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: $$\begin{align*}Ȣ1 &= 1 \\ Ȣx &= x +
Polynomial Series can be easily solved using Power Series Formulas for each term in the polynomial. However, this can be frustrating since not every Power Formula are intuitive to memorize. We would like to find a more elegant and easier-to-recall formula for computing a Polynomial Series. This can be done using matrices.
$$\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)\}$$ This will be used extensively throughout this article.
&= Ȣ\{1\}
\end{align*}$$ We will leave the lefthand-side in terms of the deritives for cleaner results later on.
When $f(x) = x^2$, $$\begin{align*}
\frac{d}{dx}(x+1)^2 &= Ȣ\{(x+1)^2-x^2\} \\
&= Ȣ\{2x+1\} \\
&= 2Ȣ\{x\} + Ȣ\{1\}
\end{align*}$$ We will notice that these terms are very similar to the expansion of $(x+1)^n$ but with the first term removed. Noticing this, we easily find that $$\begin{align*}
\frac{d}{dx}(x+1)^3 &= 3Ȣ\{x^2\} + 3Ȣ\{x\} + Ȣ\{1\} \\
\frac{d}{dx}(x+1)^4 &= 4Ȣ\{x^3\} + 6Ȣ\{x^2\} + 4Ȣ\{x\} + Ȣ\{1\} \\
\frac{d}{dx}(x+1)^5 &= 5Ȣ\{x^4\} + 10Ȣ\{x^3\} + 10Ȣ\{x^2\} + 5Ȣ\{x\} + Ȣ\{1\} \\
&\vdots \\
\frac{d}{dx}(x+1)^n &= \binom{n}{1}Ȣ\{x^{n-1}\} + \binom{n}{2}Ȣ\{x^{n-2}\} + \dots + \binom{n}{n}Ȣ\{1\}
\end{align*}$$ Though interesting, these define the $\frac{d}{dx}(x+1)^n$ in terms of $Ȣ\{x^n\}$s which is not yet very useful. What we need is precisely the other way around.
For example, if we want to solve for $Ȣ\{x^3\}$, we will treat the individual $Ȣ\{1\}, Ȣ\{x\} \dots Ȣ\{x^3\}$ like variables and solve for the system of equations of the first four equations.
$$\begin{align*}
&\begin{cases}
\frac{d}{dx}(x+1) = Ȣ\{1 \} \\
\frac{d}{dx}(x+1)^2 = 2Ȣ\{x\} + Ȣ\{1\} \\
\frac{d}{dx}(x+1)^3 = 3Ȣ\{x^2\} + 3Ȣ\{x\} + Ȣ\{1\} \\
\frac{d}{dx}(x+1)^4 = 4Ȣ\{x^3\} + 6Ȣ\{x^2\} 4Ȣ\{x\} + Ȣ\{1\}
\end{cases} \\ \\
&\begin{cases}
\frac{d}{dx}(x+1) = Ȣ\{1\} \\
\frac{d}{dx}(x+1)^2 - \frac{d}{dx}(x+1) = 2Ȣ\{x\} \\
\frac{d}{dx}(x+1)^3 - \frac{d}{dx}(x+1) = 3Ȣ\{x^2\} + 3Ȣ\{x^2\} \\
\frac{d}{dx}(x+1)^4 - \frac{d}{dx}(x+1) = 4Ȣ\{x^3\} + 6Ȣ\{x^2\} + 4Ȣ\{x\}
\end{cases} \\ \\
&\begin{cases}
\frac{d}{dx}(x+1) = Ȣ\{1\} \\
\frac{1}{2}\frac{d}{dx}(x+1)^2 - \frac{1}{2}\frac{d}{dx}(x+1) = Ȣ\{x\} \\
\frac{d}{dx}(x+1)^3 - \frac{3}{2}\frac{d}{dx}(x+1)^2 +\frac{1}{2}\frac{d}{dx}(x+1) = 3Ȣ\{x^2\} \\
\frac{d}{dx}(x+1)^4 -2\frac{d}{dx}(x+1)^2 + \frac{d}{dx}(x+1) = 4Ȣ\{x^3\} + 6Ȣ\{x^2\}
\end{cases} \\ \\
&\begin{cases}
\frac{d}{dx}(x+1) = Ȣ\{1\} \\
\frac{1}{2}\frac{d}{dx}(x+1)^2 - \frac{1}{2}\frac{d}{dx}(x+1) = Ȣ\{x\} \\
\frac{1}{3}\frac{d}{dx}(x+1)^3 - \frac{1}{2}\frac{d}{dx}(x+1)^2 + \frac{1}{6}\frac{d}{dx}(x+1) = Ȣ\{x^2\} \\
\frac{d}{dx}(x+1)^4 -2\frac{d}{dx}(x+1)^3 + \frac{d}{dx}(x+1)^2 = 4Ȣ\{x^3\}
\end{cases} \\ \\
&\begin{cases}
\frac{d}{dx}(x+1) = Ȣ\{1\} \\
\frac{1}{2}\frac{d}{dx}(x+1)^2 - \frac{1}{2}\frac{d}{dx}(x+1) = Ȣ\{x\} \\
\frac{1}{3}\frac{d}{dx}(x+1)^3 - \frac{1}{2}\frac{d}{dx}(x+1)^2 + \frac{1}{6}\frac{d}{dx}(x+1) = Ȣ\{x^2\} \\
\frac{1}{4}\frac{d}{dx}(x+1)^4 -\frac{1}{2}\frac{d}{dx}(x+1)^3 + \frac{1}{4}\frac{d}{dx}(x+1)^2 = Ȣ\{x^3\}
\end{cases}
\end{align*}$$ Though it is extremely messy, it is possible to calculate $Ȣ\{x^3\} = \frac{1}{4}\frac{d}{dx}(x+1)^4 -\frac{1}{2}\frac{d}{dx}(x+1)^3 + \frac{1}{4}\frac{d}{dx}(x+1)^2$ by hand.
We can associate this result back with the cubic series:
$$\sum_{i=m}^ni^3 = \frac{1}{4}((n+1)^4 - m^4) - \frac{1}{2}((n+1)^3-m^3) + \frac{1}{4}((n+1)^2-m^2)$$ When $m=0$, we have our familiar Power Series Formula for a cubic.
\begin{cases}
&4Ȣ\{x^3\} + &6Ȣ\{x^2\}+ &4Ȣ\{x\} + & Ȣ\{1\} = &\frac{d}{dx}(x+1)^4 \\
& &3Ȣ\{x^2\} + &3Ȣ\{x\} + & Ȣ\{1\} = &&\frac{d}{dx}(x+1)^3 \\
& & &2Ȣ\{x\} + & Ȣ\{1\} = &&&\frac{d}{dx}(x+1)^2 \\
& & & &Ȣ\{1\} = &&&&\frac{d}{dx}(x+1)
\end{cases}
\end{align*} \\ $$ is equivalent to this matrix: $$ \left[
\left. \begin{matrix}
4 & 6 & 4 & 1\\
0 & 3 & 3 & 1\\
0 & 0 & 2 & 1\\
0 & 0 & 0 & 1
\end{matrix} \right|
\begin{matrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{matrix} \right] $$ Beforing going any further, we notice that the left-hand matrix looks very familiar. It looks just like the famous Pascal's Triangle unsider-down and with the first 1s removed from each row. From here on, we will call these matrices Pascal Matrix of size $n\times n$, $P_n$.
So, to solve for the lefthand-matrix, we simply need to multiply its inverse to both sides which happens to be ${P_4}^{-1}$. This will result in the system that looks like
$$ \left[ \mathbf{I} \vert {P_4}^{-1} \right] $$ This shows that we can easily solve for $Ȣ\{x^n\}$ by simply taking the inverse of $P_{n+1}$. To find $Ȣ\{x^3\}$, we calculate ${P_4}^{-1}$: $$\begin{bmatrix}
4&6&4&1\\
0&3&3&1\\
0&0&2&1\\
0&0&0&1
\end{bmatrix} ^{-1} = \begin{bmatrix}
\frac{1}{4} & -\frac{1}{2} & \frac{1}{4} & 0 \\
0 & \frac{1}{3} & -\frac{1}{2} & \frac{1}{6} \\
0 & 0 & \frac{1}{2} & -\frac{1}{2} \\
0&0&0&1
\end{bmatrix}
$$ The first row yields the coeffficents of $\frac{d}{dx}(x+1)^n$s for $Ȣ\{x^3\}$.
$$Ȣ\{x^3\} = \frac{1}{4}\frac{d}{dx}(x+1)^4 - \frac{1}{2}\frac{d}{dx}(x+1)^3 + \frac{1}{4}\frac{d}{dx}(x+1)^2
$$ Similarly, we can also find the coefficients for $Ȣ\{x^2\}$, $Ȣ\{x\}$, and $Ȣ\{1\}$ and also be found on second, third, and fourth rows respectively.
$$\begin{align*}
Ȣ\{x^2\} &= \frac{1}{3}\frac{d}{dx}(x+1)^3 - \frac{1}{2}\frac{d}{dx}(x+1)^2 + \frac{1}{6}\frac{d}{dx}(x+1) \\
Ȣ\{x\} &= \frac{1}{2}\frac{d}{dx}(x+1)^2 + \frac{1}{2}\frac{d}{dx}(x+1) \\
Ȣ\{1\} &= \frac{d}{dx}(x+1)
\end{align*}$$ We can solve for any number of $Ȣ\{x^n\}$, which are just the Power Series Formula, at once by solving for a single Inverse of a Matrix!
Let's say we want to compute the series formula for $f(x) = 4x^3 -12x^2 +24$. We can find the individual $Ȣ\{x^n\}$ with the ${P_4}^{-1}$, and we can combine these terms to yield $f$ by multiplying Coefficient Row-Vector in front of it:$$
Ȣ\{f(x)\} = \begin{bmatrix}
4 & -12 & 0 & 24
\end{bmatrix}
\begin{bmatrix}
\frac{1}{4} & -\frac{1}{2} & \frac{1}{4} & 0 \\
0 & \frac{1}{3} & -\frac{1}{2} & \frac{1}{6} \\
0 & 0 & \frac{1}{2} & -\frac{1}{2} \\
0&0&0&1
\end{bmatrix} =
\begin{bmatrix}
1&-6&7&22
\end{bmatrix}
$$ Written in full, we find that
$$Ȣ\{4x^3 -12x^2 +24\} = \frac{d}{dx}(x+1)^4 - 6\frac{d}{dx}(x+1)^3 + 7\frac{d}{dx}(x+1)^2 + 22\frac{d}{dx}(x+1)$$ Now, we can integrate $Ȣ\{f(x)\}$ to get our Series Formula for the polynomial. Since the function is defined in terms of deritives, integrating them will not change the coefficients at all– this is why we kept them in terms of deritives in the beginning.
Therefore, we find that
$$ \begin{align*}
\sum_{i=m}^n f(i) &= \left. (x+1)^4 - 6(x+1)^3 +7(x+1)^2+22(x+1) \right \vert_{m-1}^{n} \\
&= ((n+1)^4-m^4) - 6((n+1)^3-m^3)+7((n+1)^2-m^2) +22((n+1)-m) \\
&= \begin{bmatrix}
1&-6&7&22
\end{bmatrix}\begin{bmatrix}
(n+1)^4-m^4 \\
(n+1)^3-m^3\\
(n+1)^2-m^2\\
(n+1)-m
\end{bmatrix}
\end{align*}$$ Even this last step can be represented as multiplication of the coefficient matrix with Boundary Vector of the series!
Combing all of these steps, computing the Series of $f(x)$ is as simple as calculating the following matrix product:
$$\sum_{i=m}^n 4i^3 -12i^2 +24 = \begin{bmatrix}
4 & -12 & 0 & 24
\end{bmatrix}{P_4}^{-1}\begin{bmatrix}
(n+1)^4-m^4 \\
(n+1)^3-m^3\\
(n+1)^2-m^2\\
(n+1)-m
\end{bmatrix}$$ Of course, there is nothing special with this specific polynomial and this form can be genralied for any polynomial with coefficient $c_k$:
$$\begin{align*}
\sum_{i=m}^n c_kx^k+c_{k-1}x^{k-1}+\dots + c_0 &= \begin{bmatrix} c_k & c_{k-1}&\dots&c_0\end{bmatrix}\begin{bmatrix}
\binom{k+1}{1}&\binom{k}{2}&\dots&\binom{k+1}{k+1}\\
0&\binom{k}{1}&\dots&\binom{k}{k}\\
\vdots&\vdots&\dots&\vdots\\
0&0&\dots&\binom{1}{1}
\end{bmatrix}^{-1}\begin{bmatrix}
(n+1)^{k+1}-m^{k+1}\\
(n+1)^k-m^k\\
\vdots\\
(n+1)-m
\end{bmatrix} \\ \\
&= \begin{bmatrix} c_k & c_{k-1}&\dots&c_0\end{bmatrix}{P_{k+1}}^{-1}\begin{bmatrix}
(n+1)^{k+1}-m^{k+1}\\
(n+1)^k-m^k\\
\vdots\\
(n+1)-m
\end{bmatrix}
\end{align*}$$ This might look even more frightening at first, but just remember that these are composed of three simple parts: (coefficients)(Inverse of Pascal)(Boundaries). These individual parts are simpler to remember and even more intuitive than memorizing individual Power Formulas.
In short, we can do this by adding additional Rows to the Coefficient Vector or adding additional Columns to the Boundary Vector.
For example, let's say we want to compute these two following series:
$$\sum_{i=10}^{30} 60i^3 + 24i^2 + 20 \\
\sum_{i=15}^{50} 30i^2 + 40i
$$ The actual numbers are not important. We will need two Coefficient Vectors
$$\begin{bmatrix}
60&24&0&20
\end{bmatrix}, \begin{bmatrix}
0&30&40&0
\end{bmatrix}$$ and two Boundary Vectors to compute the two series.
$$\begin{bmatrix}
31^4-10^4 \\
31^3 - 10^3\\
31^2 - 10^2\\
31 - 10
\end{bmatrix},
\begin{bmatrix}
51^4 - 15^4 \\
51^3 - 15^3 \\
51^2 - 15^2\\
51-15
\end{bmatrix}
$$ Of course, we have seen above how to use these vectors to solve for individual series, but in that case, we will have to do two separate matrix multiplication for each series.
$$
\sum_{i=10}^{30} 60i^3 + 24i^2 + 20 = \begin{bmatrix}
60&24&0&20
\end{bmatrix} {P_4}^{-1}\begin{bmatrix}
31^4-10^4 \\
31^3 - 10^3\\
31^2 - 10^2\\
31 - 10
\end{bmatrix} \\
\sum_{i=15}^{50} 30i^2 + 40i = \begin{bmatrix}
0&30&40&0
\end{bmatrix} {P_4}^{-1}\begin{bmatrix}
51^4 - 15^4 \\
51^3 - 15^3 \\
51^2 - 15^2\\
51-15
\end{bmatrix} $$ Though simple, it may be combsersome to type in twice and more so to calculate twice.
Instead, we could simply combine these vectors into one single Coefficient Vector and one single Boundary Vector and go through the multiplication only once:
$$ \begin{bmatrix}
60&24&0&20 \\
0&30&40&0
\end{bmatrix} {P_{4}}^{-1}
\begin{bmatrix}
(31^4-10^4 )&( 51^4 - 15^4 )\\
(31^3 - 10^3 )& (51^3 - 15^3 )\\
(31^2 - 10^2 )&( 51^2 - 15^2)\\
(31 - 10) & (51-15)
\end{bmatrix} = A = \begin{bmatrix}
a_{1,1} & a_{1,2} \\
a_{2,1} & a_{2,2}
\end{bmatrix}$$ It is not difficult to see that this will yield a $2\times 2$ matrix which carries every combination of the coefficients and the boundaries of both series. This is obviously because none of the rows and columns interefere with the calcualtion of the other rows/columns. In this case, we will find our desired summations at the diagonals of $A$.
$$\sum_{i=10}^{30} 60i^3 + 24i^2 + 20 = a_{1,1}\\
\sum_{i=15}^{50} 30i^2 + 40i = a_{2,2}
$$The actual values, again, aren't important and thus won't be calculated for this article. Readers may calculate themselves and confirm that this is indeed true.
In general, we can compute as many combinations of polynomial series as one likes using this form. $$\begin{bmatrix}
\sum_{i=b_1}^{a_1}c_{1,n}i^n+ \dots+c_{1,0} & \dots & \sum_{i=b_j}^{a_j}c_{1,n}i^n+\dots+c_{1,0} \\
\vdots & \vdots & \vdots\\
\sum_{i=b_n}^{a+1}c_{k,n}i^n+\dots +c_{k,0}& \dots & \sum_{i=b_j}^{a_j}c_{k,n}i^n + \dots + c_{k,0}
\end{bmatrix} = \begin{bmatrix}
c_{1,n} & c_{1,n-1} & \dots & c_{1,0} \\
\vdots & \vdots & \vdots & \vdots \\
c_{k,n} & c_{k,n-1} & \dots & c_{k,0}
\end{bmatrix}
{P_{n+1}}^{-1}\begin{bmatrix}
(b_{1}+1)^{n+1}-a_{1}^{n+1} & \dots & (b_{j}+1)^{n+1}-a_{j}^{n+1} \\
(b_{1}+1)^{n}-a_{1}^{n} &\dots & (b_{j}+1)^{n}-a_{j}^{n} \\
\vdots & \vdots & \vdots \\
(b_{1}+1)-a_{1} & \dots & (b_{j}+1)-a_{j} \\
\end{bmatrix}
$$ Admittedly, this is quite daunting to look at at first glance, but what this form is trying to tell is that you can simply append the necessary coefficients or boundary vectors to compute any combinations of polynomials series in one calulcation. This might be beneficial if there are list of possible combinations of coefficients and boundaries, in which case all values can be computed then simply used as a look-up table.
The most obvious location to look for Bernoulli's Numbers is the constant term of each Power Series Formula. These numbers go as follows:
$$ 1, \frac{-1}{2}, \frac{1}{6}, 0 ,\frac{1}{30}, \dots$$ Remember, we have found a method of computing the coefficients of the Power Formula using Pascal Matrix. These constant terms can be found at the last column of ${P_n}^{-1}$, and indeed that is exactly what we find there.
Let's compute ${P_7}^{-1}$ as an example:
$${P_{7}}^{-1} =
\begin{bmatrix}
\frac{1}{7} & \frac{-1}{2} & \frac{1}{2} & 0 & \frac{-1}{6} & 0 & \frac{1}{42} \\
0&\frac{1}{6} & \frac{-1}{2} & \frac{5}{12} & 0 & \frac{-1}{12} & 0 \\
0&0 & \frac{1}{5} & \frac{-1}{2} & \frac{1}{3} & 0 & \frac{-1}{30} \\
0&0 & 0 & \frac{1}{4} & \frac{-1}{2} & \frac{1}{4} & 0 \\
0&0 & 0 & 0 & \frac{1}{3} & \frac{-1}{2} & \frac{1}{6} \\
0&0 & 0 & 0 & 0 & \frac{1}{2} & \frac{-1}{2} \\
0&0 & 0 & 0 & 0 & 0 & 1
\end{bmatrix}$$ Accoridng to this, the sixth Bernoulli's Number should be 0 and the seventh should be $\frac{1}{42}$. This is indeed true.
Again, in general, the Last Column of ${P_n}^{-1}$ will give us the sequence of Bernoulli's Numbers until the $n$th. This may not be the most efficient method of calculating these numbers, but certainly, this is an easy method to remember how to compute them.
Notations and Equations
We will borrow the notations and simple equations from Sequence Curve articles. There, we have extended a series to be continuous through the following identity:$$\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)\}$$ This will be used extensively throughout this article.
Power Functions
A simple yet interesting pattern emerges from the above identity when $f$ is a power function. When $f(x)=x$, $$\begin{align*} \frac{d}{dx}(x+1) &= Ȣ\{(x+1)-x\} \\&= Ȣ\{1\}
\end{align*}$$ We will leave the lefthand-side in terms of the deritives for cleaner results later on.
When $f(x) = x^2$, $$\begin{align*}
\frac{d}{dx}(x+1)^2 &= Ȣ\{(x+1)^2-x^2\} \\
&= Ȣ\{2x+1\} \\
&= 2Ȣ\{x\} + Ȣ\{1\}
\end{align*}$$ We will notice that these terms are very similar to the expansion of $(x+1)^n$ but with the first term removed. Noticing this, we easily find that $$\begin{align*}
\frac{d}{dx}(x+1)^3 &= 3Ȣ\{x^2\} + 3Ȣ\{x\} + Ȣ\{1\} \\
\frac{d}{dx}(x+1)^4 &= 4Ȣ\{x^3\} + 6Ȣ\{x^2\} + 4Ȣ\{x\} + Ȣ\{1\} \\
\frac{d}{dx}(x+1)^5 &= 5Ȣ\{x^4\} + 10Ȣ\{x^3\} + 10Ȣ\{x^2\} + 5Ȣ\{x\} + Ȣ\{1\} \\
&\vdots \\
\frac{d}{dx}(x+1)^n &= \binom{n}{1}Ȣ\{x^{n-1}\} + \binom{n}{2}Ȣ\{x^{n-2}\} + \dots + \binom{n}{n}Ȣ\{1\}
\end{align*}$$ Though interesting, these define the $\frac{d}{dx}(x+1)^n$ in terms of $Ȣ\{x^n\}$s which is not yet very useful. What we need is precisely the other way around.
Solving for Power Series Formula
What we are interested in are the individual $Ȣ\{x^n\}$ terms. Using the equalities above, it is possible to solve for $Ȣ\{x^n\}$ in terms of $\frac{d}{dx}(x+1)^n$.For example, if we want to solve for $Ȣ\{x^3\}$, we will treat the individual $Ȣ\{1\}, Ȣ\{x\} \dots Ȣ\{x^3\}$ like variables and solve for the system of equations of the first four equations.
$$\begin{align*}
&\begin{cases}
\frac{d}{dx}(x+1) = Ȣ\{1 \} \\
\frac{d}{dx}(x+1)^2 = 2Ȣ\{x\} + Ȣ\{1\} \\
\frac{d}{dx}(x+1)^3 = 3Ȣ\{x^2\} + 3Ȣ\{x\} + Ȣ\{1\} \\
\frac{d}{dx}(x+1)^4 = 4Ȣ\{x^3\} + 6Ȣ\{x^2\} 4Ȣ\{x\} + Ȣ\{1\}
\end{cases} \\ \\
&\begin{cases}
\frac{d}{dx}(x+1) = Ȣ\{1\} \\
\frac{d}{dx}(x+1)^2 - \frac{d}{dx}(x+1) = 2Ȣ\{x\} \\
\frac{d}{dx}(x+1)^3 - \frac{d}{dx}(x+1) = 3Ȣ\{x^2\} + 3Ȣ\{x^2\} \\
\frac{d}{dx}(x+1)^4 - \frac{d}{dx}(x+1) = 4Ȣ\{x^3\} + 6Ȣ\{x^2\} + 4Ȣ\{x\}
\end{cases} \\ \\
&\begin{cases}
\frac{d}{dx}(x+1) = Ȣ\{1\} \\
\frac{1}{2}\frac{d}{dx}(x+1)^2 - \frac{1}{2}\frac{d}{dx}(x+1) = Ȣ\{x\} \\
\frac{d}{dx}(x+1)^3 - \frac{3}{2}\frac{d}{dx}(x+1)^2 +\frac{1}{2}\frac{d}{dx}(x+1) = 3Ȣ\{x^2\} \\
\frac{d}{dx}(x+1)^4 -2\frac{d}{dx}(x+1)^2 + \frac{d}{dx}(x+1) = 4Ȣ\{x^3\} + 6Ȣ\{x^2\}
\end{cases} \\ \\
&\begin{cases}
\frac{d}{dx}(x+1) = Ȣ\{1\} \\
\frac{1}{2}\frac{d}{dx}(x+1)^2 - \frac{1}{2}\frac{d}{dx}(x+1) = Ȣ\{x\} \\
\frac{1}{3}\frac{d}{dx}(x+1)^3 - \frac{1}{2}\frac{d}{dx}(x+1)^2 + \frac{1}{6}\frac{d}{dx}(x+1) = Ȣ\{x^2\} \\
\frac{d}{dx}(x+1)^4 -2\frac{d}{dx}(x+1)^3 + \frac{d}{dx}(x+1)^2 = 4Ȣ\{x^3\}
\end{cases} \\ \\
&\begin{cases}
\frac{d}{dx}(x+1) = Ȣ\{1\} \\
\frac{1}{2}\frac{d}{dx}(x+1)^2 - \frac{1}{2}\frac{d}{dx}(x+1) = Ȣ\{x\} \\
\frac{1}{3}\frac{d}{dx}(x+1)^3 - \frac{1}{2}\frac{d}{dx}(x+1)^2 + \frac{1}{6}\frac{d}{dx}(x+1) = Ȣ\{x^2\} \\
\frac{1}{4}\frac{d}{dx}(x+1)^4 -\frac{1}{2}\frac{d}{dx}(x+1)^3 + \frac{1}{4}\frac{d}{dx}(x+1)^2 = Ȣ\{x^3\}
\end{cases}
\end{align*}$$ Though it is extremely messy, it is possible to calculate $Ȣ\{x^3\} = \frac{1}{4}\frac{d}{dx}(x+1)^4 -\frac{1}{2}\frac{d}{dx}(x+1)^3 + \frac{1}{4}\frac{d}{dx}(x+1)^2$ by hand.
We can associate this result back with the cubic series:
$$\sum_{i=m}^ni^3 = \frac{1}{4}((n+1)^4 - m^4) - \frac{1}{2}((n+1)^3-m^3) + \frac{1}{4}((n+1)^2-m^2)$$ When $m=0$, we have our familiar Power Series Formula for a cubic.
System of Equations using Matrices
We can more efficiently solve a system using matrices. the system above, for example, $$\begin{align*}\begin{cases}
&4Ȣ\{x^3\} + &6Ȣ\{x^2\}+ &4Ȣ\{x\} + & Ȣ\{1\} = &\frac{d}{dx}(x+1)^4 \\
& &3Ȣ\{x^2\} + &3Ȣ\{x\} + & Ȣ\{1\} = &&\frac{d}{dx}(x+1)^3 \\
& & &2Ȣ\{x\} + & Ȣ\{1\} = &&&\frac{d}{dx}(x+1)^2 \\
& & & &Ȣ\{1\} = &&&&\frac{d}{dx}(x+1)
\end{cases}
\end{align*} \\ $$ is equivalent to this matrix: $$ \left[
\left. \begin{matrix}
4 & 6 & 4 & 1\\
0 & 3 & 3 & 1\\
0 & 0 & 2 & 1\\
0 & 0 & 0 & 1
\end{matrix} \right|
\begin{matrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{matrix} \right] $$ Beforing going any further, we notice that the left-hand matrix looks very familiar. It looks just like the famous Pascal's Triangle unsider-down and with the first 1s removed from each row. From here on, we will call these matrices Pascal Matrix of size $n\times n$, $P_n$.
So, to solve for the lefthand-matrix, we simply need to multiply its inverse to both sides which happens to be ${P_4}^{-1}$. This will result in the system that looks like
$$ \left[ \mathbf{I} \vert {P_4}^{-1} \right] $$ This shows that we can easily solve for $Ȣ\{x^n\}$ by simply taking the inverse of $P_{n+1}$. To find $Ȣ\{x^3\}$, we calculate ${P_4}^{-1}$: $$\begin{bmatrix}
4&6&4&1\\
0&3&3&1\\
0&0&2&1\\
0&0&0&1
\end{bmatrix} ^{-1} = \begin{bmatrix}
\frac{1}{4} & -\frac{1}{2} & \frac{1}{4} & 0 \\
0 & \frac{1}{3} & -\frac{1}{2} & \frac{1}{6} \\
0 & 0 & \frac{1}{2} & -\frac{1}{2} \\
0&0&0&1
\end{bmatrix}
$$ The first row yields the coeffficents of $\frac{d}{dx}(x+1)^n$s for $Ȣ\{x^3\}$.
$$Ȣ\{x^3\} = \frac{1}{4}\frac{d}{dx}(x+1)^4 - \frac{1}{2}\frac{d}{dx}(x+1)^3 + \frac{1}{4}\frac{d}{dx}(x+1)^2
$$ Similarly, we can also find the coefficients for $Ȣ\{x^2\}$, $Ȣ\{x\}$, and $Ȣ\{1\}$ and also be found on second, third, and fourth rows respectively.
$$\begin{align*}
Ȣ\{x^2\} &= \frac{1}{3}\frac{d}{dx}(x+1)^3 - \frac{1}{2}\frac{d}{dx}(x+1)^2 + \frac{1}{6}\frac{d}{dx}(x+1) \\
Ȣ\{x\} &= \frac{1}{2}\frac{d}{dx}(x+1)^2 + \frac{1}{2}\frac{d}{dx}(x+1) \\
Ȣ\{1\} &= \frac{d}{dx}(x+1)
\end{align*}$$ We can solve for any number of $Ȣ\{x^n\}$, which are just the Power Series Formula, at once by solving for a single Inverse of a Matrix!
Polynomial Series using Matrices
So far, we only rediscovered the Power Series Formula but with matrices. Ultimately, we would like to define $Ȣ\{f(x)\}$ for any polynomial $f$. This can be easily achieved by adding up the correct combinations of each $Ȣ\{x^n\}$ terms which we have already found. This again can be easily done with matrices.Let's say we want to compute the series formula for $f(x) = 4x^3 -12x^2 +24$. We can find the individual $Ȣ\{x^n\}$ with the ${P_4}^{-1}$, and we can combine these terms to yield $f$ by multiplying Coefficient Row-Vector in front of it:$$
Ȣ\{f(x)\} = \begin{bmatrix}
4 & -12 & 0 & 24
\end{bmatrix}
\begin{bmatrix}
\frac{1}{4} & -\frac{1}{2} & \frac{1}{4} & 0 \\
0 & \frac{1}{3} & -\frac{1}{2} & \frac{1}{6} \\
0 & 0 & \frac{1}{2} & -\frac{1}{2} \\
0&0&0&1
\end{bmatrix} =
\begin{bmatrix}
1&-6&7&22
\end{bmatrix}
$$ Written in full, we find that
$$Ȣ\{4x^3 -12x^2 +24\} = \frac{d}{dx}(x+1)^4 - 6\frac{d}{dx}(x+1)^3 + 7\frac{d}{dx}(x+1)^2 + 22\frac{d}{dx}(x+1)$$ Now, we can integrate $Ȣ\{f(x)\}$ to get our Series Formula for the polynomial. Since the function is defined in terms of deritives, integrating them will not change the coefficients at all– this is why we kept them in terms of deritives in the beginning.
Therefore, we find that
$$ \begin{align*}
\sum_{i=m}^n f(i) &= \left. (x+1)^4 - 6(x+1)^3 +7(x+1)^2+22(x+1) \right \vert_{m-1}^{n} \\
&= ((n+1)^4-m^4) - 6((n+1)^3-m^3)+7((n+1)^2-m^2) +22((n+1)-m) \\
&= \begin{bmatrix}
1&-6&7&22
\end{bmatrix}\begin{bmatrix}
(n+1)^4-m^4 \\
(n+1)^3-m^3\\
(n+1)^2-m^2\\
(n+1)-m
\end{bmatrix}
\end{align*}$$ Even this last step can be represented as multiplication of the coefficient matrix with Boundary Vector of the series!
Combing all of these steps, computing the Series of $f(x)$ is as simple as calculating the following matrix product:
$$\sum_{i=m}^n 4i^3 -12i^2 +24 = \begin{bmatrix}
4 & -12 & 0 & 24
\end{bmatrix}{P_4}^{-1}\begin{bmatrix}
(n+1)^4-m^4 \\
(n+1)^3-m^3\\
(n+1)^2-m^2\\
(n+1)-m
\end{bmatrix}$$ Of course, there is nothing special with this specific polynomial and this form can be genralied for any polynomial with coefficient $c_k$:
$$\begin{align*}
\sum_{i=m}^n c_kx^k+c_{k-1}x^{k-1}+\dots + c_0 &= \begin{bmatrix} c_k & c_{k-1}&\dots&c_0\end{bmatrix}\begin{bmatrix}
\binom{k+1}{1}&\binom{k}{2}&\dots&\binom{k+1}{k+1}\\
0&\binom{k}{1}&\dots&\binom{k}{k}\\
\vdots&\vdots&\dots&\vdots\\
0&0&\dots&\binom{1}{1}
\end{bmatrix}^{-1}\begin{bmatrix}
(n+1)^{k+1}-m^{k+1}\\
(n+1)^k-m^k\\
\vdots\\
(n+1)-m
\end{bmatrix} \\ \\
&= \begin{bmatrix} c_k & c_{k-1}&\dots&c_0\end{bmatrix}{P_{k+1}}^{-1}\begin{bmatrix}
(n+1)^{k+1}-m^{k+1}\\
(n+1)^k-m^k\\
\vdots\\
(n+1)-m
\end{bmatrix}
\end{align*}$$ This might look even more frightening at first, but just remember that these are composed of three simple parts: (coefficients)(Inverse of Pascal)(Boundaries). These individual parts are simpler to remember and even more intuitive than memorizing individual Power Formulas.
Additional Advantage: Multiple Series at Once
The above equation provides a novel and easy-to-remember formula to compute a polynomial series, which was our original goal. Due to the nature of matrices, the above formula comes with an interesting advantage: multiple polynomial series can be computed at once.In short, we can do this by adding additional Rows to the Coefficient Vector or adding additional Columns to the Boundary Vector.
For example, let's say we want to compute these two following series:
$$\sum_{i=10}^{30} 60i^3 + 24i^2 + 20 \\
\sum_{i=15}^{50} 30i^2 + 40i
$$ The actual numbers are not important. We will need two Coefficient Vectors
$$\begin{bmatrix}
60&24&0&20
\end{bmatrix}, \begin{bmatrix}
0&30&40&0
\end{bmatrix}$$ and two Boundary Vectors to compute the two series.
$$\begin{bmatrix}
31^4-10^4 \\
31^3 - 10^3\\
31^2 - 10^2\\
31 - 10
\end{bmatrix},
\begin{bmatrix}
51^4 - 15^4 \\
51^3 - 15^3 \\
51^2 - 15^2\\
51-15
\end{bmatrix}
$$ Of course, we have seen above how to use these vectors to solve for individual series, but in that case, we will have to do two separate matrix multiplication for each series.
$$
\sum_{i=10}^{30} 60i^3 + 24i^2 + 20 = \begin{bmatrix}
60&24&0&20
\end{bmatrix} {P_4}^{-1}\begin{bmatrix}
31^4-10^4 \\
31^3 - 10^3\\
31^2 - 10^2\\
31 - 10
\end{bmatrix} \\
\sum_{i=15}^{50} 30i^2 + 40i = \begin{bmatrix}
0&30&40&0
\end{bmatrix} {P_4}^{-1}\begin{bmatrix}
51^4 - 15^4 \\
51^3 - 15^3 \\
51^2 - 15^2\\
51-15
\end{bmatrix} $$ Though simple, it may be combsersome to type in twice and more so to calculate twice.
Instead, we could simply combine these vectors into one single Coefficient Vector and one single Boundary Vector and go through the multiplication only once:
$$ \begin{bmatrix}
60&24&0&20 \\
0&30&40&0
\end{bmatrix} {P_{4}}^{-1}
\begin{bmatrix}
(31^4-10^4 )&( 51^4 - 15^4 )\\
(31^3 - 10^3 )& (51^3 - 15^3 )\\
(31^2 - 10^2 )&( 51^2 - 15^2)\\
(31 - 10) & (51-15)
\end{bmatrix} = A = \begin{bmatrix}
a_{1,1} & a_{1,2} \\
a_{2,1} & a_{2,2}
\end{bmatrix}$$ It is not difficult to see that this will yield a $2\times 2$ matrix which carries every combination of the coefficients and the boundaries of both series. This is obviously because none of the rows and columns interefere with the calcualtion of the other rows/columns. In this case, we will find our desired summations at the diagonals of $A$.
$$\sum_{i=10}^{30} 60i^3 + 24i^2 + 20 = a_{1,1}\\
\sum_{i=15}^{50} 30i^2 + 40i = a_{2,2}
$$The actual values, again, aren't important and thus won't be calculated for this article. Readers may calculate themselves and confirm that this is indeed true.
In general, we can compute as many combinations of polynomial series as one likes using this form. $$\begin{bmatrix}
\sum_{i=b_1}^{a_1}c_{1,n}i^n+ \dots+c_{1,0} & \dots & \sum_{i=b_j}^{a_j}c_{1,n}i^n+\dots+c_{1,0} \\
\vdots & \vdots & \vdots\\
\sum_{i=b_n}^{a+1}c_{k,n}i^n+\dots +c_{k,0}& \dots & \sum_{i=b_j}^{a_j}c_{k,n}i^n + \dots + c_{k,0}
\end{bmatrix} = \begin{bmatrix}
c_{1,n} & c_{1,n-1} & \dots & c_{1,0} \\
\vdots & \vdots & \vdots & \vdots \\
c_{k,n} & c_{k,n-1} & \dots & c_{k,0}
\end{bmatrix}
{P_{n+1}}^{-1}\begin{bmatrix}
(b_{1}+1)^{n+1}-a_{1}^{n+1} & \dots & (b_{j}+1)^{n+1}-a_{j}^{n+1} \\
(b_{1}+1)^{n}-a_{1}^{n} &\dots & (b_{j}+1)^{n}-a_{j}^{n} \\
\vdots & \vdots & \vdots \\
(b_{1}+1)-a_{1} & \dots & (b_{j}+1)-a_{j} \\
\end{bmatrix}
$$ Admittedly, this is quite daunting to look at at first glance, but what this form is trying to tell is that you can simply append the necessary coefficients or boundary vectors to compute any combinations of polynomials series in one calulcation. This might be beneficial if there are list of possible combinations of coefficients and boundaries, in which case all values can be computed then simply used as a look-up table.
Bonus Fact: Calculating Bernoulli's Number with Pascal Matrix
Bernoulli's Numbers are basically the original attempt at discovering the Power Series Formula. (Though named after Bernoulli's, it was actually studied by other mathematicians before him). With their roots in computing series for power functions, it will be exciting to find a link between Bernoulli's Numbers and our novel method of calculating series. This link is found in the Pascal's Matrix and more specifically in its inverse.The most obvious location to look for Bernoulli's Numbers is the constant term of each Power Series Formula. These numbers go as follows:
$$ 1, \frac{-1}{2}, \frac{1}{6}, 0 ,\frac{1}{30}, \dots$$ Remember, we have found a method of computing the coefficients of the Power Formula using Pascal Matrix. These constant terms can be found at the last column of ${P_n}^{-1}$, and indeed that is exactly what we find there.
Let's compute ${P_7}^{-1}$ as an example:
$${P_{7}}^{-1} =
\begin{bmatrix}
\frac{1}{7} & \frac{-1}{2} & \frac{1}{2} & 0 & \frac{-1}{6} & 0 & \frac{1}{42} \\
0&\frac{1}{6} & \frac{-1}{2} & \frac{5}{12} & 0 & \frac{-1}{12} & 0 \\
0&0 & \frac{1}{5} & \frac{-1}{2} & \frac{1}{3} & 0 & \frac{-1}{30} \\
0&0 & 0 & \frac{1}{4} & \frac{-1}{2} & \frac{1}{4} & 0 \\
0&0 & 0 & 0 & \frac{1}{3} & \frac{-1}{2} & \frac{1}{6} \\
0&0 & 0 & 0 & 0 & \frac{1}{2} & \frac{-1}{2} \\
0&0 & 0 & 0 & 0 & 0 & 1
\end{bmatrix}$$ Accoridng to this, the sixth Bernoulli's Number should be 0 and the seventh should be $\frac{1}{42}$. This is indeed true.
Again, in general, the Last Column of ${P_n}^{-1}$ will give us the sequence of Bernoulli's Numbers until the $n$th. This may not be the most efficient method of calculating these numbers, but certainly, this is an easy method to remember how to compute them.
omg king shit this was so informational wow!!!!!!!
ReplyDeletePascal want to talk with you...
ReplyDelete