Transforming polynomials

In mathematics, a polynomial is a function of the form $a\_0\; +\; a\_1\; x\; +\; cdots\; +\; a\_\{n-1\}\; x^n$.This article will discuss various polynomial transformations.

Reciprocals of the roots

Suppose we have some polynomial $a\_0\; +\; a\_1\; x\; +\; cdots\; +\; a\_\{n-1\}\; x^n$ and we are to find the sum of the reciprocals of the roots. Initially we may think that we are faced with the task of finding the roots and then their reciprocals and summing them, but there is an easier way.

For every root $r\_n$ of f(x) the following holds, $fleft(\; frac\{1\}\{frac\{1\}\{r\_n\; ight)\; =\; 0$. Looking at this statement further we see that plugging in x for $frac\{1\}\{r\_n\}$ we get a polynomial whose roots are the reciprocals of the roots of f(x). Nevertheless this function will not be a polynomial, as its highest power will be $frac\{1\}\{x^n\}$. To turn this into a polynomial all we have to do is multiply it by $x^n$. Note the order of the coefficents, earlier the polynomial was $a\_0\; +\; a\_1\; x\; +\; cdots\; +\; a\_\{n-1\}\; x^n$, now the coefficents have simply reversed so we get $a\_\{n-1\}\; +\; a\_\{n-2\}\; x\; +\; cdots\; +\; a\_0\; x^n$. The sum of the roots of this polynomial will be $a\_1$, which was the product of the roots taken n-1 at a time fot the original polynomial.

Constant multiples of the roots

To find say a function that has k times the roots of f(x) we can either again find all the roots, multiply then by k and then multiply them togher appropriately to find some function for which they are the roots or we can think of it the smart way. We can make the substution $frac\{x\}\{2\}\; =\; x$ into f(x).

Roots that differ by a constant

Suppose we have some polynomial $a\_0\; +\; a\_1\; x\; +\; cdots\; +\; a\_\{n-1\}\; x^n$ and we want to find some polynomial that has roots that are k units greater than the roots of f(x). One obvious approach would be to find the roots of f(x) and then multiply out a polynomial that has k greater than those roots. Also notice that this can be easily accomplished by thinking of the graph of the function. If it has roots at $r\_1,\; r\_2\; ldots\; r\_n$ than we can simply shift the function over k units which will give us a function that has roots that are k units greater than the roots of f(x). so for "x" in "f"("x") substutite "x" − "k".

The pattern

Suppose we have some one to one onto function "m"("x") and some function "f"("x") that has roots $r\_0,\; r\_1,\; r\_2,\; r\_3,\; ldots\; ,\; r\_n$ and we are faced with the problem of finding a function that has roots

:$m(r\_0),\; m(r\_1),\; m(r\_2),\; ldots,\; m(r\_n).$

Given that the function "m"("x") is both one to one and onto we can find $m^\{-1\}(x)$. Now to find a function that has roots $m(r\_0),\; m(r\_1),\; m(r\_2),\; ldots,\; m(r\_n)$ we only have to make the substution $x=\; m^\{-1\}(x)$. This function will have roots at $m(r\_0),\; m(r\_1),\; m(r\_2),\; ldots,\; m(r\_n)$ because "f"("x") only has roots at $r\_0,\; r\_1,\; r\_2,\; r\_3,\; ldots\; ,\; r\_n$ and plugging in $m(r\_n)$ into the function will be the same as pluggiing in $r\_n$ into f(x) because $m^\{-1\}(x)$ is a function such that $m^\{-1\}(m(x))\; =\; x$.