### Method NO. 4 : The polynomial argument

19Dec08

The polynomial argument is a very useful fact which can be used in order to prove, or justify a statement.

It bases on simple fact that non-zero polynomial of k degree, can have no more than k solutions, futhermore the difference of two polynomials of k degree can also have max of k solutions if only those polynomials weren’t identical, then they have infinite numbers of solutions.

So how to use that fact?

Let me show an example, with use of extended Newtons binominal for all real numbers defined as:

${r \choose k} = \frac{r(r-1)...(r-k+1)}{(k(k-1)...(1))} \quad r \in \mathbb{R}, k \in \mathbb{N}$

We will try to prove one of it’s properties using the properties of the normal binominal and the polynomial argument.

First, the folowing equality called the rule of absoption works for all real numbers $r, k \neq 0$: ${r \choose k} = \frac{r}{k}{{r-1} \choose {k-1}}$. We will it into other form: $k{r \choose k} = r{{r-1} \choose {k-1}} \quad *$

Next you can simply check the property of symmetry of the normal binominal isn’t extendable to our extended definition of the binomial: ${r \choose k} \neq {r \choose {n-k}}$. However symmetry stays for all positive $n$ values, so will call this propetry using symbol **.

Now the proper part, we’ll prove that $(r-k){r \choose k} = r{{r-1} \choose {k}}$ for all real numbers $r$

Proof:

$(r-k){r \choose k} =^{**} (r-k){r \choose {r-k}} =^{*} r{{r-1} \choose {r-k-1}} =^{**} r{{r-1} \choose k}$

Now probably most of people would disagree, since for proving something for all real numbers I used a property which works only for the positive integers. Well they are right, the missing part is the polynominal argument: Look at the left and right part of equotation like on a polynomial of degree $k+1$ then using using the property of correctness of this theorem for infinite number of arguments $r \in \mathbb{N}$, we can tell that both of this polynomials have to be identical ■

Quite clever isn’t it?

#### 5 Responses to “Method NO. 4 : The polynomial argument”

1. 1 kauser

” like on a polynomial of degree K+1″
why this is of degree k+1 not k .
from r(r-1 k)

• I know that might have look confusing at first, but if you look at the definition of ${r \choose k}$ then you’ll notice that the degree of this polynomial with respect to r depends on k (not r-1 as one might have thought). Therefore the degree of the polynomial on both sides is k+1.

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