Previously, we have explored methods to compute the essence of power functions $Ȣx^n$ which involves solving a large system of linear equations. This method is equivalent to solving for the inverse of a large $n\times n$ matrix where entries are values of pascal's triangle. Though the matrix method allow us to solve for large number of essences at once, it does not extend easily to solve for next iterations of essence coefficients. Rather than reusing the values we have already solved for, we will have to solve for inverse of a separate larger matrix again. Here we will introduce an iterative method for solving for these coefficients. Chapter 0: Recap Let us first remind ourselves of the definition of essence. For a function $f(x)$, we want to find the transformation $Ȣf(x)$ such that we are able to 'smooth out' its series: $$\sum_{i=a}^b f(i) = \int_{a-1}^b Ȣf(x) dx$$ For example, we can solve for the following functions: $$\begin{align*}Ȣ1 &= 1 \\ Ȣx &= x +
Polynomial Series can be easily solved using Power Series Formulas for each term in the polynomial. However, this can be frustrating since not every Power Formula are intuitive to memorize. We would like to find a more elegant and easier-to-recall formula for computing a Polynomial Series. This can be done using matrices. Notations and Equations We will borrow the notations and simple equations from Sequence Curve articles . There, we have extended a series to be continuous through the following identity: $$\sum_{i=m}^n f(i) = \int_{m-1}^nȢ\{f(x)\}dx $$ The $Ȣ\{f(x)\}$ acts as the rate of change of a series relative to its bounds. $$\frac{d}{dt} \sum_{i=m}^nf(i) = Ȣ\{f(n)\}\frac{dn}{dt} - Ȣ\{f(m-1)\}\frac{dm}{dt}$$ Letting $t=m=n$, we find $$\frac{d}{dt} \sum_{i=t}^tf(i) = \frac{d}{dt}f(t)= Ȣ\{f(t)\} - Ȣ\{f(t-1)\} $$ Remebering that $Ȣ$ Transformation is Linear, we can derive a simple identity of $$\frac{d}{dx}f(x+1) = Ȣ\{f(x+1)\} - Ȣ\{f(x)\} = Ȣ\{f(x+1)-f(x)\}$$ This will be use