Playing around with coefficients

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.

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