Need help deriving formula for a sum/polynomial stuff

**qtpipi** · March 3rd 2009, 10:18 PM

OK, I have no idea what this is asking... I don't know what to do. Can someone help please?? thanks.

**Grandad** · March 6th 2009, 12:44 PM

Hello qtpipi

Originally Posted by qtpipi

OK, I have no idea what this is asking... I don't know what to do. Can someone help please?? thanks.

First of all, let me write out what the question is asking. The first $\Sigma$ expression means:

$f(x) = c_0 + c_1x+c_2x^2 + \dots + c_nx^n$

And to say it has zeros $\alpha_1, \dots, \alpha_n$ means

$f(\alpha_1) = f(\alpha_2) = \dots = f(\alpha_n) = 0$

And then, we are to derive a formula for

$\frac{1}{\alpha_1} + \dots + \frac{1}{\alpha_n}$

So, using the Remainder Theorem, $(x-\alpha_1), \dots, (x-\alpha_n)$ are factors of $f(x)$ .

Since there are $n$ of these factors, and $f(x)$ is a polynomial of degree $n$ , this means that we can write $f(x)$ as:

$f(x) = c(x-\alpha_1)(x-\alpha_2)\dots (x-\alpha_n)$

(This is what the expression $f(x) = c\prod (x-\alpha_i)$ means.)

Now the coefficient of $x^n$ is $c_n$ . So $c = c_n$ . And the coefficient of $x^{n-1}$ in the expansion of $c\prod(x-\alpha_i)$ is $-c\sum \alpha_i$ which is $c_{n-1}$ . So:

$\sum \alpha_i = -\frac{c_{n-1}}{c_n}$

So how do we find $\sum\frac{1}{\alpha_i}$ ?

Replace $x$ by $\tfrac{1}{y}$ :

$f(\tfrac{1}{y}) = c_0 + c_1\tfrac{1}{y} + \dots + c_n(\tfrac{1}{y})^n$ which has zeros at

$y = \tfrac{1}{\alpha_1}, \dots, \tfrac{1}{\alpha_n}$

and $c_0 + c_1\tfrac{1}{y} + \dots + c_n(\tfrac{1}{y})^n = 0 \Rightarrow c_0y^n + c_1y^{n-1} + \dots + c_n = 0$

and, using the same method as before, the sum of the roots of this polynomial is $-\frac{\text{coefficient of } y^{n-1}}{\text{coefficient of } y^{n}}$

$\Rightarrow \sum_{i=0}^n \frac{1}{\alpha_i} = -\frac{c_1}{c_0}$

Grandad

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