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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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Sequence Curve: Essence of Power Function with Induction Formula and Sequence Flatlining


Last time, we introduced the concept of Sequence Curve. This time we will explore how the Essence Transformation will actually work. 
To start this, we will consider the simplest function: linear function $f(x) = x$.
$$\sum_{i=n}^m i = \int_{n-1}^m Ȣ\{x\} dx$$

How can we find $Ȣ\{x\}$?

For this, we can utilize the well known Arithmetic Series formula to simplify this problem.
$$\sum_{i=n}^m i = \frac{m+1 - n}{2}(m+n) $$
This lends to that
$$\frac{m+1-n}{2}(m+n) = \int_{n-1}^m Ȣ\{x\} dx $$
From this relation we will be able to solve for $Ȣ\{x\}$.

Let’s consider the case when $n$ is a constant $c$. 
Then it must be case that
$$ \int_{c-1}^m Ȣ\{x\} dx = \frac{m+1-c}{2}(m+c) $$
From First Fundamental Theorem of Calculus, we find that
$$\begin{align*} \frac{d}{dm}\int_{c-1}^m Ȣ\{x\} dx & = \frac{d}{dm}\frac{m+1-c}{2}(m+c) \\
Ȣ\{m\} & = \frac{m+1-c}{2} + \frac{m+c}{2} \\
& = m + \frac{1}{2}
\end{align*}$$

Let's see this in action to demonstrate that this works:
$$\begin{align*} \sum_{i=3}^7 i & = \frac{7+1-3}{2}(7+3)  & = \int_{3-2}^7 x + \frac{1}{2} dx\\
3 + 4 + 5 + 6 + 7 & = \frac{5}{2}(10) & = \left[ \frac{x^2}{2} + \frac{x}{2} \right]^7_2 \\
& & = \frac{49}{2} + \frac{7}{2} - \frac{4}{2} - \frac{2}{2}\\
25 &= 25 &= 25 \end{align*}$$

We can continue pairing Induction with Essences, but that will limit us to get only as far as Inductions we already know. What we are trying to achieve through Essence Transformation is to find a systematic approach of rather than relying on Induction Theory. 

Sequence Flatlining


Let's try to solve for $Ȣ\{x\}$ in a different method that does not involve using induction formulas.
Consider the case when $m=n$:
$$\sum_{i=n}^m i = \sum_{i=n}^n i = n$$
Our Essence Transformation should still hold true even for these cases, leading us to
$$\sum_{i=n}^n i = n = \int_{n-1}^n Ȣ\{x\} dx$$
For the sake of ease, let's define $Ȣ\{x\} = f(x)$.
Like with the previous method, we apply FTC and find that
$$\begin{align*}
n = & \int_{n-1}^n f(x) dx\\
1 = & f(n)-f(n-1) \\
0 = & f'(n) - f'(n-1)
\end{align*}$$
At this point I want to introduce a lemma:

Naught Curve Difference Postulate

If the difference of function $f$ between $x$ and $x+c$ is 0, then it must be that $f$ is constant or that $f$ oscillates with period of $c$.
Given that postulate, and since there is no reason to believe that $f'$ would be oscillating, we can assume that $f'(x) = c$. (We will discuss how to solve for $Ȣ\{x\}$ without making such assumptions later on, but for now know that this assumption still gives us consistent results)
Since, $f'(x)$ is a constant, it must follow that
$$f'(x) = c_1\\
f(x) = c_1x + c_2\\
\int f(x) dx = \frac{c_1}{2}x^2 + c_2x + c_3$$
Let us now substitute these into the identities above, we find that
$$\begin{align*} 1 = & f(n) - f(n-1)\\
= & c_1n + c_2 - c_1(n-1) - c_2\\
= & c_1n - c_1n + c_1\\
= & c_1 \end{align*}$$ And then $$\begin{align*} n = & \int_{n-1}^nf(x)dx\\
= & \int_{n-1}^n x+c_2 dx\\
= & \frac{1}{2}n^2 + c_2n - \frac{1}{2}(n-1)^2 - c_2(n-1)\\
= & \frac{1}{2}(n^2 - (n-1)^2) +  c_2n - c_2(n-1)\\
= & \frac{1}{2}(2n - 1) + c_2\\
n = & n - \frac{1}{2} + c_2\\
\frac{1}{2} = & c_2
 \end{align*}$$
Combing them together we find that
$$f(x) = Ȣ\{x\} = x+\frac{1}{2}$$ which is the same result we got from deriving $Ȣ\{x\}$ from Arithmetic Series formula. $\blacksquare$

Notice that this method relies differentiating the sequence formula until it eventually reaches 0, at which point we apply Naught Difference Postulate and substitute values backwards to solve for Essence. This will work for all Integer Power Series.
We can think of this process as 'flatlining' the sequence; as morbid as it sounds, it captures the spirit of the technique pretty well.

Let's use this method on next simplest Sequence $x^2$.
$$\begin{align*} \sum_{i=n}^ni^2 =  n^2 = & \int_{n-1}^nf(x)dx\\
 2n = & f(n) - f(n-1)\\
 2 = & f'(n) - f'(n-1)\\
 0 = & f''(n) - f''(n-1)\\
& \therefore f''(x) = c_1 \text{, by Naught Difference Postulate}\\ \\
& \therefore f'(x) = c_1x + c_2\\
\therefore 2 = & c_1n + c_2 - c_1(n-1) - c_2\\
2 =& c_1\\ \\
& \therefore f(x) = x^2 + c_2x + c_3\\
\therefore 2n = & n^2 + c_2n + c_3 - (n-1)^2 - c_2(n-1) - c_3\\
= & 2n - 1 + c_2\\
1 =& c_2 \\ \\
& \therefore F(x) = \frac{1}{3}x^3 + \frac{1}{2}x^2 + c_3x + c_4\\
\therefore n^2 =& \frac{1}{3}n^3 + \frac{1}{2}n^2 + c_3n - \frac{1}{3}(n-1)^3 - \frac{1}{2}(n-1)^2 - c_3(n-1) \\
= & \frac{1}{3}(n^3 - (n-1)^3) + \frac{1}{2}(n^2 - (n-1)^2) + c_3 \\
= & \frac{1}{3}(3n^2 - 3n + 1) + \frac{1}{2}(2n -1) + c_3 \\
n^2 = & n^2 - n + \frac{1}{2} + n - \frac{1}{2} + c_3\\
\frac{1}{2} - \frac{1}{3} = & c_3\\
\frac{1}{6} = & \\ \\
& \therefore f(x) = Ȣ\{x^2\} = x^2 + x + \frac{1}{6}\\ \blacksquare
\end{align*}$$

Surprisingly, this result, once you integrate it, is also consistent with known induction formula $\frac{n(n+1)(2n+1)}{6}$.

We can continue on with Sequence Flatlining technique and discover that $$\begin{align*} Ȣx^3 = & x^3+\frac{3x^2}{2}+\frac{x}{2} \\
Ȣx^4 = & x^4 + 2x^3 + x^2 - \frac{1}{30}\\
Ȣx^5 = & x^5 + \frac{5x^4}{2} + \frac{5x^3}{3} - \frac{x}{6}\\
Ȣx^6 = & x^6 + 3x^5 + \frac{5x^4}{2} - \frac{x^2}{2}+\frac{1}{42}\\
Ȣx^7 = & x^7 + \frac{7x^6}{2} + \frac{7x^5}{2}-\frac{7x^3}{6}+\frac{x}{6}\\
Ȣx^8 = &x^8 + 4x^7 + \frac{14x^6}{3}- \frac{7x^4}{3}+\frac{2x^2}{3} - \frac{1}{30} \\
Ȣx^9 = &x^9 + \frac{9x^8}{2}+ 6x^7- \frac{21x^5}{5}+ 2x^3 - \frac{3x}{10}\\
Ȣx^{10} = &x^{10} + 5x^9 +\frac{15x^8}{2}-7x^6+ 5x^4-\frac{3x^2}{2}+\frac{5}{66}
\end{align*}$$ and so on and so on.

Again, let's demonstrate that Essence Transformation works as intended:
$$\begin{align*} \sum_{i=11}^{18}i^5 = & 11^5 + 12^5 + 13^5 + 14^5 + 15^5 + 16^5 + 17^5 + 18^5 \\
& \vdots \\
= & 6,436,376 \\ \\
\int_{11-1}^{18} x^5 + \frac{5x^4}{2} + \frac{5x^3}{3} - \frac{x}{6} dx = & \left[  \frac{x^6}{6} + \frac{x^5}{2} + \frac{5x^4}{12}-\frac{x^2}{12} \right]_{10}^{18}\\
= & (\frac{18^6}{6} + \frac{18^5}{2} + \frac{5*18^4}{12}-\frac{18^2}{12})- (\frac{10^6}{6} + \frac{10^5}{2} + \frac{5*10^4}{12}-\frac{10^2}{12})\\
= & 6,657,201 - 220,825\\
= & 6,436,376
\end{align*}$$
Pretty cool, right?
Notice that as the range gets bigger, calculating the series through Essence would be much simpler and easier than summing up individual terms of the sequence.

Note, also, that through Essence definition of series, something like $\sum_{i=1}^{1.5}i$ becomes computable. This was not possible from original definition of series as incrementing by a unit from lower bound $1$ will never arrive us at the upper bound $1.5$.
But using our new definition of series we find that
$$\begin{align*}\sum_{i=1}^{1.5}i = & \int_{0}^{1.5}x+\frac{1}{2} dx \\
 = & \frac{15}{8} \end{align*}$$

One interesting point to note is that for Power Sequences $Ȣf(x)$ results in curve with same degree as $f(x)$. This means that solving for Essence of Power Sequence can essentially be reduced to finding patterns in each terms' coefficients.
The general expression for the coefficients will be studied in further depth in future articles.

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