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: $$\begin{align*}Ȣ1 &= 1 \\ Ȣx &= x +
Last time, we introduced the concept of Sequence Curve. This time we will explore how the Essence Transformation will actually work. To start this, we will consider the simplest function: linear function $f(x) = x$. $$\sum_{i=n}^m i = \int_{n-1}^m Ȣ\{x\} dx$$ How can we find $Ȣ\{x\}$? For this, we can utilize the well known Arithmetic Series formula to simplify this problem. $$\sum_{i=n}^m i = \frac{m+1 - n}{2}(m+n) $$ This lends to that $$\frac{m+1-n}{2}(m+n) = \int_{n-1}^m Ȣ\{x\} dx $$ From this relation we will be able to solve for $Ȣ\{x\}$. Let’s consider the case when $n$ is a constant $c$. Then it must be case that $$ \int_{c-1}^m Ȣ\{x\} dx = \frac{m+1-c}{2}(m+c) $$ From First Fundamental Theorem of Calculus, we find that $$\begin{align*} \frac{d}{dm}\int_{c-1}^m Ȣ\{x\} dx & = \frac{d}{dm}\frac{m+1-c}{2}(m+c) \\ Ȣ\{m\} & = \frac{m+1-c}{2} + \frac{m+c}{2} \\ & = m + \frac{1}{2} \end{align*}$$ Let's see this in action t