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: $$\begin{align*}Ȣ1 &= 1 \\ Ȣx &= x +...
Last time, we covered Geometric Series and before that we covered Arithmetic Series.
Now we will try to find formulas for more unusual Series such as $\sum\sin$ and $\sum\cos$.
$$\sum_{i=n}^m \sin(i) = \int_{n-1}^mȢ\{\sin(x)\}dx, \\
\sum_{i=n}^m \cos(i) = \int_{n-1}^mȢ\{\cos(x)\}dx $$
$$\begin{align*}
Ȣ\{\sin(x+1)-\sin(x)\}&=\frac{d}{dx}\sin(x+1)\\
Ȣ\{\sin(1)\cos(x)+\cos(1)\sin(x) - \sin(x)\}&=\cos(x+1)\\
Ȣ\{\sin(1)\cos(x) + (\cos(1)-1)\sin(x) \}&=\\
\sin(1)Ȣ\{\cos(x)\}+(\cos(1)-1)Ȣ\{\sin(x)\}&=\\
\therefore Ȣ\{\sin(x)\} &= \frac{\cos(x+1) - \sin(1)Ȣ\{\cos(x)\}}{\cos(1)-1}
\end{align*}$$ and $$\begin{align*}
Ȣ\{\cos(x+1)-\cos(x)\}&= \frac{d}{dx}\cos(x+1)\\
Ȣ\{\cos(1)\cos(x)-\sin(1)\sin(x)-\cos(x)\}& = -\sin(x+1) \\
Ȣ\{-\sin(1)\sin(x) + (\cos(1)-1)\cos(x)\}&= \\
-\sin(1)Ȣ\{\sin(x)\}+(\cos(1)-1)Ȣ\{\cos(x)\}&=\\
\therefore Ȣ\{\sin(x)\}&=\frac{(\cos(1)-1)Ȣ\{\cos(x)\}+\sin(x+1)}{\sin(1)} \end{align*}$$ We thus find two definitions of Essence of Sine each in terms of Essence of Cosine. We can use substitution to solve for Essence of Cosine.
$$\begin{align*}\frac{(\cos(1)-1)Ȣ\{\cos(x)\}+\sin(x+1)}{\sin(1)} &= \frac{\cos(x+1) - \sin(1)Ȣ\{\cos(x)\}}{\cos(1)-1} \\
(\cos(1)-1)^2Ȣ\{\cos(x)\} + (\cos(1)-1)\sin(x+1) &= \sin(1)\cos(x+1) - \sin^2(1)Ȣ\{\cos(x)\} \\
( (\cos(1)-1)^2+\sin^2(1) )Ȣ\{\cos(x)\} & = \sin(1)\cos(x+1) - (\cos(1)-1)\sin(x+1) \\
( \cos^2(1) + 1 - 2\cos(1) + \sin^2(1) )Ȣ\{\cos(x)\} & = \sin(1)\cos(x+1) - \cos(1)\sin(x+1) + \sin(x+1) \\
(2 - 2\cos(1))Ȣ\{\cos(x)\} &= \sin(1 - (x+1) ) + \sin(x+1) \\
2(1-\cos(1))Ȣ\{\cos(x)\} &= \sin(-x) + \sin(x+1) \\
\therefore Ȣ\{\cos(x) \} = \frac{\sin(x+1)-\sin(x)}{2(1-\cos(1))}
\end{align*}$$ Now knowing Essence of Cosine, we can substitute this back into one of the definitions for Essence of Sine. We will use the second definition in order to easily cancel out the $1-\cos(1)$ in the denominator. $$\begin{align*}
\because Ȣ\{\sin(x)\}&=\frac{(\cos(1)-1)Ȣ\{\cos(x)\}+\sin(x+1)}{\sin(1)} \\
&= \frac{-(1-\cos(1))\frac{\sin(x+1)-\sin(x)}{2(1-\cos(1))}+\sin(x+1)}{\sin(1)} \\
& = \frac{-\frac{1}{2}\sin(x+1) + \frac{1}{2}\sin(x) + \sin(x+1)}{\sin(1)} \\
& = \frac{\frac{1}{2}\sin(x+1) + \frac{1}{2}\sin(x)}{\sin(1)} \\
\therefore Ȣ\{\sin(x)\} & = \frac{\sin(x+1) + \sin(x)}{2\sin(1)}
\end{align*}$$ We arrived at simple equations for both Essences, but these are not the only possible definitions.
Using the Interchangeability of Essence and Differentiation, $$\frac{d}{dx}Ȣ\{f(x)\} = Ȣ\{\frac{d}{dx}f(x)\}$$ we can find that $$\begin{align*}
Ȣ\{\frac{d}{dx}\sin(x)\} & = \frac{d}{dx}\frac{\sin(x+1) + \sin(x)}{2\sin(1)} \\
\thereforeȢ\{\cos(x)\} & = \frac{\cos(x+1) + \cos(x)}{2\sin(1)} \\ \\
Ȣ\{\frac{d}{dx}\cos(x)\} & = \frac{d}{dx}\frac{\sin(x+1)-\sin(x)}{2(1-\cos(1))} \\
-Ȣ\{\sin(x)\} &= \frac{\cos(x+1) - \cos(x)}{2(1-\cos(1))} \\
\thereforeȢ\{\sin(x)\} &= \frac{-\cos(x+1) + \cos(x)}{2(1-\cos(1))} \end{align*}$$ With this, we conclude the Essence formulas for both:
$$\begin{align*}
Ȣ\{\sin(x)\} & = \frac{\sin(x+1) + \sin(x)}{2\sin(1)} & =\frac{-\cos(x+1) + \cos(x)}{2(1-\cos(1))} \\
Ȣ\{\cos(x)\} & = \frac{\cos(x+1) + \cos(x)}{2\sin(1)} & = \frac{\sin(x+1)-\sin(x)}{2(1-\cos(1))} \\
&&\blacksquare
\end{align*}$$
We note an interesting symmetry with the two definitions of each Sine and Cosine: when $\sin(1)$ is in the denominator, it is as if it is taking the average of the current and the next term of the sequence; when $1-\cos(1)$ is at the denominator, it is taking the difference of the current and next term of the other trigonometric function.
These simple pattern lends itself to be easily memorized for future uses. Personally, the first definitions with $\sin(1)$ seems to be more easily memorized and used. $$\sum_{i=n}^mf(i) = \int_{n-1}^mȢ\{f(x)\}dx$$ By definition of Essences, we find that $$\sum_{i=n}^m \sin(i) = \int_{n-1}^m\frac{\sin(x+1)+\sin(x)}{2\sin(1)}dx = \frac{\cos(n)+\cos(n-1)-\cos(m+1)-\cos(m)}{2\sin(1)} \\
\sum_{i=n}^m\cos(i) = \int_{n-1}^m \frac{\cos(x+1)+\cos(x)}{2\sin(1)}dx = \frac{\sin(m+1)+\sin(m) - \sin(n) - \sin(n-1)}{2\sin(1)}
$$ or if you wish to use alternate definition, $$\sum_{i=n}^m\sin(i) = \int_{n-1}^m\frac{-\cos(x+1)+\cos(x)}{2(1-\cos(1))}dx = \frac{-\sin(m+1) + \sin(m) + \sin(n) - \sin(n-1)}{2(1-\cos(1))} \\
\sum_{i=n}^m\cos(i) = \int_{n-1}^m\frac{\sin(x+1)-\sin(x)}{2(1-\cos(1))}dx = \frac{\cos(n)-\cos(n-1) - \cos(m+1) - \cos(m)}{2(1-\cos(1))}
$$
By similar process we can calculate $Ȣ_x\{\sin(nx)\}$ and $Ȣ_x\{\cos(nx)\}$ where $Ȣ_x$ denotes Essence with respect to $x$.
With these we can find that
$$
\begin{align*}
\sum_{i=n}^m\sin(r*i) & = \int_{n-1}^m\frac{r(\sin(rx+r) + \sin(rx) )}{2\sin(r)} dx &&= \int_{n-1}^m \frac{r(-\cos(rx+r) + \cos(rx))}{2(1-\cos(r))} dx \\
& = \frac{\cos(rn)+\cos(rn-r) - \cos(rm+r) - \cos(rm)}{2\sin(r)} &&= \frac{ -\sin(rm+r) + \sin(rm) + \sin(rn) - \sin(rn-r) }{2(1-\cos(r))} \\
\sum_{i=n}^m \cos(r*i) &= \int_{n-1}^m \frac{ r(\cos(rx+r) +\cos(rx) ) }{2\sin(r)} dx &&= \int_{n-1}^m \frac{ r( \sin(rx+r) - \sin(rx) ) }{2(1-\cos(r))}dx \\
&= \frac{ \sin(rm+r)+\sin(rm) - \sin(rn) - \sin(rn-r) }{2\sin(r)} &&= \frac{ \cos(rn) -\cos(rn-r) -\cos(rm+r) + \cos(rm) }{2(1-\cos(r))}
\end{align*} $$ What fascinates me the most is that with these formulas we are able to condense a long sum of sine and cosine, which has no obvious reduction rules by themselves, into a concise single equation. Also, the added benefit of being able to calculate non-integer boundary series.
In addition to those, $\cos(\pi x)$ is a popular continuous substitute for $(-1)^x$ and so knowing its Essence will help us interpolate between more discontinuous functions later on.
Now we will try to find formulas for more unusual Series such as $\sum\sin$ and $\sum\cos$.
$$\sum_{i=n}^m \sin(i) = \int_{n-1}^mȢ\{\sin(x)\}dx, \\
\sum_{i=n}^m \cos(i) = \int_{n-1}^mȢ\{\cos(x)\}dx $$
Essence Identity
$$Ȣ\{f(x+1) - f(x)\} = \frac{d}{dx}f(x+1) $$ Last time we were able to calculate Essence of Exponentials with help of this Identity (also called Essence Formula). This will also help us calculate Essence for Sine and Cosine, though through a more complicated process. Unlike Exponentials, the difference cannot be grouped into a single term. To compute $Ȣ\{\sin(x)\}$ and $Ȣ\{\cos(x)\}$, we will have to solve them simultaneously side-by-side.$Ȣ\{\sin(x)\} \text{ and }Ȣ\{\cos(x)\}$
Firstly, let us use Essence Identity to find that$$\begin{align*}
Ȣ\{\sin(x+1)-\sin(x)\}&=\frac{d}{dx}\sin(x+1)\\
Ȣ\{\sin(1)\cos(x)+\cos(1)\sin(x) - \sin(x)\}&=\cos(x+1)\\
Ȣ\{\sin(1)\cos(x) + (\cos(1)-1)\sin(x) \}&=\\
\sin(1)Ȣ\{\cos(x)\}+(\cos(1)-1)Ȣ\{\sin(x)\}&=\\
\therefore Ȣ\{\sin(x)\} &= \frac{\cos(x+1) - \sin(1)Ȣ\{\cos(x)\}}{\cos(1)-1}
\end{align*}$$ and $$\begin{align*}
Ȣ\{\cos(x+1)-\cos(x)\}&= \frac{d}{dx}\cos(x+1)\\
Ȣ\{\cos(1)\cos(x)-\sin(1)\sin(x)-\cos(x)\}& = -\sin(x+1) \\
Ȣ\{-\sin(1)\sin(x) + (\cos(1)-1)\cos(x)\}&= \\
-\sin(1)Ȣ\{\sin(x)\}+(\cos(1)-1)Ȣ\{\cos(x)\}&=\\
\therefore Ȣ\{\sin(x)\}&=\frac{(\cos(1)-1)Ȣ\{\cos(x)\}+\sin(x+1)}{\sin(1)} \end{align*}$$ We thus find two definitions of Essence of Sine each in terms of Essence of Cosine. We can use substitution to solve for Essence of Cosine.
$$\begin{align*}\frac{(\cos(1)-1)Ȣ\{\cos(x)\}+\sin(x+1)}{\sin(1)} &= \frac{\cos(x+1) - \sin(1)Ȣ\{\cos(x)\}}{\cos(1)-1} \\
(\cos(1)-1)^2Ȣ\{\cos(x)\} + (\cos(1)-1)\sin(x+1) &= \sin(1)\cos(x+1) - \sin^2(1)Ȣ\{\cos(x)\} \\
( (\cos(1)-1)^2+\sin^2(1) )Ȣ\{\cos(x)\} & = \sin(1)\cos(x+1) - (\cos(1)-1)\sin(x+1) \\
( \cos^2(1) + 1 - 2\cos(1) + \sin^2(1) )Ȣ\{\cos(x)\} & = \sin(1)\cos(x+1) - \cos(1)\sin(x+1) + \sin(x+1) \\
(2 - 2\cos(1))Ȣ\{\cos(x)\} &= \sin(1 - (x+1) ) + \sin(x+1) \\
2(1-\cos(1))Ȣ\{\cos(x)\} &= \sin(-x) + \sin(x+1) \\
\therefore Ȣ\{\cos(x) \} = \frac{\sin(x+1)-\sin(x)}{2(1-\cos(1))}
\end{align*}$$ Now knowing Essence of Cosine, we can substitute this back into one of the definitions for Essence of Sine. We will use the second definition in order to easily cancel out the $1-\cos(1)$ in the denominator. $$\begin{align*}
\because Ȣ\{\sin(x)\}&=\frac{(\cos(1)-1)Ȣ\{\cos(x)\}+\sin(x+1)}{\sin(1)} \\
&= \frac{-(1-\cos(1))\frac{\sin(x+1)-\sin(x)}{2(1-\cos(1))}+\sin(x+1)}{\sin(1)} \\
& = \frac{-\frac{1}{2}\sin(x+1) + \frac{1}{2}\sin(x) + \sin(x+1)}{\sin(1)} \\
& = \frac{\frac{1}{2}\sin(x+1) + \frac{1}{2}\sin(x)}{\sin(1)} \\
\therefore Ȣ\{\sin(x)\} & = \frac{\sin(x+1) + \sin(x)}{2\sin(1)}
\end{align*}$$ We arrived at simple equations for both Essences, but these are not the only possible definitions.
Using the Interchangeability of Essence and Differentiation, $$\frac{d}{dx}Ȣ\{f(x)\} = Ȣ\{\frac{d}{dx}f(x)\}$$ we can find that $$\begin{align*}
Ȣ\{\frac{d}{dx}\sin(x)\} & = \frac{d}{dx}\frac{\sin(x+1) + \sin(x)}{2\sin(1)} \\
\thereforeȢ\{\cos(x)\} & = \frac{\cos(x+1) + \cos(x)}{2\sin(1)} \\ \\
Ȣ\{\frac{d}{dx}\cos(x)\} & = \frac{d}{dx}\frac{\sin(x+1)-\sin(x)}{2(1-\cos(1))} \\
-Ȣ\{\sin(x)\} &= \frac{\cos(x+1) - \cos(x)}{2(1-\cos(1))} \\
\thereforeȢ\{\sin(x)\} &= \frac{-\cos(x+1) + \cos(x)}{2(1-\cos(1))} \end{align*}$$ With this, we conclude the Essence formulas for both:
$$\begin{align*}
Ȣ\{\sin(x)\} & = \frac{\sin(x+1) + \sin(x)}{2\sin(1)} & =\frac{-\cos(x+1) + \cos(x)}{2(1-\cos(1))} \\
Ȣ\{\cos(x)\} & = \frac{\cos(x+1) + \cos(x)}{2\sin(1)} & = \frac{\sin(x+1)-\sin(x)}{2(1-\cos(1))} \\
&&\blacksquare
\end{align*}$$
We note an interesting symmetry with the two definitions of each Sine and Cosine: when $\sin(1)$ is in the denominator, it is as if it is taking the average of the current and the next term of the sequence; when $1-\cos(1)$ is at the denominator, it is taking the difference of the current and next term of the other trigonometric function.
These simple pattern lends itself to be easily memorized for future uses. Personally, the first definitions with $\sin(1)$ seems to be more easily memorized and used. $$\sum_{i=n}^mf(i) = \int_{n-1}^mȢ\{f(x)\}dx$$ By definition of Essences, we find that $$\sum_{i=n}^m \sin(i) = \int_{n-1}^m\frac{\sin(x+1)+\sin(x)}{2\sin(1)}dx = \frac{\cos(n)+\cos(n-1)-\cos(m+1)-\cos(m)}{2\sin(1)} \\
\sum_{i=n}^m\cos(i) = \int_{n-1}^m \frac{\cos(x+1)+\cos(x)}{2\sin(1)}dx = \frac{\sin(m+1)+\sin(m) - \sin(n) - \sin(n-1)}{2\sin(1)}
$$ or if you wish to use alternate definition, $$\sum_{i=n}^m\sin(i) = \int_{n-1}^m\frac{-\cos(x+1)+\cos(x)}{2(1-\cos(1))}dx = \frac{-\sin(m+1) + \sin(m) + \sin(n) - \sin(n-1)}{2(1-\cos(1))} \\
\sum_{i=n}^m\cos(i) = \int_{n-1}^m\frac{\sin(x+1)-\sin(x)}{2(1-\cos(1))}dx = \frac{\cos(n)-\cos(n-1) - \cos(m+1) - \cos(m)}{2(1-\cos(1))}
$$
By similar process we can calculate $Ȣ_x\{\sin(nx)\}$ and $Ȣ_x\{\cos(nx)\}$ where $Ȣ_x$ denotes Essence with respect to $x$.
$$\begin{align*}
\because \frac{d}{dx} f(x+1) &= Ȣ\{ f(x+1)-f(x) \} \\ \frac{d}{dx}\sin(n(x+1)) & = Ȣ\{\sin(n(x+1)) - \sin(nx) \} \\
n\cos(n(x+1)) & = Ȣ\{ \sin(nx+n) - \sin(nx) \} \\
n\cos(nx+n) &= Ȣ\{ \sin(n)\cos(nx) + \cos(n)\sin(nx) - \sin(nx) \} \\
&= \sin(n)Ȣ\{\cos(nx)\} + (\cos(n)-1)Ȣ\{\sin(nx)\} \\
\therefore Ȣ\{\sin(nx)\} &= \frac{n\cos(nx+n) - \sin(n)Ȣ\{\cos(nx)\}}{\cos(n) - 1} \\ \\
\frac{d}{dx}\cos(n(x+1)) &= Ȣ\{\cos(n(x+1)) - \cos(nx) \} \\
-n\sin(n(x+1)) &= Ȣ\{\cos(nx+n) - \cos(nx) \} \\
-n\sin(nx+n) &= Ȣ\{\cos(n)\cos(nx) - \sin(n)\sin(nx) - \cos(nx) \} \\
-\sin(nx + n) &= (\cos(n)-1)Ȣ\{\cos(nx)\} - \sin(n)Ȣ\{\sin(nx)\} \\
\therefore Ȣ\{\sin(nx)\} &= \frac{n\sin(nx+n) + (\cos(n)-1)Ȣ\{\cos(nx)\}}{\sin(n)}
\end{align*} \\ $$ $$
\begin{align*}
\therefore \frac{n\cos(nx+n) - \sin(n)Ȣ\{\cos(nx)\}}{\cos(n) - 1} &= \frac{n\sin(nx+n) + (\cos(n)-1)Ȣ\{\cos(nx)\}}{\sin(n)} \\ n\sin(n)\cos(nx+n) - \sin^2(n)Ȣ\{\cos(nx)\} & = n(\cos(n)-1)\sin(nx+n) + (\cos(n)-1)^2Ȣ\{\cos(nx)\} \\
n(\sin(n)\cos(nx+n) - (\cos(n)-1)\sin(nx+n) ) &= (\cos(n)-1)^2 + \sin^2(n))Ȣ\{\cos(nx)\} \\
\because (\cos(\theta)-1)^2 = 2(1-\cos(\theta))-\sin^2(\theta) \\
n(\sin(n)\cos(nx+n) - \cos(n)\sin(nx+n) + \sin(nx+n) ) &= 2(1-\cos(n))Ȣ\{\cos(nx)\} \\
n(\sin(n-(nx+n)) + \sin(nx+n)) &= \\
n(\sin(nx+n) - \sin(nx)) & =\\ \\
\therefore Ȣ_x\{\cos(nx)\} = \frac{n(\sin(nx+n) - \sin(nx))}{2(1-\cos(n))} \\
\end{align*} $$ $$
\begin{align*}
\because Ȣ\{\sin(nx)\} & = \frac{n\sin(nx+n) + (\cos(n)-1)Ȣ\{\cos(nx)\}}{\sin(n)} \\
& = \frac{n\sin(nx+n) - (1-\cos(n))\frac{n(\sin(nx+n) - \sin(nx))}{2(1-\cos(n))}}{\sin(n)} \\
& = \frac{n( 2\sin(nx+n) - \sin(nx+n) + \sin(nx) )}{2\sin(n)} \\ \\
\therefore Ȣ_x\{\sin(nx)\} &= \frac{n(\sin(nx+n) + \sin(nx) )}{2\sin(n)}
\end{align*} $$ $$
\begin{align*}
\because Ȣ\{\frac{d}{dx}f(x) \} & = \frac{d}{dx}Ȣ\{f(x)\} \\
\therefore Ȣ\{ \frac{d}{dx} \sin(nx) \} &= \frac{d}{dx}Ȣ\{\sin(nx) \} \\
Ȣ\{ n\cos(nx) \} & = \frac{d}{dx} frac{n(\sin(nx+n) + \sin(nx) )}{2\sin(n)} \\
nȢ\{\cos(nx)\} &= \frac{n^2(\cos(nx+n) + \cos(nx) )}{2\sin(n)} \\
\therefore Ȣ\{\cos(nx) \} &= \frac{n(\cos(nx+n) + \cos(nx) )}{2\sin(n)} \\ \\
\therefore Ȣ\{\frac{d}{dx} \cos(nx)\} &= \frac{d}{dx}Ȣ\{\cos(nx)\} \\
Ȣ\{-n\sin(nx)\} & = \frac{d}{dx}\frac{n(\sin(nx+n) - \sin(nx))}{2(1-\cos(n))} \\
-nȢ\{\sin(nx) \} &= \frac{n^2(\cos(nx+n) - \cos(nx))}{2(1-\cos(n))} \\
\therefore Ȣ\{\sin(nx)\} &= \frac{n(-\cos(nx+n) + \cos(nx))}{2(1-\cos(n))}
\end{align*}$$ $$
\begin{align*}
Ȣ_x\{\sin(nx)\} &= \frac{n(\sin(nx+n) + \sin(nx) )}{2\sin(n)} &= \frac{n(-\cos(nx+n) + \cos(nx))}{2(1-\cos(n))} \\
Ȣ_x\{\cos(nx) \} &= \frac{n(\cos(nx+n) + \cos(nx) )}{2\sin(n)} &= \frac{n(\sin(nx+n) - \sin(nx))}{2(1-\cos(n))} \\
&&\blacksquare \end{align*} $$
Notice that this is consistent with original definitions above when $n=1$.`\because \frac{d}{dx} f(x+1) &= Ȣ\{ f(x+1)-f(x) \} \\ \frac{d}{dx}\sin(n(x+1)) & = Ȣ\{\sin(n(x+1)) - \sin(nx) \} \\
n\cos(n(x+1)) & = Ȣ\{ \sin(nx+n) - \sin(nx) \} \\
n\cos(nx+n) &= Ȣ\{ \sin(n)\cos(nx) + \cos(n)\sin(nx) - \sin(nx) \} \\
&= \sin(n)Ȣ\{\cos(nx)\} + (\cos(n)-1)Ȣ\{\sin(nx)\} \\
\therefore Ȣ\{\sin(nx)\} &= \frac{n\cos(nx+n) - \sin(n)Ȣ\{\cos(nx)\}}{\cos(n) - 1} \\ \\
\frac{d}{dx}\cos(n(x+1)) &= Ȣ\{\cos(n(x+1)) - \cos(nx) \} \\
-n\sin(n(x+1)) &= Ȣ\{\cos(nx+n) - \cos(nx) \} \\
-n\sin(nx+n) &= Ȣ\{\cos(n)\cos(nx) - \sin(n)\sin(nx) - \cos(nx) \} \\
-\sin(nx + n) &= (\cos(n)-1)Ȣ\{\cos(nx)\} - \sin(n)Ȣ\{\sin(nx)\} \\
\therefore Ȣ\{\sin(nx)\} &= \frac{n\sin(nx+n) + (\cos(n)-1)Ȣ\{\cos(nx)\}}{\sin(n)}
\end{align*} \\ $$ $$
\begin{align*}
\therefore \frac{n\cos(nx+n) - \sin(n)Ȣ\{\cos(nx)\}}{\cos(n) - 1} &= \frac{n\sin(nx+n) + (\cos(n)-1)Ȣ\{\cos(nx)\}}{\sin(n)} \\ n\sin(n)\cos(nx+n) - \sin^2(n)Ȣ\{\cos(nx)\} & = n(\cos(n)-1)\sin(nx+n) + (\cos(n)-1)^2Ȣ\{\cos(nx)\} \\
n(\sin(n)\cos(nx+n) - (\cos(n)-1)\sin(nx+n) ) &= (\cos(n)-1)^2 + \sin^2(n))Ȣ\{\cos(nx)\} \\
\because (\cos(\theta)-1)^2 = 2(1-\cos(\theta))-\sin^2(\theta) \\
n(\sin(n)\cos(nx+n) - \cos(n)\sin(nx+n) + \sin(nx+n) ) &= 2(1-\cos(n))Ȣ\{\cos(nx)\} \\
n(\sin(n-(nx+n)) + \sin(nx+n)) &= \\
n(\sin(nx+n) - \sin(nx)) & =\\ \\
\therefore Ȣ_x\{\cos(nx)\} = \frac{n(\sin(nx+n) - \sin(nx))}{2(1-\cos(n))} \\
\end{align*} $$ $$
\begin{align*}
\because Ȣ\{\sin(nx)\} & = \frac{n\sin(nx+n) + (\cos(n)-1)Ȣ\{\cos(nx)\}}{\sin(n)} \\
& = \frac{n\sin(nx+n) - (1-\cos(n))\frac{n(\sin(nx+n) - \sin(nx))}{2(1-\cos(n))}}{\sin(n)} \\
& = \frac{n( 2\sin(nx+n) - \sin(nx+n) + \sin(nx) )}{2\sin(n)} \\ \\
\therefore Ȣ_x\{\sin(nx)\} &= \frac{n(\sin(nx+n) + \sin(nx) )}{2\sin(n)}
\end{align*} $$ $$
\begin{align*}
\because Ȣ\{\frac{d}{dx}f(x) \} & = \frac{d}{dx}Ȣ\{f(x)\} \\
\therefore Ȣ\{ \frac{d}{dx} \sin(nx) \} &= \frac{d}{dx}Ȣ\{\sin(nx) \} \\
Ȣ\{ n\cos(nx) \} & = \frac{d}{dx} frac{n(\sin(nx+n) + \sin(nx) )}{2\sin(n)} \\
nȢ\{\cos(nx)\} &= \frac{n^2(\cos(nx+n) + \cos(nx) )}{2\sin(n)} \\
\therefore Ȣ\{\cos(nx) \} &= \frac{n(\cos(nx+n) + \cos(nx) )}{2\sin(n)} \\ \\
\therefore Ȣ\{\frac{d}{dx} \cos(nx)\} &= \frac{d}{dx}Ȣ\{\cos(nx)\} \\
Ȣ\{-n\sin(nx)\} & = \frac{d}{dx}\frac{n(\sin(nx+n) - \sin(nx))}{2(1-\cos(n))} \\
-nȢ\{\sin(nx) \} &= \frac{n^2(\cos(nx+n) - \cos(nx))}{2(1-\cos(n))} \\
\therefore Ȣ\{\sin(nx)\} &= \frac{n(-\cos(nx+n) + \cos(nx))}{2(1-\cos(n))}
\end{align*}$$ $$
\begin{align*}
Ȣ_x\{\sin(nx)\} &= \frac{n(\sin(nx+n) + \sin(nx) )}{2\sin(n)} &= \frac{n(-\cos(nx+n) + \cos(nx))}{2(1-\cos(n))} \\
Ȣ_x\{\cos(nx) \} &= \frac{n(\cos(nx+n) + \cos(nx) )}{2\sin(n)} &= \frac{n(\sin(nx+n) - \sin(nx))}{2(1-\cos(n))} \\
&&\blacksquare \end{align*} $$
With these we can find that
$$
\begin{align*}
\sum_{i=n}^m\sin(r*i) & = \int_{n-1}^m\frac{r(\sin(rx+r) + \sin(rx) )}{2\sin(r)} dx &&= \int_{n-1}^m \frac{r(-\cos(rx+r) + \cos(rx))}{2(1-\cos(r))} dx \\
& = \frac{\cos(rn)+\cos(rn-r) - \cos(rm+r) - \cos(rm)}{2\sin(r)} &&= \frac{ -\sin(rm+r) + \sin(rm) + \sin(rn) - \sin(rn-r) }{2(1-\cos(r))} \\
\sum_{i=n}^m \cos(r*i) &= \int_{n-1}^m \frac{ r(\cos(rx+r) +\cos(rx) ) }{2\sin(r)} dx &&= \int_{n-1}^m \frac{ r( \sin(rx+r) - \sin(rx) ) }{2(1-\cos(r))}dx \\
&= \frac{ \sin(rm+r)+\sin(rm) - \sin(rn) - \sin(rn-r) }{2\sin(r)} &&= \frac{ \cos(rn) -\cos(rn-r) -\cos(rm+r) + \cos(rm) }{2(1-\cos(r))}
\end{align*} $$ What fascinates me the most is that with these formulas we are able to condense a long sum of sine and cosine, which has no obvious reduction rules by themselves, into a concise single equation. Also, the added benefit of being able to calculate non-integer boundary series.
In addition to those, $\cos(\pi x)$ is a popular continuous substitute for $(-1)^x$ and so knowing its Essence will help us interpolate between more discontinuous functions later on.
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