I was once thinking about how polynomials could be thought of infinite dimensional vectorspace with the basis vectors being different powers of x i.e. \(x^0, x^1, x^2, ...\). And then I thought about the vector notation of a polynomial.

Say a polynomial \(f(x) = 3x^2 + 2x - 7\) be a vector of 3-dimensional vectorspace with basis vectors \(x^0, x^1, x^2\) which in the vector notation is written as \(\left( \begin{array}{r} 3\\ 2 \\ -7 \\ \end{array} \right)\)

So, I started to think how can one extract the coefficients of a polynomial without any manual operation (reminded me of calculator programming).

To reiterate the problem, given a polynomial of a finite (known) degree, what could be a performed that yields each cofficient of the polynomial? So if \(f(x) = ax^2 + bx + c\) is a polynomial, I want 3 formula or methods (\(A(f(x)), B(f(x))\) and \(C(f(x))\)) or a single formula with a parameter, that yields \(a, b\) and \(c\).

After few minutes of thinking, I found out that finding \(a\) is really simple.

Here,

\[f(x) = ax^2 + bx + c\]

So,

\[f'(x) = 2ax + b\]

and,

\[f''(x) = 2a\]

Hence,

\[a = A(f(x)) = \frac{f''(x)}{2} = \frac{d^2}{dx^2}\left(\frac{f(x)}{2}\right)\]

At this point I knew finding other coefficients would also be very simple and immediately I found the trick to do so.

Using the expression for \(a\) and the above expression for first derivative \(f'(x)\), we get,

\[b=B(f(x)=f'(x)-2ax=f'(x)-2x\frac{d^2}{dx^2}\left(\frac{f(x)}{2}\right)\]

Similarly, to find the coefficient \(c\),

\[c = f(x)-ax^2-bx\]

Or,

\[c = f(x)-x^2\frac{d^2}{dx^2}\left(\frac{f(x)}{2}\right)-x\left(f'(x)-2x\frac{d^2}{dx^2}\left(\frac{f(x)}{2}\right)\right)\]

So, that is it. Except, it is not. I thought that was it at first glance and I was okay with the result. But the introduction of the chains of derivatives in the equation reminded me of the Taylor expansion of any function, particularly at \(x=0\) (which is called the Maclalurin Series). The Maclaurin series is given as:

\[f(x) = f(0) + xf'(0) + x^2\frac{f''(0)}{2!} + x^3\frac{f'''(0)}{3!} + ... +x^n\frac{f^n(0)}{n!} + ...\]

For a \(n^{th}\) order polynomial, the higher order derivatives (after n) vanish, and the coefficients of different powers of x are represented by the higher order derivatives at \(x=0\) divided by corresponding factorial term. In our case,

\[c=\frac{f(0)}{0!}\] \[b=\frac{f'(0)}{1!}\]

and

\[a=\frac{f''(0)}{2!}\]

So, finally I came to more elegant looking stuffs which was already there in my knowledge but had rusted for a while. Overall, I felt good and it was like trying to run an old engine again and repairing it.

Last Updated: Tuesday, 9 Oct, 2019
Author: Madhav Humagain (scimad)