpipiFirst of all, let me write out what the question is asking. The first $\displaystyle \Sigma$ expression means:
$\displaystyle f(x) = c_0 + c_1x+c_2x^2 + \dots + c_nx^n$
And to say it has zeros $\displaystyle \alpha_1, \dots, \alpha_n$ means
$\displaystyle f(\alpha_1) = f(\alpha_2) = \dots = f(\alpha_n) = 0$
And then, we are to derive a formula for
$\displaystyle \frac{1}{\alpha_1} + \dots + \frac{1}{\alpha_n}$
So, using the Remainder Theorem, $\displaystyle (x-\alpha_1), \dots, (x-\alpha_n)$ are factors of $\displaystyle f(x)$.
Since there are $\displaystyle n$ of these factors, and $\displaystyle f(x)$ is a polynomial of degree $\displaystyle n$, this means that we can write $\displaystyle f(x)$ as:
$\displaystyle f(x) = c(x-\alpha_1)(x-\alpha_2)\dots (x-\alpha_n)$
(This is what the expression $\displaystyle f(x) = c\prod (x-\alpha_i)$ means.)
Now the coefficient of $\displaystyle x^n$ is $\displaystyle c_n$. So $\displaystyle c = c_n$. And the coefficient of $\displaystyle x^{n-1}$ in the expansion of $\displaystyle c\prod(x-\alpha_i)$ is $\displaystyle -c\sum \alpha_i$ which is $\displaystyle c_{n-1}$. So:
$\displaystyle \sum \alpha_i = -\frac{c_{n-1}}{c_n}$
So how do we find $\displaystyle \sum\frac{1}{\alpha_i}$?
Replace $\displaystyle x$ by $\displaystyle \tfrac{1}{y}$:
$\displaystyle f(\tfrac{1}{y}) = c_0 + c_1\tfrac{1}{y} + \dots + c_n(\tfrac{1}{y})^n$ which has zeros at
$\displaystyle y = \tfrac{1}{\alpha_1}, \dots, \tfrac{1}{\alpha_n}$
and $\displaystyle c_0 + c_1\tfrac{1}{y} + \dots + c_n(\tfrac{1}{y})^n = 0 \Rightarrow c_0y^n + c_1y^{n-1} + \dots + c_n = 0$
and, using the same method as before, the sum of the roots of this polynomial is $\displaystyle -\frac{\text{coefficient of } y^{n-1}}{\text{coefficient of } y^{n}}$
$\displaystyle \Rightarrow \sum_{i=0}^n \frac{1}{\alpha_i} = -\frac{c_1}{c_0}$
Grandad