# Polynomial representation

Why is the polynomial $P(x)$ represented as

$$P(x) = a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots+ a_2 x^2 + a_1 x + a_0 \text{ ?}$$

A polynomial can be $5x^4 + 3x^3 + 7x^2 + 10x -2$ and it is not necessarily $a_n x^n + a_{n-1} x^{n-1}+\cdots$ I.e if $a_n x^n$ is $5x^4$, doesn't $a_{n-1} x^{n-1}$ mean $4x^3$?

• $a_0, \ldots a_n$ are to be treated as numbers in their own right, independent of the others. It is false in general that $a_n - 1 = a_{n-1}$. In this case, $a_4 = 5$, $a_3 = 3$, $a_2 = 7$, etc Commented Jul 24, 2014 at 2:00
• Got it. Thanks.
– Minu
Commented Jul 24, 2014 at 2:08

$$a_{n-1}$$ is not the same as $$a_n -1$$

The $-1$ is in the subscript in $a_{n-1}$.

Think of it as a function $a$, and we are looking at $a(1), a(2), \dots, a(n-1), a(n)$, so the polynomial is

$$a(n) x^n + a(n-1) x^{n-1} + \dots + a(1) x + a(0)$$

So for instance, if we take $a(m) = m^2$ (can also be written as $a_{m} = m^2$), then for $n=3$ the polynomial will be

$$9x^3 + 4x^2 + x$$

• I think the function notation will only confuse things more Commented Jul 24, 2014 at 2:04
• @Mathmo123: Don't know. Let's see Minu's response. (You might be right though) Commented Jul 24, 2014 at 2:04
• Got it. Although, they need not be a function right? They are just independent numbers listed as $a_1, a_2, a_3 ...$ Etc.
– Minu
Commented Jul 24, 2014 at 2:09
• @Minu: Yes, that is correct. Commented Jul 24, 2014 at 2:29

A polynomial is defined by its coefficients and associated exponents, creating a $2$-dimensional space.

Note that it is possible to define a polynomial by coefficients and "exponents" but in a different fashion, as in binomial terms:

$${x\choose n}=\frac{x!}{n!(x-n)!}=\frac {\prod_{i=0}^{n-1}(x-i)}{n!}$$

Then a polynomial $P(x)$ is defined as

$$P(x)=\sum_{i=0}^na_i{x\choose i}$$

Note that the $a_i$ in this definition are different but easily translated to the $a_i$ from the "usual" definition.