# I Newton's formula for the sums of powers of roots?

1. Aug 29, 2016

### Clara Chung

Please take a look of the photo. In the middle part, it says For each I, by division and gets the following results. Please further explain to me how to get the result by division. The photo is attached.

Attempt: 1. Using f(x)=α0(x-α1)(x-α2)...(x-αn) form. If it is divided by x-αi , there should be no αi occurs in the results.
2. Using fundamental theorem of algebra. a1/a0=α1+α2+α3+α4+..+αn, αi is lost so it is (a1-a0αi)x^(n-2),not(a1+a0αi)x^(n-2), so this is not the method. Then how should i divide it? Please help

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2. Sep 2, 2016

### stevendaryl

Staff Emeritus
The definition of $f(x)$ is this:

$f(x) = a_0 x^n + a_1 x^{n-1} + ... + a_n$

We want to divide by $x-\alpha_j$. To make this easy, let's rewrite the first term as follows:

$a_0 x^n = a_0 x \cdot x^{n-1} = a_0 (x-\alpha_j + \alpha_j) \cdot x^{n-1} = a_0(x-\alpha_j) x^{n-1} + a_0 \alpha_j x^{n-1}$

The term $a_0 \alpha_j x^{n-1}$ is of order $x^{n-1}$, so it can be combined with the term $a_1 x^{n-1}$. So $f(x)$ can be written as:

$f(x) = a_0 (x-\alpha_j) x^{n-1} + [a_0 \alpha_j + a_1] x^{n-1} + a_2 x^{n-2} + ...$

We can similarly rewrite the second term:

$[a_0 \alpha_j + a_1] x^{n-1} = [a_0 \alpha_j + a_1](x - \alpha_j) \cdot x^{n-2} + [a_0 \alpha_j + a_1]\alpha_j x^{n-2}$

The second term, $[a_0 \alpha_j + a_1]\alpha_j x^{n-2}$, can be combined with the term $a_2 x^{n-2}$. So we can rewrite $f(x)$ yet again as:

$f(x) = a_0 (x-\alpha_j) x^{n-1} + [a_0 \alpha_j + a_1] (x-\alpha_j) x^{n-2} + [[a_0 \alpha_j + a_1]\alpha_j + a_2] x^{n-2} + ...$
$= a_0 (x-\alpha_j) x^{n-1} + [a_0 \alpha_j + a_1] (x-\alpha_j) x^{n-2} + [a_0 (\alpha_j)^2 + a_1 \alpha_j + a_2] x^{n-2} + ...$

You can continue this pattern to get:

$f(x) = a_0 (x-\alpha_j) x^{n-1} + [a_0 \alpha_j + a_1] (x-\alpha_j) x^{n-2} + [a_0 (\alpha_j)^2 + a_1 \alpha_j + a_2] (x-\alpha_j) x^{n-3} + ... + [a_0 (\alpha_j)^{n-1} + a_1 (\alpha_j)^{n-2} + a_2 (\alpha_j)^{n-3} + ... + a_{n-1}](x-\alpha_j) x^0 + [a_0 (\alpha_j)^n + a_1 (\alpha_j)^{n-1} + a_2 (\alpha_j)^{n-2} + ... + a_n](x-\alpha_j) x^{-1}$

The very last term is zero, because we have the coefficient:
$[a_0 (\alpha_j)^n + a_1 (\alpha_j)^{n-1} + a_2 (\alpha_j)^{n-2} + ... + a_n]$

which is just equal to $f(\alpha_j)$. By definition, $\alpha_j$ is one of the zeros of $f(x)$. So we can ignore the last term, to get:

$f(x) = a_0 (x-\alpha_j) x^{n-1} + [a_0 \alpha_j + a_1] (x-\alpha_j) x^{n-2} + [a_0 (\alpha_j)^2 + a_1 \alpha_j + a_2] (x-\alpha_j) x^{n-3} + ... + [a_0 (\alpha_j)^{n-1} + a_1 (\alpha_j)^{n-2} + a_2 (\alpha_j)^{n-3} + ... + a_{n-1}](x-\alpha_j) x^0$

Now having rewritten $f(x)$ in this way, we can easily divide by $x-\alpha_j$, since every term is multiplied by that. So we get:

$\frac{f(x)}{x-\alpha_j} = a_0 x^{n-1} + [a_0 \alpha_j + a_1] x^{n-2} + [a_0 (\alpha_j)^2 + a_1 \alpha_j + a_2] x^{n-3} + ... + [a_0 (\alpha_j)^{n-1} + a_1 (\alpha_j)^{n-2} + a_2 (\alpha_j)^{n-3} + ... + a_{n-1}] x^0$

3. Sep 2, 2016

### Clara Chung

Thanks for solving the problem that perplexed me so long. The answer is very clear.