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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 +...
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Curvature in Terms of r', r'', and Their Magnitude

 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 $$\begin{align*} T &= \frac{{r}'}{|{r}'|} \\  \\
T' &= \left(  \frac{r '}{|r'|} \right)'  \\
&= \frac{|r'|r'' + | r '|' r'}{| r '|^2} \\ 
|T'| &= \frac{\left |  | r '| r'' + | r' |' r' \right |}{| {r}'|^2} \\ \\
N &= \frac{T'}{| T' |} \\
&= T' \frac{ 1 }{ |T'| } \\
&=  \frac{ | {r}'|  {r}'' + | r' |'  r' }{ | {r}'|^2 }  \frac{ | {r}' |^2 }{ \left |  | {r}'| {r}'' + | {r}'|' {r}' \right | } \\
&= \frac{| {r}'| {r}'' + | {r}'|' {r}'}{\left |  | {r}'|  {r}'' + | {r}'|' {r}' \right |} \\ \\
B &= T\times N \\ 
&= \frac{ {r}'}{| {r}'|} \times \frac{| {r}'|r'' + | {r}'|'r'}{\left |  | {r}'| r'' + | {r}'|' r' \right |} \\
&= \frac{1}{|r'|} \frac{ 1 }{ \left |  | {r}'| r'' + | {r}'|' r' \right | } \left( r' \times (|r'| r'' - |r'|' r') \right), \text{ Factor our Scalars.} \\
&= \frac{1}{|r'|} \frac{ 1 }{ \left |  | {r}'| r'' + | {r}'|' r' \right | } \left( |r'| \, r' \times r'' - |r''| \, r' \times r' \right), \text{Distribute out cross product. }\\
&=  \frac{1}{|r'|} \frac{ 1 }{ \left |  | {r}'| r'' + | {r}'|' r' \right | } \left( |r'| \, r' \times r'' \right) \\
&= \frac{ r' \times r'' }{ \left |  | {r}'| r'' + | {r}'|' r' \right | }
\end{align*}$$

Notice that $$|T| = 1 \\ \therefore |T|^2 = T\cdot T = 1$$

$$\begin{align*}
\because T \cdot T &= 1 \\
\therefore (T\cdot T)'  &= 0\\
T'\cdot T + T\cdot T' &= 0\\
2T\cdot T'  &= 0\\
T\cdot \frac{T'}{| T' |} &= 0\\
\therefore T\cdot N &= 0 
\end{align*}$$ This shows that $T$ and $N$ are orthogonal. 

$$\therefore |B| = \sin(\theta_{T,N})|T||N| = |T||N|= 1$$

$$\begin{align*}
\because |B| &= \frac{ \left |  r' \times r''  \right | }{ \left | |r'| r'' - |r'|' r' \right | } \\ \\
\therefore \left | |r'| r'' - |r'|' r' \right | &=  \left |  r' \times r''  \right | \\
&= \sin(\theta_{r', r''})|r'||r''|
\end{align*}$$

By definition of cross product, where $\theta_{r', r''}$ is angle between the vectors $r'$ and $r''$.

$$\begin{align*}
\therefore \sin(\theta_{r', r''})|r''| &= \left |  r'' - |r'|' \frac{r'}{|r'|} \right | \\
&=  \left |  r'' - |r'|' T \right | \\
\end{align*}$$

Arclength is defined as $$s(t) = \int_{a}^t | {r}'| dt \\ \therefore s' = | {r}' |$$

Curvature is originally defined as $$\begin{align*} 
\kappa &= \left | \frac{dT}{ds} \right | \\
&= \left | \frac{dT}{dt} \frac{dt}{ds}  \right | \\
&= \left | \frac{T'}{s'} \right | \\
\therefore \kappa &= \frac{ \left | T' \right | }{ \left | {r}' \right | }
\end{align*}$$

We can define the second derivatives of the curve using the terms above.

$$\begin{align*}
r'' &= (r')' \\
&= \left( |r'|T \right)' \\
&= |r'|' T + |r'| T' \\
&= |r'|' T + |r'| |T'| \frac{T'}{|T'|} \\
&= |r'|' T + |r'| |T'| N \\
&= |r'|' T + |r'| ^2\frac{|T'|}{|r'|} N \\
r'' &= |r'|' T + |r'| ^2\kappa N \\ \\
\end{align*}$$

$$\therefore |r'| ^2\kappa N = r'' - |r'|' T $$

Remember that $$\begin{align*}
\sin(\theta_{r', r''}) |r''| &= \left |  r'' - |r'|' T \right | \\
&= \left | |r'|^2 \kappa N \right | \\
&= |r'|^2 \kappa
\end{align*}$$

So we can redefine Curvature as

$$\kappa = \frac{ \sin(\theta_{r' , r''} ) |r''| }{ |r'|^2 }$$

Where $\theta_{r',r''} \in [0,\pi)$ is the angle between the two vectors $r'$ and $r''$.

Notice that because $\theta_{r',r''} \in [0,\pi)$, 
$$\sin(\theta_{r',r''}) \geq 0$$  We also know from properites of dot product and trig identity that $$\begin{align*}
\because \cos(\theta_{r',r''}) &= \frac{r'\cdot r''}{|r'||r''|} \\
\because \sin^2(\theta) + \cos^2(\theta) &= 1 \\ \\
\therefore \sin^2(\theta_{r',r''}) &= 1 - \cos^2(\theta_{r',r''}) \\
&= 1 - \left( \frac{ r'\cdot r'' }{ |r'||r''| } \right)^2
\end{align*}$$

Notice that because $\sin(\theta_{r',r''}) \geq 0$, we can solve for $\sin(\theta_{r',r''}) $ by rooting both sides. $$\sin(\theta_{r',r''})  = \sqrt{1 - \left( \frac{ r'\cdot r'' }{ |r'||r''| } \right)^2}$$

Finally, we find that

$$\begin{align*}
\kappa &= \frac{ |r''| }{ |r'|^2 } \sqrt{1 - \left( \frac{ r'\cdot r'' }{ |r'||r''| } \right)^2} \\ \\
&= \frac{1}{|r'|^3} \sqrt{ (|r'||r''|)^2 - (r' \cdot r'')^2 } 
\end{align*}$$ QED

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