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Power Essence Coefficients and Bernoulli Numbers

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 +

Reduced Column Echelon Form: Image, Kernel, and their Dimensions

In Linear Algebra, Vector Transformations are analogous to Functions of Real Numbers: both take in an input and yields a specific output. Main difference appears in the fact that inputs of Transformations, Vectors, is multiple-dimensional construct, rather than being a single real number. Real Function:$$  f(x) = y, \\ \text{where } x\in \mathbb R, \\ y \in \mathbb R $$ Vector Transformation/Matrices$$ T(\vec{x}) = A\vec{x} = \vec{y}, \\ \text{where } \vec{x} \in \mathbb R ^m, \\ \vec{y} \in \mathbb R ^ n, \\ \text{and $A$ is of order $n \times m$} $$ Similar to how algebra students learn to solve for the Zeroes and Range of a Function, linear algebra students learn to solve for the linear-algebraic counterparts of the zeroes and range known as Kernel and Image. Unlike in a real number functions where zeroes and range can be described by real number intervals, Kernels and Images exists as multidimensional Sub-Space of $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively. Solving f