2
$\begingroup$

Usually, we define a polynomial as

$a_n x^n + \cdots + a_1 x + a_0$

where $x$ is called indeterminate.

Would it be better to define it as

$a_n x^n + \cdots + a_1 x + a_0 x^0$

where $x^0$ means the identity element in the structure which $x$ belongs to.

For the study of the polynomial itself, I think these two definition make no difference. But when you treat a polynomial (expression) as a polynomial function, and if we use the second definition, we can just substitute all the $x$ with some variable (eg. some square matrix), rather than define it in a "adhoc" way, i.e. if we are using the first definition, we have to explicitly define the function to add an identity element to the constant term $a_0$ to make $a_1 x$ and $a_0$ addable.

Is the second definition equivalents to the first one?

If yes, then is it true that the authors of those texts actually means def 2 when the define the polynomial using def 1, they just omit the identity element?

If not, why? Would it be nicer to eliminate the non-addable $a_1x + a_0$ with $a_1x + a_0 x^0$, (even though we really don't need to add these two terms together when we are studying the knowledge of polynomial itself)?

$\endgroup$
13
  • $\begingroup$ It really doesn't matter. Quite often, when written in sigma notation, we write it as $\sum_{i=0}^n a_ix^i$, even if we otherwise write $a+bx+cx^2$. I often use this notation for one of the reasons we define $0^0=1$ :) $\endgroup$ – Thomas Andrews Aug 7 '14 at 3:33
  • 1
    $\begingroup$ Please give an example of what you mean by the second requires "adding an identity... to make addable" $\endgroup$ – Bill Dubuque Aug 7 '14 at 3:34
  • 1
    $\begingroup$ Did you mean $x^0$ instead of $x_0$ in that last line? $\endgroup$ – Thomas Andrews Aug 7 '14 at 3:35
  • $\begingroup$ You can also define a polynomial as a sequence $(a_0,a_1,\dots)$ where only finitely many $a_i$ are non-zero. Then $(1,0,0,\dots)$ is not $x^0$, where $x=(0,1,0,0,\dots)$, but rather $(1,0,0,0,0,\dots)$ is just the identity. $\endgroup$ – Thomas Andrews Aug 7 '14 at 3:37
  • $\begingroup$ @BillDubuque Suppose that $x$ is a matrix, then $a_1 x + a_0$ is not a valid expression, but if you "add" the identity matrix $I$ to $a_0$, you get $a_0I$, which makes $a_1 x + a_0 I$ a valid expression. $\endgroup$ – Not an ID Aug 7 '14 at 4:09
2
$\begingroup$

It really doesn't matter which way you define polynomial. Another way is to consider all sequences:

$$(a_0,a_1,\dots,a_n,\dots)$$ where only finitely many $a_i$ are non-zero.

Then we add series point-wise, and we find their products by the Cauchy product.

Then $(1,0,0,0,\dots)$ is the multiplicative identity, and $(0,1,0,0,\dots)$ is $x$. So $1$ isn't really $x^0$, and $x^2$ just means $x\cdot x$.

$\endgroup$
3
  • $\begingroup$ For the study of the polynomial itself, it really doesn't matter. It is when you treat it as function that causes a problem, if you think def 2 is equivalent to def 1, then you will run into trouble with this formula $p(\lambda) = det(\lambda I - A)$, where $p$ is the characteristic polynomial of matrix $A$. because substitution is allowed, we can substitute the $\lambda$ in both sides with $A$, then you get "zero (that is the number 0) = ZERO (that is the zero matrix)". (to be continued) $\endgroup$ – Not an ID Aug 7 '14 at 4:27
  • $\begingroup$ ... Please see this question for detail. $\endgroup$ – Not an ID Aug 7 '14 at 4:28
  • $\begingroup$ @NotanID No, there is no problem when treating it as a function, as long as you accept that $0^0=1$. $\endgroup$ – Thomas Andrews Aug 7 '14 at 4:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.