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# Newton, once again...

Another year has passed and the COVID-19 pandemic is yet to be gone. During this time, I’ve been daydreaming about my good old days when I was still going to school without any restriction.

It was during these moments that I remembered one thing I’ve always wanted to prove to myself. And, once again, it involved Isaac Newton.

## The elementary problem

There was a time when, during maths classes, the teacher would always have asked questions like this one:

Given $p(x)=x^2-4x+3$ and $\lambda_1, \lambda_2$ such that $p(\lambda_1)=p(\lambda_2)=0$ (i.e. $\lambda_1, \lambda_2$ are the roots of the polynomial $p(x)$), compute $\lambda_1^2+\lambda_2^2$.

It was not difficult to answer this problem and there were two main approaches to this problem.

### Computing roots first

We can compute $\lambda_1$ and $\lambda_2$ using the quadratic formula. We get $\lambda_1=\frac{4 +\sqrt{(-4)^2 -4\cdot1\cdot3}}{2\cdot1}=3$ and $\lambda_2=\frac{4 - \sqrt{(-4)^2 -4\cdot1\cdot3}}{2\cdot1}=1$.

Hence, $\lambda_1^2+\lambda_2^2=3^2+1^2=10$.

The problem given by this kind of approach is that we have to find the roots of $p(x)$ in terms of sums, products and radicals, which is not always possible for polynomials with degree greater than $4$ by the Abel-Ruffini theorem.

Furthermore, we have to compute the result of the sum of all the roots raised to a certain power, which may be difficult to compute if we had irrational roots.

### Vieta’s formulae

We know by the factor theorem that $(x-\lambda_1)(x-\lambda_1)=x^2-(\lambda_1 + \lambda_2)x+\lambda_1\lambda_2$. Therefore, $\lambda_1 + \lambda_2 = 4$ and $\lambda_1\lambda_2=3$.

We have to compute $\lambda_1^2 + \lambda_2^2$, which can be rearranged as $(\lambda_1 + \lambda_2)^2 - 2\lambda_1\lambda_2$ and therefore is equal to $4^2 - 2 \cdot 3 = 10$.

In general, given $f(x)=a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$, we have that $a_m = \displaystyle (-1)^{n-m} a_n \sum_{\mathrm{i>j>\cdots>k}} \underbrace{\lambda_i \lambda_j \cdots \lambda_k}_{(n - m)\ \text{times}}$ where $\lambda_w$ is a root of $f(x)$. These relationships are called Vieta’s formulae.

The benefit we get in using this approach is that we do not have to compute the roots of the polynomial $p(x)$.

## Generalizing the problem

What if the problem asked to compute $\lambda_1^3 + \lambda_2^3$? We can once again factor $\lambda_1^3 + \lambda_2^3$ as $(\lambda_1 + \lambda_2)(\lambda_1^2 + \lambda_2^2 - \lambda_1\lambda_2)$, getting $4(10 - 3) = 28$. We can check the result using the roots we had already computed: $\lambda_1^3 + \lambda_2^3 = 3^3 + 1^3 = 28$.

It is worth noting that, in order to calculate $\lambda_1^3 + \lambda_2^3$, we had to make use of $\lambda_1^2 + \lambda_2^2$ and $\lambda_1 + \lambda_2$.

So, is there a generalization of this kind of relationship between the sums of powers of the roots of a polynomial? Of course there is, and it is the Newton-Girard identity. What I was seeking in the first place was its proof, that I’m now going to explain.

### Newton-Girard identity

Newton’s identity states that given $e_m = |\frac{a_{n-m}}{a_n}|$ and $P_k = \lambda_1^k + \lambda_2^k + \ldots + \lambda_n^k = \displaystyle \sum_{i=1}^n \lambda_i^k$, the following identity holds:

$m e_m = \displaystyle \sum_{i=1}^{m} (-1)^{i-1} e_{m-i} P_i$

We can even rearrange this identity by noticing that $e_0$ must always be equal to $1$ and get:

$m e_m = \displaystyle \sum_{i=1}^{m-1} (-1)^{i-1} e_{m-i} P_i + (-1)^{m-1} P_m$

Hence, we obtain:

$P_m = (-1)^m \left( \displaystyle \sum_{i=1}^{m-1} (-1)^{i-1} e_{m-i} P_i - m e_m \right)$

This proves the relationship between all the sums of the powers of the roots of a polynomial we wanted to prove to ourselves earlier and gives us a powerful tool to compute the result of the kind of problems we discussed earlier.

#### Proof of Newton-Girard identity

The following proof makes use of calculus, Taylor-Maclaurin series expansion, Vieta’s formulas, the $n$-th derivative of the natural logarithm and Leibniz’s rule.1

##### First step: defining a polynomial $p(x)$

The first thing we do is defining a polynomial $p(x) = \displaystyle \prod_{i=1}^n \left(1 + \lambda_i x \right)$. It is worth noting that $p(x)$ is equal to $e_0 + e_1 x + e_2 x^2 + \cdots + e_n x^n$. This expression can also be expressed as an infinite sum, given that $e_k = 0$ for $k > n$, therefore giving us the following identity:

$p(x) = \displaystyle \sum_{i=0}^\infty e_i x^i$
##### Second step: defining the function $g(x)$

Then, we take the natural logarithm of the polynomial $p(x)$, getting:

$\ln(p(x)) = \displaystyle \sum_{i=1}^n \ln(1+\lambda_i x)$

Hence, we compute the first derivative of each side of the equation and we define it as $g(x)$:

$g(x)=\dfrac{\dot p(x)}{p(x)}=\displaystyle \sum_{i=1}^n \frac{\lambda_i}{1 + \lambda_i x}$

It is easily provable by induction that the following identity holds:

$\overset{i}{\dot g}(x)=(-1)^i i! \displaystyle \sum_{j=1}^n \left[ \left( \frac{\lambda_j}{1+\lambda_j x} \right) ^{i+1} \right]$

If we set $x$ to be equal to $0$, we get:

$\overset{i}{\dot g}(0)=(-1)^i i! \displaystyle \sum_{j=1}^n \lambda_j^{i+1}=(-1)^i P_{i+1}$
##### Third step: comparing $p(x)$ with its own Taylor-Maclaurin series

The polynomial $p(x)$ is also representable by its own Taylor-Maclaurin series (i.e. its Taylor series with $a$ set to be equal to $0$), which is $\displaystyle \sum_{i=0}^\infty \frac{\overset{i}{\dot p}(0)}{i!} x^i$.

Each term of the series must coincide with its corresponding term we found in the first step.

Hence, we get:

$e_i = \frac{\overset{i}{\dot p}(0)}{i!} \Longrightarrow \overset{i}{\dot p}(0) = e_i i!$
##### Fourth step: taking the $i$-th derivative of $p(x)$

Due to the fact that we defined $g(x)$ as $\frac{\dot p(x)}{p(x)}$, we also know that $\dot p(x) = p(x) g(x)$. Taking the $n$-th derivative using the Leibniz’s rule, we get:

$\overset{i}{\dot p}(x) = \displaystyle \sum_{j=1}^{i} \binom{i-1}{j-1} \overset{i-j}{\dot p}(x) \overset{j-1}{\dot g}(x)$

Hence, setting $x$ to be equal to $0$:

$\overset{i}{\dot p}(0) = \displaystyle \sum_{j=1}^{i} \binom{i-1}{j-1} \overset{i-j}{\dot p}(0) \overset{j-1}{\dot g}(0)$
##### Final step: comparing the $i$-th derivative of $p(0)$

If we compare the $i$-th derivative of $p(0)$ we found in the third step and the one we found in the fourth step, we get:

$e_i i! = \displaystyle \sum_{j=1}^{i} \binom{i-1}{j-1} \overset{i-j}{\dot p}(0) \overset{j-1}{\dot g}(0)$

Hence, by replacing all the terms with the corresponding ones we found in the third step and the second step and simplifying the expression we get, the following identity holds:

$i e_i = \displaystyle \sum_{j=1}^{i} (-1)^{j-1} e_{i-j} P_j$ $\tag*{\blacksquare}$

#### Application to the stated problem

We had to compute $P_2 = \lambda_1^2 + \lambda_2^2$ given that $\lambda_1$ and $\lambda_2$ are the roots of $p(x)=x^2-4x+3$.

Comparing the coefficients of $x$ in $p(x)$, we get $e_0 = 1$, as we’d already said earlier, $e_1 = 4$ and $e_2=3$.

Using the Newton-Girard identity, we notice that:

$2 e_2 = e_1 P_1 - e_0 P_2 \Longrightarrow P_2 = e_1 P_1 - 2 e_2$

$P_1$ is the same as $e_1$, hence, rearranging the expression, we obtain:

$P_2 = {e_1}^2 - 2 e_2 = 4^2 - 2 \cdot 3 = 10$

We can follow the same steps to compute $P_3 = \lambda_1^3 + \lambda_2^3$:

$P_3 = 3 e_3 - e_2 P_1 + e_1 P_2$

Due to the fact that $e_k = 0$ for $k > n$, $e_3$ must be equal to $0$. Therefore, we obtain:

$P_3 = - e_2 P_1 + e_1 P_2 = - 3 \cdot 4 + 4 \cdot 10 = 28$

In conclusion, we have found a generalized solution for the kind of problems we’d stated in the first place. Newton, we did it, once again!

1. The concept of the shown proof is taken from ProofWiki, which I recommend you to visit.