1

So my text book defined polynomials as expressions of the form: $$\mathcal{P}(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_{2}x^{2}+a_{1}x^{1}+a_{0}x^{0}$$

And it even called $a_{0}$ a coefficient. And in one exercise they ask to calculate $\mathcal{Q}(0)$ where $\mathcal{Q}(x)=6x^2+x-7$ but that's impossible since according to the definition $\mathcal{P}(0)$ is undefined since $0^{0}$ is undefined. So we should rather define polynomials as $$\mathcal{P}(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_{2}x^{2}+a_{1}x^{1}+a_{0}$$ where $a_{0}$ is a constant.

What do you think?

user126208
  • 33
  • 2
    $0^0$ is only undefined if it hasn't been defined. It is not uncommon that it is defined (as $1$). – Daniel Fischer Feb 04 '14 at 19:35
  • 2
    Here $,x^0 $ denotes $1$. So evaluating it yields $1$. – Bill Dubuque Feb 04 '14 at 19:35
  • 2
    A common convention in discrete mathematics (as opposed to say calculus) is to take $0^0 = 1$. – vonbrand Feb 04 '14 at 19:35
  • $x^0$ denotes $x$ to the zeroth power which is undefined in case of $0$. And $0^{0}$ is undefined because $0/0$ is undefined since $$0^{0}=0^{1-1}=\frac{0^1}{0^1}undefined!$$ – user126208 Feb 04 '14 at 19:45
  • @user126208 You seem to be confusing the powering operation in a ring/monoid with an (real-valued) exponential function. They are not the same. The power function in a monoid is defined recursively by $, a^0 = 1,,\ a^{n+1} = a a^n.,$ In particular, in your case $,x^0 = 1,$ by definition. So the "value" $0^0$ never arises when evaluating polynomials. Also the monomials $,x^n,$ are not generally functions, e.g. $,x^2 = x,$ as functions on $,\Bbb Z/2,$ but not as formal polynomials. – Bill Dubuque Feb 04 '14 at 20:23

1 Answers1

1

Your textbook defines a polynomial as: $$\mathcal{P}(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_{2}x^{2}+a_{1}x^{1}+\color{red}{a_{0}x^{0}}$$

Normally, a polynomial is defined as: $$\mathcal{P}(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_{2}x^{2}+a_{1}x^{1}+\color{red}{a_{0}}$$

Where, as you may notice, the striking difference is the omission of $x^0$, which we're allowed to do because we define $x^0$ as $1$. (Remember that $ax =a$ iff $x=1$ or $a=0$.) Consequently, in this context, we define $0^0 = 1$.

Newb
  • 17,672
  • 13
  • 67
  • 114