Now we generalise this to *n* variables: . It’s clear what the corresponding building blocks of symmetric polynomials would be:

- ;
- ;
- ;
- …
- .

We call these *e _{i}*‘s the

**elementary symmetric polynomials**in the

*x*‘s. Note that each

_{i}*e*is the coefficient of

_{i}*T*in the expansion of the following polynomial:

^{i}It’s easy to generalise the results in the earlier posts to derive a recurrence relation in the sum :

(*)

But there’s a slight hitch in using this recurrence relation: we must know the first *n* sums before applying it! Thus, we must already have the relations for . Some of these are quite easy:

- ;
- ;
- .

But from the third powers onwards, it gets much trickier. Turns out there’s a good trick to this:

**Example 1**. Express the polynomial in terms of the elementary symmetric polynomials , , .

**Solution**. First, we reduce the number of variables via the substitution *d* = 0. Then *P*, *Q* and *R* are just the usual elementary symmetric polynomials in *a*, *b*, *c*. And our desired target polynomial is just . We already know how to express this in terms of *P*, *Q* and *R* :

Now we lift this back to the 4-variable case and consider the difference:

.

Since Δ is zero when *d*=0, the polynomial Δ must be a multiple of *d*. By symmetry it must be a multiple of *a*, *b* and *c*. Thus Δ is a multiple of *abcd*, which can only be zero since it has degree 3. Thus . ♦

And why stop here? Using the same reasoning, we can show that the equality holds in the 5-variable case. And so on. This allows us to generalise the recurrence (*) to include all *S _{k}*.

Newton’s Identities. Let be the sum of the k-th powers of the variables x_{i}. Then we have:

- ;
- ;
- ;
- …
- .

These allow us to recursively calculate all *S _{k}*.

Now we’re ready to solve tougher problems.

**Example 2**. (APMO 2003 Q1) The following polynomial has eight real and positive roots. Find all possible values of *f*.

**Solution**. Denote the roots by *x _{i}*, where

*i*= 1,…,8. Then and . Hence . Next, we use the root-mean-square & arithmetic-mean inequality (a special case of Cauchy-Schwarz inequality:

with equality if and only if all the *x _{i }*are equal. But LHS = 1/2 = RHS, so indeed equality holds, so each

*x*= 1/2 and

_{i}*f*= 1/256. ♦

**Example 3**. Given the following set of linear simultaneous equations in *x*, *y*, *z*, *w*, find the sum *x*+*y*+*z*+*w*.

**Solution**. Let *x*, *y*, *z*, *w* be fixed values. Consider the following equation in *T*:

The conditions of the problem imply that are solutions. Upon clearing denominators and simplifying, we get a quartic equation in the form of . Since the degree of the polynomial is 4, its roots are precisely 4, 16, 36, 64. On the other hand, the sum of roots is the negative of the coefficient of *T*^{3}, which is 84+*x*+*y*+*z*+*w*; so *x*+*y*+*z*+*w* = 36. ♦

Finally, some exercises for you to hone your skills.

**Exercise 1**. Solve the following simultaneous equations:

**Exercise 2**. Factor the following polynomials:

**Exercise 3**. For a given complex *a*, prove that for any solution to the following simultaneous equations:

two of the *x _{i}*‘s are equal. [ No, it’s not a typo; we really mean 3 equations in 4 variables. ]

**Exercise 4**. (USAMO 1977 Q3) (*) Prove that if *x*, *y*, *z*, *w* are roots of the quartic polynomial , then *xy*, *xz*, *xw*, *yz*, *yw*, *zw* are roots of the sextic polynomial .

**Exercise 5**. Let *n* be a positive integer. Prove that the only complex solutions to the following simultaneous equations:

are .