Sungjai's Recreational Mathematics

The search for a basis for the orthogonal complement of a subspace has often involved pure guesswork or many tedious computations of removing orthogonal projections from a known basis for the entire vector space. Here we propose an alternative faster method of producing a basis for the orthogonal complement of an image of a matrix.  Let $A$ be $n \times m$ matrix over $\mathbb{C}$. Then such matrix maps from $\mathbb{C}^m$ to $\mathbb{C}^n$. $$A:\mathbb{C}^m \rightarrow \mathbb{C}^n$$ (Choice of the field $\mathbb{C}$ is to allow us to use the standard inner product $<\cdot, \cdot>$ to define orthogonality. This, of course, works also for $\mathbb{R}$ ) Then its image $im\,A$ and the orthogonal complement of the image $im\,A ^\perp$ are both subspaces of the codomain $\mathbb{C}^n$ such that $$\mathbb{C}^n = im\, A \oplus im\, A^\perp$$ A Basis for Image of $A$ Our first task will be to determine a basis for the image. I have already discussed this in a separate lengt

 This article is intended to provide a simpler formula for the curvature in terms of easy-to-solve $r'$, $r''$, $| r' |$, and $|r'' |$.  Let curve $C$ be defined by $r(t)$ such that $$r:[a,b] \subseteq \mathbb{R}  \rightarrow \mathbb{R}^n $$ Then we can define Tagent, Normal, and Binormal Unit Vectors to be $$\begin{align*} T &= \frac{{r}'}{|{r}'|} \\  \\ T' &= \left(  \frac{r '}{|r'|} \right)'  \\ &= \frac{|r'|r'' + | r '|' r'}{| r '|^2} \\  |T'| &= \frac{\left |  | r '| r'' + | r' |' r' \right |}{| {r}'|^2} \\ \\ N &= \frac{T'}{| T' |} \\ &= T' \frac{ 1 }{ |T'| } \\ &=  \frac{ | {r}'|  {r}'' + | r' |'  r' }{ | {r}'|^2 }  \frac{ | {r}' |^2 }{ \left |  | {r}'| {r}'' + | {r}'|' {r}' \right | } \\ &= \frac{| {r}'| {r}'' + | {r

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

Partitions are number of ways an integer can be represented as sum of positive integers. We can ask, for example, what is the partition of 5? If we write out every possible combination, $$\begin{align*}  5 &= 1+1+1+1+1 \\  &= 1+1+1+2\\ &= 1+1+3\\ &= 1+4\\ &= 1+2+2\\ &= 2+3\\  &= 5 \end{align*} $$ we can see that partition of 5 is 7. One will immediately notice, however, that this is not the most efficient approach to answering the question: not only does partition grow quickly, attempting to organize and not miss or repeat an algebraic expression becomes messy and impractical. Chapter 1: Partition Tree A cleaner, but still not the best, approach would be to use tree diagrams to represent the partitions of a number. In a Partition Tree of $n$, every path to a terminating node will add up to the number $n$. And also, every child of a node (every nodes below a given parent node) will always be greater than or equal to the parent node in order to not