Well...
I haven't seen much symmetric polynomials in Olympiads, but still I am learning, because I found them cute. And I am basically using this blog as my notes :P
What are symmetric polynomials?
One can understand this with examples. If we are considering over 3 variables, x_1,x_2,x_3 then
\sum_{sym}x_1^2\cdot x_2^3\cdot x_3=x_1^2\cdot x_2^3\cdot x_3+x_1^2\cdot x_3^3\cdot x_2+x_2^2\cdot x_1^3\cdot x_3+x_2^2\cdot x_3^3\cdot x_1+x_3^2\cdot x_1^3\cdot x_2.
See? 3! terms! Let's take one more example with again over 3 variables, x_1,x_2,x_3 then
\sum_{sym}x_1^2\cdot x_2^2= x_1^2\cdot x_2^2+x_1^2\cdot x_3^2+x_2^2\cdot x_1^2+x_2^2\cdot x_3^2+x_3^2\cdot x_1^2+x_3^2\cdot x_2^2
Wait.. why 2 times ? So basically what happens in symmetrictric sums, is we go through all n! possible permutations. So, here we have a^2\cdot b^2\cdot c^0 as like the "general" form type, right? Now, list down all the 3!=6 permutations of x_1,x_2,x_3, and put them in the general form, and then we sum it up.
So all possible permutations and then putting in the general form we get ,
{x_1,x_2,x_3}\rightarrow x_1^2\cdot x_2^2
{x_1,x_3,x_2}\rightarrow x_1^2\cdot x_3^2
{x_2,x_1,x_3}\rightarrow x_2^2\cdot x_1^2
{x_2,x_3,x_1}\rightarrow x_2^2\cdot x_3^2
{x_3,x_1,x_2}\rightarrow x_3^2\cdot x_1^2
{x_3,x_2,x_1}\rightarrow x_3^2\cdot x_3^2
Okie! Summing them and we will get the symmetric sums. So this makes sense I hope?
One thing to be noted is to not get confused between cyclic sums and symmetric sums.
To make things clear, Cyclic sums have only n terms whereas in symmetric sums, we have n! terms.
Let's go through another example , which I hope will perhaps clear this thing.
So considering over 4 variables, we get \sum_{cyc}x_1^2\cdot x_2=x_1^2\cdot x_2+x_2^2\cdot x_3+x_3^2\cdot x_4+x_4^2\cdot x_1.
See 4 terms rights? and we are summing it in "cycle" types.
Well, if you are still confused then go through OTIS excerpts, it's really nice!
Elementary Symmetric Polynomials:
Nothing fancy here, just viettas with no signs type.
The elementary symmetric polynomials in n variables x_1,x_2,\dots x_n can be defined as
\sigma_1= \sum x_i
\sigma_2= \sum_{i<j} x_i\cdot x_j
\sigma_3=\sum_{i<j<k} x_i\cdot x_j\cdot x_k
and so on with \sigma_n= x_1\cdot x_2 \dots x_n
Fundamental theorem of symmetric polynomials:
The fundamental theorem of symmetric polynomials states that any symmetric polynomial in n variables can be expressed by a polynomial expression in terms of these elementary symmetric polynomials (uniquely).
And now we are going to take a lot of examples to verify this theorem !!! Oh, and why am I doing that? Again, I like to understand what's happening :P.
a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)={\sigma_1}^2-2\cdot \sigma_2
a^3+b^3=(a+b)(a^2+b^2-ab)=(a+b)({a+b}^2-3ab)=\sigma_1\cdot (\sigma_1^2-3\sigma_2)=\sigma_1 ^3-3\cdot \sigma_1\cdot \sigma_2
a^3+b^3+c^3=(a+b+c)^3 -3(\sum_{sym}a^2b)-6(abc) =(a+b+c)^3-3(a+b+c)(ab+bc+ca)+9abc-6abc=\sigma_1^3-3\cdot \sigma_1\cdot \sigma_2+3\sigma_3
Okie this much was fine.. let's try a^4+b^4+c^4. So,
a^4+b^4+c^4=(a+b+c)^4- 4(a^2+b^2+c^2)(ab+ca+bc)- 8(abc)(a+b+c) =\sigma_1^4 -4({\sigma_1}^2-2\cdot \sigma_2)\cdot \sigma_2-8\sigma_3\cdot \sigma_1
These were simple.. now what about \sum_{sym}x_1^2x_2x_3 ? Note that,
2 \sum_{sym}x_1^2x_2x_3=2(\sum_{cyc}x_1x_2)^2- 2(\sum_{cyc}x_1^2x_2^2)=2\sigma_2^2-(x_1^2+x_2^2+x_3^2)^2 + (x_1^4+x_2^4+x_3^4) = 2\sigma_2^2- [{\sigma_1}^2-2\cdot \sigma_2]^2+[\sigma_1^4 -4({\sigma_1}^2-2\cdot \sigma_2)\cdot \sigma_2-8\sigma_3\cdot \sigma_1]
Okie.. this looks fine to me.. but too tiresome.. T_T. I might have made some mistake :(
Newton Sums:
It's given in AOPS wiki.
So we can basically find out any x_1^d+\dots+x_n^d, recursively using this
a^n+b^n+c^n=(a+b+c)(a^{n-1}+b^{n-1}+c^{n-1})-(ab+bc+ca)(a^{n-2}+b^{n-2}+c^{n-2})+abc(a^{n-3}+b^{n-3}+c^{n-3}).
So this guarantees that we can express in the elementary symmetric polynomials form, however I am not quite sure about the uniqueness..
Yeah so that's it for today! I will do some problem solving now!
Sunaina💜
Nice Post! (:
ReplyDeleteTry proving the uniqueness :D
ReplyDeleteNice
ReplyDelete