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: $$\begin{align*}Ȣ1 &= 1 \\ Ȣx &= x +...
Last time, we explored Essence of Power Functions. This time, we will try to calculate for Essence of Exponentials in a similar way.
$$\sum_{i=n}^m r^i = \int_{n-1}^mȢ\{r^x\}dx$$
We will need to find another way to approach Essences.
$$\sum_{i=n}^m f(i) = \int_{n-1}^mȢ\{f(x)\}dx \\
\sum_{i=n}^n f (i) = f(n) = \int_{n-1}^nȢ\{f(x)\}dx \\
\therefore \frac{d}{dn}f(n) = Ȣ\{f(n)\} - Ȣ\{f(n-1)\}$$ Rather than continuing to differentiate this like in Sequence Flatlining, we stop here.
By Linearity of Essence (see Sequence Curve: Essential Properties of Essence), we can find that
$$\frac{d}{dn}f(n) = Ȣ\{ f(n) - f(n-1) \} \, \blacksquare$$
This Formula becomes fundamental to calculating Essences from here on.
Let's use this to solve for Exponential Essences.
\because \frac{d}{dx}f(x) &= Ȣ\{f(x) - f(x-1)\} \\
\therefore \frac{d}{dx}r^x &= Ȣ\{ r^x - r^{x-1} \} \\
\ln(r)r^x & = Ȣ\{ r^x - \frac{r^x}{r}\} \\
& = Ȣ\{ r^x(1-\frac{1}{r}) \}, \text{ and due to Linearity of Essence,} \\
& = (1-\frac{1}{r})Ȣ\{r^x\} \\
\therefore Ȣ\{r^x\} &= \frac{\ln(r)r^x}{ 1-\frac{1}{r} } \blacksquare
\end{align*}$$ So we discover that
$$\sum_{i=n}^m r^i = \int_{n-1}^m \frac{\ln(r)r^x}{1-\frac{1}{r}} dx$$
From this definition, we can derive Finite and Infinite Geometric Series formulas:
\sum_{i=0}^nr^x & = \int_{-1}^n\frac{\ln(r)r^x}{1-\frac{1}{r}} dx \\
& = \left [ \frac{r^x}{1- \frac{1}{r} } \right ] ^n _{-1} \\
& = \frac{r^n - \frac{1}{r}}{1-\frac{1}{r}} \\
& = \frac{r^{n+1}-1}{r-1} \\
& \blacksquare
\end{align*}$$
\sum_{i=0}^\infty r^i & = \int_{-1}^\infty \frac{\ln(r) r^x}{1-\frac{1}{r}} dx\\
& = \left [ \frac{r^x}{1-\frac{1}{r}} \right ]_{-1}^\infty \\
& = \lim_{n \rightarrow \infty} \frac{r^n - \frac{1}{r}}{1 - \frac{1}{r}} \\
& = \frac{-\frac{1}{r}}{1 - \frac{1}{r}}, |r| < 1 \\
& = \frac{1}{1-r} \\
& \blacksquare
\end{align*}$$
So far, we have covered the two types of series taught in typical Calc classes. Next, we will explore more complex and unusual series with help of Essence Formula, such as $\sin(x)$ and $\cos(x)$ and $xe^x$, etc.
$$\sum_{i=n}^m r^i = \int_{n-1}^mȢ\{r^x\}dx$$
Sequence Flatlining
We were able to find Essence of Power functions using Sequence Flatlining. The basic idea behind Sequence Flatlining is to differentiate the equation for sequence until you reach 0 (Check out Sequence Curve: Essence of Power Function for more). This method, however, will not work for Exponential Functions because they will never differentiate to 0.We will need to find another way to approach Essences.
Essence Formula
We start similarly to beginning of Sequence Flatlining:$$\sum_{i=n}^m f(i) = \int_{n-1}^mȢ\{f(x)\}dx \\
\sum_{i=n}^n f (i) = f(n) = \int_{n-1}^nȢ\{f(x)\}dx \\
\therefore \frac{d}{dn}f(n) = Ȣ\{f(n)\} - Ȣ\{f(n-1)\}$$ Rather than continuing to differentiate this like in Sequence Flatlining, we stop here.
By Linearity of Essence (see Sequence Curve: Essential Properties of Essence), we can find that
$$\frac{d}{dn}f(n) = Ȣ\{ f(n) - f(n-1) \} \, \blacksquare$$
Let's use this to solve for Exponential Essences.
Exponential Essences
$$\begin{align*}\because \frac{d}{dx}f(x) &= Ȣ\{f(x) - f(x-1)\} \\
\therefore \frac{d}{dx}r^x &= Ȣ\{ r^x - r^{x-1} \} \\
\ln(r)r^x & = Ȣ\{ r^x - \frac{r^x}{r}\} \\
& = Ȣ\{ r^x(1-\frac{1}{r}) \}, \text{ and due to Linearity of Essence,} \\
& = (1-\frac{1}{r})Ȣ\{r^x\} \\
\therefore Ȣ\{r^x\} &= \frac{\ln(r)r^x}{ 1-\frac{1}{r} } \blacksquare
\end{align*}$$ So we discover that
$$\sum_{i=n}^m r^i = \int_{n-1}^m \frac{\ln(r)r^x}{1-\frac{1}{r}} dx$$
Finite Geometric Series
$$\begin{align*}\sum_{i=0}^nr^x & = \int_{-1}^n\frac{\ln(r)r^x}{1-\frac{1}{r}} dx \\
& = \left [ \frac{r^x}{1- \frac{1}{r} } \right ] ^n _{-1} \\
& = \frac{r^n - \frac{1}{r}}{1-\frac{1}{r}} \\
& = \frac{r^{n+1}-1}{r-1} \\
& \blacksquare
\end{align*}$$
Infinite Geometric Series
$$\begin{align*}\sum_{i=0}^\infty r^i & = \int_{-1}^\infty \frac{\ln(r) r^x}{1-\frac{1}{r}} dx\\
& = \left [ \frac{r^x}{1-\frac{1}{r}} \right ]_{-1}^\infty \\
& = \lim_{n \rightarrow \infty} \frac{r^n - \frac{1}{r}}{1 - \frac{1}{r}} \\
& = \frac{-\frac{1}{r}}{1 - \frac{1}{r}}, |r| < 1 \\
& = \frac{1}{1-r} \\
& \blacksquare
\end{align*}$$
So far, we have covered the two types of series taught in typical Calc classes. Next, we will explore more complex and unusual series with help of Essence Formula, such as $\sin(x)$ and $\cos(x)$ and $xe^x$, etc.
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