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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 $4x^3$?

Michael Hardy
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Minu
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    $a_0, \ldots a_n$ 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 – Mathmo123 Jul 24 '14 at 02:00
  • Got it. Thanks. – Minu Jul 24 '14 at 02:08

2 Answers2

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$$a_{n-1}$$ 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

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

Aryabhata
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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.

abiessu
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