3

If we have two polynomials, say for example, $a_0+a_1x+a_2x^2 = 1+2x+3x^2$ for all $x$, why must it follow that $a_0 = 1,a_2 = 2,$ and $a_3 = 3$? Can we generalize this for any polynomials and infinite sum polynomials as well?

Puzzled417
  • 6,956

4 Answers4

1

One way to think about it is that a polynomial is characterized by the name of the variable, the degree, and the list of coefficients. In other words, if I said, "I'm thinking of a polynomial in the variable $x$, of degree $2$, with constant term $1$, linear coefficient $2$, and quadratic coefficient $3$", you could guess which polynomial I was thinking about in one try. I gave you all the information necessary to write down $1 + 2x + 3x^2$.

To make it slightly more sophisticated, two polynomials are equal if and only if they have the same variable, degree, and coefficients. So if $$ a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \dots + a_n x^n = 1 + 2x + 3x^2 $$ then $a_0 = 1$, $a_1 = 2$, $a_3 = 3$, and $a_4 = a_5 = \dots = a_n = 0$.

Yes, it does generalize. For any ring $R$, the ring $R[x]$ stands for polynomials in $x$ with coefficients from $R$. By definition, the powers of $x$—$1$, $x$, $x^2$, $x^3$, etc.—are independent over $R$, so two polynomials are equal iff their coefficients are equal. The ring $R[[x]]$ consists of infinite series in $x$, and has the same property.

Matthew Leingang
  • 26,112
0

Yes.

Evaluate $a_0+a_1x+a_2x^2 = 1+2x+3x^2$ at $x=0$ and conclude that $a_0=1$.

Subtract $a_0$ from both sides, divide by $x$ and evaluate at $x=0$ to conclude that $a_1=2$.

Repeat.

This procedure works in general, for both polynomials and formal power series.

lhf
  • 216,483
  • 2
    If you divide by $x$ then you get an equality that is true only if $x \neq 0$. Thus, you cannot evaluate the resulting equality at $x = 0$ as you suggested. What am I missing? – s7amuser Jun 07 '20 at 17:27
  • 1
    @s7amuser Good catch. This subtlety is what you are missing. That surely does deserve explicit justification since it often puzzles readers in this and analogous contexts. – Bill Dubuque Dec 07 '22 at 19:12
0

If equal, subtract them from each other and you know the answer is zero, perfect! So that means $0 = (a_0 - 1) + (a_1 - 2)x + (a_2 - 3) x^2$, but you already picked up on that. But, and here's where it all happens, there's a LOT of flexibility with what $x$ is. What is it telling you when $x = 0$, for example? And if that's the case, about $a_1$? I think you see the pattern.

A rural reader
  • 2,507
0

This follows from the fact that if a polynomial is identically vanishing, then it mist be the null polynomial, all of whose coefficients vanish.

Now transpose all terms to one side to have a polynomial which vanishes identically. Setting the coefficients of this polynomial each to $0$ gives you your result.

Allawonder
  • 13,327